CBSE Class 12 Mathematics Three Dimensional Geometry Assignment Set 02

(i) Direction Ratios and Direction Cosines

Level-I

Question 1. Write the direction-cosines of the line joining the points (1,0,0) and (0,1,1) [CBSE 2011]
Answer:
First, we find the direction ratios of the line segment connecting the points \( P(1, 0, 0) \) and \( Q(0, 1, 1) \).
The direction ratios are calculated as:
\( a = x_2 - x_1 = 0 - 1 = -1 \)
\( b = y_2 - y_1 = 1 - 0 = 1 \)
\( c = z_2 - z_1 = 1 - 0 = 1 \)
Now, we find the magnitude of these direction ratios:
\( \sqrt{a^2 + b^2 + c^2} = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3} \)
The direction cosines \( (l, m, n) \) are given by dividing each direction ratio by this magnitude:
\( l = \frac{a}{\sqrt{a^2 + b^2 + c^2}} = -\frac{1}{\sqrt{3}} \)
\( m = \frac{b}{\sqrt{a^2 + b^2 + c^2}} = \frac{1}{\sqrt{3}} \)
\( n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} = \frac{1}{\sqrt{3}} \)
Therefore, the direction cosines of the line are \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \) or \( \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) \).
In simple words: To find the direction cosines, first subtract the coordinates of the two points to get the direction ratios. Then, divide each ratio by the total length of the vector to make it a unit vector.

Exam Tip: Be sure to write both possible sets of direction cosines, as a line has two opposite directions and both are correct.

Question 2. Find the direction cosines of the line passing through the following points (-2,4,-5), (1,2,3).
Answer:
Let the given points be \( A(-2, 4, -5) \) and \( B(1, 2, 3) \).
The direction ratios of the line \( AB \) are:
\( a = 1 - (-2) = 3 \)
\( b = 2 - 4 = -2 \)
\( c = 3 - (-5) = 8 \)
Next, we calculate the magnitude:
\( \sqrt{a^2 + b^2 + c^2} = \sqrt{3^2 + (-2)^2 + 8^2} = \sqrt{9 + 4 + 64} = \sqrt{77} \)
The direction cosines of the line are obtained by dividing the direction ratios by this magnitude:
\( l = \frac{3}{\sqrt{77}}, \quad m = -\frac{2}{\sqrt{77}}, \quad n = \frac{8}{\sqrt{77}} \)
Thus, the direction cosines are \( \left(\frac{3}{\sqrt{77}}, -\frac{2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\right) \).
In simple words: Subtract the first point's values from the second to find the ratios, and then divide by the square root of the sum of their squares to get the final cosines.

Exam Tip: Carefully perform the subtraction with negative values to avoid simple sign errors, such as writing \( 3 - 5 \) instead of \( 3 - (-5) \).

Question 3. Write the direction cosines of a line equally inclined to the three coordinate axes
Answer:
Let the line make angles \( \alpha, \beta, \gamma \) with the coordinate axes. Since it is equally inclined, we have:
\( \alpha = \beta = \gamma \)
This means their cosines are also equal:
\( \cos \alpha = \cos \beta = \cos \gamma \)
Using the fundamental identity for direction cosines:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)
Substituting the equal values gives:
\( 3 \cos^2 \alpha = 1 \)

\(\implies \cos^2 \alpha = \frac{1}{3} \)

\(\implies \cos \alpha = \pm \frac{1}{\sqrt{3}} \)
Therefore, the direction cosines of the line are \( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \).
In simple words: When a line is tilted by the same angle towards all three axes, its direction cosines are all equal, and each one is equal to plus or minus one divided by the square root of three.

Exam Tip: Remember to write the \( \pm \) sign, because there are multiple lines (passing through different octants) that are equally inclined to the axes.

Level-II

Question 1. Write the direction cosines of a line parallel to the line \( \frac{3-x}{3} = \frac{y+2}{-2} = \frac{z+2}{6} \)
Answer:
First, we rewrite the given equation of the line in standard symmetric form:
\( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \)
Transforming the term \( \frac{3-x}{3} \) by multiplying the numerator and denominator by \( -1 \), we get:
\( \frac{x-3}{-3} = \frac{y+2}{-2} = \frac{z+2}{6} \)
Thus, the direction ratios of the given line are \( a = -3 \), \( b = -2 \), and \( c = 6 \).
Since a parallel line shares the same direction ratios, its direction cosines are calculated using the magnitude:
\( \sqrt{a^2 + b^2 + c^2} = \sqrt{(-3)^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \)
The direction cosines are:
\( l = \frac{-3}{7}, \quad m = \frac{-2}{7}, \quad n = \frac{6}{7} \)
Therefore, the direction cosines of the parallel line are \( \left(-\frac{3}{7}, -\frac{2}{7}, \frac{6}{7}\right) \) or \( \left(\frac{3}{7}, \frac{2}{7}, -\frac{6}{7}\right) \).
In simple words: First rewrite the line's equation so that \( x, y, z \) have positive coefficients. Read off the numbers at the bottom as direction ratios, and then divide them by their total length (which is 7).

Exam Tip: Be very careful with the standard form. A common trap is directly taking the denominator \( 3 \) as a direction ratio without converting \( 3-x \) to \( x-3 \) first.

Question 2. Write the direction ratios of a line parallel to the line \( \frac{5-x}{3} = \frac{y+7}{-2} = \frac{z+2}{6} \)
Answer:
To find the direction ratios, we must first express the equation of the line in standard form:
\( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \)
Changing \( \frac{5-x}{3} \) to standard form gives \( \frac{x-5}{-3} \).
The symmetric form is:
\( \frac{x-5}{-3} = \frac{y+7}{-2} = \frac{z+2}{6} \)
The denominators represent the direction ratios of the line, which are \( (-3, -2, 6) \).
Since parallel lines have proportional direction ratios, the direction ratios of the parallel line can be written as \( (-3, -2, 6) \) or \( (3, 2, -6) \).
In simple words: Rewrite the first fraction to make the \( x \) positive. The numbers on the bottom are the direction ratios.

Exam Tip: Direction ratios are not unique; any scalar multiple of a set of direction ratios is also a valid set of direction ratios for the same line.

Question 3. If the equation of a line AB is \( \frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+6}{3} \), find the direction cosines.
Answer:
The equation of the line \( AB \) is already in standard form:
\( \frac{x-3}{2} = \frac{y+2}{-1} = \frac{z+6}{3} \)
By comparing this with the standard symmetric equation of a line, we get the direction ratios as:
\( a = 2, \quad b = -1, \quad c = 3 \)
Next, we calculate the magnitude of these ratios:
\( \sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \)
Dividing the ratios by the magnitude gives the direction cosines \( (l, m, n) \):
\( l = \frac{2}{\sqrt{14}}, \quad m = -\frac{1}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}} \)
Thus, the direction cosines of the line are \( \left(\frac{2}{\sqrt{14}}, -\frac{1}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right) \).
In simple words: The numbers under the variables are the direction ratios. Divide each by their square-root-sum length to convert them into direction cosines.

Exam Tip: Double-check that the coefficient of \( x, y, \) and \( z \) in the numerators is positive 1 before extracting the direction ratios.

Question 4. Find the direction cosines of a line, passing through origin and lying in the first octant, making equal angles with the three coordinate axis.
Answer:
Since the line is equally inclined to the three coordinate axes, we have:
\( \alpha = \beta = \gamma \)

\(\implies \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)

\(\implies 3 \cos^2 \alpha = 1 \)

\(\implies \cos \alpha = \pm \frac{1}{\sqrt{3}} \)
Because the line lies in the first octant, all coordinates of its points are positive, which means the angles \( \alpha, \beta, \gamma \) are acute. Therefore, the direction cosines must be positive:
\( l = \cos \alpha = \frac{1}{\sqrt{3}} \)
\( m = \cos \beta = \frac{1}{\sqrt{3}} \)
\( n = \cos \gamma = \frac{1}{\sqrt{3}} \)
Thus, the direction cosines are \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \).
In simple words: Since the line is in the first octant and is tilted equally to all axes, all its direction cosines must be equal and positive, which is one over the square root of three.

Exam Tip: Pay close attention to the phrase "lying in the first octant." This specifies that the direction cosines must be positive, so the negative solutions are discarded.

(ii) Cartesian and Vector Equation of a Line in Space & Conversion of One into Another Form

Level-I

Question 1. Write the vector equation of the line \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2} \). [CBSE 2011]
Answer:
First, we rewrite the Cartesian equation of the line in standard form:
\( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{-2} \)
By comparing this with the standard symmetric equation \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), we can identify:
- The point through which the line passes: \( (x_1, y_1, z_1) = (5, -4, 6) \). Its position vector is \( \vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k} \).
- The direction ratios of the line: \( (a, b, c) = (3, 7, -2) \). The vector parallel to the line is \( \vec{b} = 3\hat{i} + 7\hat{j} - 2\hat{k} \).
The vector equation of the line is given by \( \vec{r} = \vec{a} + \lambda \vec{b} \):
\( \vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} - 2\hat{k}) \)
where \( \lambda \) is a real parameter.
In simple words: Write the line in standard form so you can find the starting point and the direction vector. The vector equation is then the starting point vector plus a variable times the direction vector.

Exam Tip: Be careful with the numerator \( 6-z \). You must factor out a negative sign to make it \( z-6 \), which also changes the sign of its denominator to \( -2 \).

Question 2. Write the equation of a line parallel to the line \( \frac{x-2}{-3} = \frac{y+3}{2} = \frac{z+5}{6} \) and passing through the point (1,2,3).
Answer:
The given line is:
\( \frac{x-2}{-3} = \frac{y+3}{2} = \frac{z+5}{6} \)
The direction ratios of this line are \( (-3, 2, 6) \).
Since the required line is parallel to this line, it will have the same direction ratios: \( a = -3 \), \( b = 2 \), and \( c = 6 \).
The required line passes through the point \( (x_1, y_1, z_1) = (1, 2, 3) \).
Using the standard Cartesian equation of a line:
\( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \)
Substituting the values, we get the Cartesian equation:
\( \frac{x-1}{-3} = \frac{y-2}{2} = \frac{z-3}{6} \)
Alternatively, in vector form, the equation is:
\( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-3\hat{i} + 2\hat{j} + 6\hat{k}) \)
In simple words: Since the new line is parallel to the old one, it points in the exact same direction. Use the bottom numbers from the old line, and the coordinates of the new point, to write the new equation.

Exam Tip: Unless the question specifies a particular format (vector or Cartesian), writing the Cartesian form is generally safer and standard.

Question 3. Express the equation of the plane \( \vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k}) \) in the Cartesian form.
Answer:
Note that the given equation is in the form \( \vec{r} = \vec{a} + \lambda \vec{b} \), which represents a line in space rather than a plane.
Here, the position vector of the point through which the line passes is:
\( \vec{a} = \hat{i} - 2\hat{j} + \hat{k} \)
This corresponds to the coordinates \( (x_1, y_1, z_1) = (1, -2, 1) \).
The direction vector parallel to the line is:
\( \vec{b} = 2\hat{i} + \hat{j} + 2\hat{k} \)
This corresponds to the direction ratios \( (a, b, c) = (2, 1, 2) \).
Substituting these into the Cartesian symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), we get:
\( \frac{x-1}{2} = \frac{y+2}{1} = \frac{z-1}{2} \)
In simple words: This equation is for a line. The first set of coordinates \( (1, -2, 1) \) is the starting point, and the second set \( (2, 1, 2) \) is the direction. We substitute these to get the Cartesian fractions.

Exam Tip: Sometimes exam papers contain minor typos (like calling a line a "plane"). Answer the question based on the actual mathematical structure of the equation shown.

Question 4. Express the equation of the plane \( \vec{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) + 4 = 0 \) in the Cartesian form.
Answer:
The given vector equation of the plane is:
\( \vec{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k}) + 4 = 0 \)
To convert this into Cartesian form, we substitute the general position vector \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \):
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - 3\hat{j} + \hat{k}) + 4 = 0 \)
Taking the dot product, we multiply the corresponding components:
\( 2x - 3y + z + 4 = 0 \)
Therefore, the Cartesian equation of the plane is \( 2x - 3y + z + 4 = 0 \).
In simple words: Replace \( \vec{r} \) with \( (x, y, z) \) and do the dot product to get a simple linear equation.

Exam Tip: This conversion is straightforward; just ensure that the signs of the coefficients match the components of the normal vector exactly.

(iii) Co-planer and Skew Lines

Level-II

Question 1. Find whether the lines \( \vec{r} = (\hat{i} - \hat{j} - \hat{k}) + \lambda(2\hat{i} + \hat{j}) \) and \( \vec{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} + \hat{j} - \hat{k}) \) intersect or not. If intersecting, find their point of intersection.
Answer:
Any point on the first line can be written in terms of parameter \( \lambda \) as:
\( (x, y, z) = (1 + 2\lambda, -1 + \lambda, -1) \)
Any point on the second line can be written in terms of parameter \( \mu \) as:
\( (x, y, z) = (2 + \mu, -1 + \mu, -\mu) \)
If the two lines intersect, these coordinates must be equal for some values of \( \lambda \) and \( \mu \):
1) \( 1 + 2\lambda = 2 + \mu \)
2) \( -1 + \lambda = -1 + \mu \)
3) \( -1 = -\mu \)
From equation (3), we get:
\( \mu = 1 \)
From equation (2), we get:
\( \lambda = \mu = 1 \)
Let us verify if these values of \( \lambda \) and \( \mu \) satisfy equation (1):
L.H.S. = \( 1 + 2(1) = 3 \)
R.H.S. = \( 2 + (1) = 3 \)
Since L.H.S. = R.H.S., the system of equations is consistent, which means the lines intersect.
Substituting \( \lambda = 1 \) (or \( \mu = 1 \)) into the coordinate equations gives the point of intersection:
\( x = 1 + 2(1) = 3 \)
\( y = -1 + 1 = 0 \)
\( z = -1 \)
Thus, the point of intersection is \( (3, 0, -1) \).
In simple words: Write out the coordinates of any point on both lines in terms of their parameters. Set these equal to each other and solve. Since a single pair of numbers works for all three axes, the lines do cross, and we find the crossing point by plugging those numbers in.

Exam Tip: Always verify your solution for the parameters by substituting them back into the third equation. If they do not satisfy it, the lines are skew and do not intersect.

Question 2. Show that the four points (0,-1,-1), (4,5,1), (3,9,4) and (-4,4,4) are coplanar. Also, find the equation of the plane containing them.
Answer:
Let the given points be \( A(0, -1, -1) \), \( B(4, 5, 1) \), \( C(3, 9, 4) \), and \( D(-4, 4, 4) \).
The equation of any plane passing through \( A(0, -1, -1) \) is:
\( a(x - 0) + b(y + 1) + c(z + 1) = 0 \)

\(\implies ax + b(y + 1) + c(z + 1) = 0 \quad \text{--- (1)} \)
If this plane also passes through \( B(4, 5, 1) \) and \( C(3, 9, 4) \), these points must satisfy equation (1):
For \( B(4, 5, 1) \):
\( a(4) + b(5 + 1) + c(1 + 1) = 0 \)

\(\implies 4a + 6b + 2c = 0 \)

\(\implies 2a + 3b + c = 0 \quad \text{--- (2)} \)
For \( C(3, 9, 4) \):
\( a(3) + b(9 + 1) + c(4 + 1) = 0 \)

\(\implies 3a + 10b + 5c = 0 \quad \text{--- (3)} \)
Solving equations (2) and (3) using the cross-multiplication method:
\( \frac{a}{(3)(5) - (1)(10)} = \frac{-b}{(2)(5) - (1)(3)} = \frac{c}{(2)(10) - (3)(3)} \)
\( \frac{a}{15 - 10} = \frac{-b}{10 - 3} = \frac{c}{20 - 9} \)
\( \frac{a}{5} = \frac{b}{-7} = \frac{c}{11} = k \)
Let \( a = 5 \), \( b = -7 \), and \( c = 11 \). Substituting these coefficients into equation (1):
\( 5x - 7(y + 1) + 11(z + 1) = 0 \)
\( 5x - 7y - 7 + 11z + 11 = 0 \)
\( 5x - 7y + 11z + 4 = 0 \quad \text{--- (4)} \)
Now, we check if the fourth point \( D(-4, 4, 4) \) lies on this plane by substituting its coordinates into equation (4):
L.H.S. = \( 5(-4) - 7(4) + 11(4) + 4 \)
L.H.S. = \( -20 - 28 + 44 + 4 = -48 + 48 = 0 = \text{R.H.S.} \)
Since the fourth point satisfies the plane's equation, the four points are coplanar.
The equation of the plane containing them is \( 5x - 7y + 11z + 4 = 0 \).
In simple words: First find the equation of the plane that passes through any three of the points. Then, plug in the fourth point's coordinates to see if it is on the same plane. Since the result is zero, all four points lie on the exact same flat surface.

Exam Tip: When using cross-multiplication to solve for coefficients, be very careful with the middle variable's sign to ensure correct coordinates for the normal vector.

Question 3. Show that the lines \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \) and \( \frac{x - 4}{5} = \frac{y - 1}{2} = z \) intersect. Find their point of intersection.
Answer:
Let any point on the first line be represented by parameter \( r_1 \):
\( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = r_1 \)
\( x = 2r_1 + 1, \quad y = 3r_1 + 2, \quad z = 4r_1 + 3 \)
Let any point on the second line be represented by parameter \( r_2 \) (noting that \( z \) can be written as \( \frac{z-0}{1} \)):
\( \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1} = r_2 \)
\( x = 5r_2 + 4, \quad y = 2r_2 + 1, \quad z = r_2 \)
For the two lines to intersect, we equate their respective coordinates:
1) \( 2r_1 + 1 = 5r_2 + 4 \implies 2r_1 - 5r_2 = 3 \)
2) \( 3r_1 + 2 = 2r_2 + 1 \implies 3r_1 - 2r_2 = -1 \)
3) \( 4r_1 + 3 = r_2 \)
Substituting equation (3) into equation (2):
\( 3r_1 - 2(4r_1 + 3) = -1 \)
\( 3r_1 - 8r_1 - 6 = -1 \)
\( -5r_1 = 5 \)

\(\implies r_1 = -1 \)
Using this in equation (3):
\( r_2 = 4(-1) + 3 = -1 \)
Now, let us verify if \( r_1 = -1 \) and \( r_2 = -1 \) satisfy equation (1):
L.H.S. = \( 2(-1) - 5(-1) = -2 + 5 = 3 \)
R.H.S. = \( 3 \)
Since L.H.S. = R.H.S., the lines intersect.
Substituting \( r_1 = -1 \) into the coordinate equations for the first line:
\( x = 2(-1) + 1 = -1 \)
\( y = 3(-1) + 2 = -1 \)
\( z = 4(-1) + 3 = -1 \)
Thus, the point of intersection is \( (-1, -1, -1) \).
In simple words: Represent the coordinates of both lines using helper variables \( r_1 \) and \( r_2 \). Solve for these variables using the coordinates on two of the axes, and then verify the third. Since everything matches, the lines cross at the point \( (-1, -1, -1) \).

Exam Tip: Be sure to write \( z \) as \( \frac{z-0}{1} \) to make it easy to identify its denominator as \( 1 \) and its numerator coordinate as \( 0 \).

Level-III

Question 1. Show that the lines \( \frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5} \) and \( \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5} \) are coplanar. Also find the equation of the plane.
Answer:
Two lines \( \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \) and \( \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2} \) are coplanar if:
\( \left| \begin{matrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{matrix} \right| = 0 \)
For the given lines:
- Point on the first line: \( (x_1, y_1, z_1) = (-3, 1, 5) \) and direction ratios \( (a_1, b_1, c_1) = (-3, 1, 5) \).
- Point on the second line: \( (x_2, y_2, z_2) = (-1, 2, 5) \) and direction ratios \( (a_2, b_2, c_2) = (-1, 2, 5) \).
Let us calculate the determinant:
\( \left| \begin{matrix} -1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{matrix} \right| = \left| \begin{matrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{matrix} \right| \)
Expanding along the first row:
\( = 2((1)(5) - (5)(2)) - 1((-3)(5) - (5)(-1)) + 0 \)
\( = 2(5 - 10) - 1(-15 + 5) \)
\( = 2(-5) - 1(-10) = -10 + 10 = 0 \)
Since the determinant is \( 0 \), the lines are coplanar.
Now, the equation of the plane containing these lines is given by:
\( \left| \begin{matrix} x - x_1 & y - y_1 & z - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{matrix} \right| = 0 \)
\( \left| \begin{matrix} x + 3 & y - 1 & z - 5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{matrix} \right| = 0 \)
Expanding along the first row:
\( (x + 3)((1)(5) - (5)(2)) - (y - 1)((-3)(5) - (5)(-1)) + (z - 5)((-3)(2) - (1)(-1)) = 0 \)
\( (x + 3)(5 - 10) - (y - 1)(-15 + 5) + (z - 5)(-6 + 1) = 0 \)
\( -5(x + 3) + 10(y - 1) - 5(z - 5) = 0 \)
Dividing by \( -5 \):
\( (x + 3) - 2(y - 1) + (z - 5) = 0 \)
\( x + 3 - 2y + 2 + z - 5 = 0 \)
\( x - 2y + z = 0 \)
Therefore, the equation of the plane is \( x - 2y + z = 0 \).
In simple words: To show the two lines are in the same plane, we check if the determinant of the vectors between their points and their directions is zero. To find the plane's equation, we write down the general formula using the determinant of the direction values and the coordinates of one point.

Exam Tip: Be sure to divide the final linear equation by common factors (such as dividing by \( -5 \)) to express the equation of the plane in its simplest possible form.

Question 2. The points A(4,5,10) , B(2,3,4) and C(1,2,-1) are three vertices of a parallelogram ABCD. Find the vector equation of the sides AB and BC and also find the coordinates of D.
Answer:
1) Vector equation of the side AB:
The line \( AB \) passes through the point \( A(4, 5, 10) \) and has the direction of vector \( \vec{AB} \):
Position vector of \( A \): \( \vec{a} = 4\hat{i} + 5\hat{j} + 10\hat{k} \)
Direction vector \( \vec{AB} = (2-4)\hat{i} + (3-5)\hat{j} + (4-10)\hat{k} = -2\hat{i} - 2\hat{j} - 6\hat{k} \)
Thus, the vector equation of the side \( AB \) is:
\( \vec{r} = (4\hat{i} + 5\hat{j} + 10\hat{k}) + \lambda(-2\hat{i} - 2\hat{j} - 6\hat{k}) \)
This can also be written in simplified form as:
\( \vec{r} = (4\hat{i} + 5\hat{j} + 10\hat{k}) + \lambda(\hat{i} + \hat{j} + 3\hat{k}) \)

2) Vector equation of the side BC:
The line \( BC \) passes through the point \( B(2, 3, 4) \) and has the direction of vector \( \vec{BC} \):
Position vector of \( B \): \( \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
Direction vector \( \vec{BC} = (1-2)\hat{i} + (2-3)\hat{j} + (-1-4)\hat{k} = -\hat{i} - \hat{j} - 5\hat{k} \)
Thus, the vector equation of the side \( BC \) is:
\( \vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) + \mu(-\hat{i} - \hat{j} - 5\hat{k}) \)
This can also be written in simplified form as:
\( \vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) + \mu(\hat{i} + \hat{j} + 5\hat{k}) \)

3) Coordinates of D:
In a parallelogram \( ABCD \), the diagonals \( AC \) and \( BD \) bisect each other at the same midpoint.
Midpoint of diagonal \( AC \):
\( \left(\frac{4+1}{2}, \frac{5+2}{2}, \frac{10-1}{2}\right) = \left(\frac{5}{2}, \frac{7}{2}, \frac{9}{2}\right) \)
Let the coordinates of \( D \) be \( (x, y, z) \). The midpoint of diagonal \( BD \) is:
\( \left(\frac{2+x}{2}, \frac{3+y}{2}, \frac{4+z}{2}\right) \)
Equating the coordinates of the midpoints:
\( \frac{2+x}{2} = \frac{5}{2} \implies 2 + x = 5 \implies x = 3 \)
\( \frac{3+y}{2} = \frac{7}{2} \implies 3 + y = 7 \implies y = 4 \)
\( \frac{4+z}{2} = \frac{9}{2} \implies 4 + z = 9 \implies z = 5 \)
Therefore, the coordinates of the vertex \( D \) are \( (3, 4, 5) \).
In simple words: The equation of each side is found by taking the first point's vector and adding a parameter times the difference vector between the two points. To find the fourth vertex \( D \), we use the property that the diagonals of a parallelogram have the exact same center point.

Exam Tip: Using the midpoint property of diagonals is the fastest and most elegant way to find the fourth vertex of a parallelogram in 3D geometry.

Question 3. Find the equations of the line which intersects the lines \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) and \( \frac{x+2}{1} = \frac{y-3}{2} = \frac{z+1}{4} \) and passes through the point (1,1,1).
Answer:
Let the required line pass through the point \( P(1, 1, 1) \) and have direction ratios \( (a, b, c) \).
The equation of this line is:
\( \frac{x-1}{a} = \frac{y-1}{b} = \frac{z-1}{c} \quad \text{--- (1)} \)
This line intersects the first given line:
\( L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \)
Since these two lines intersect, they must be coplanar. The coplanarity condition is:
\( \left| \begin{matrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a & b & c \\ a_1 & b_1 & c_1 \end{matrix} \right| = 0 \)
Here, \( (x_1, y_1, z_1) = (1, 1, 1) \) and \( (x_2, y_2, z_2) = (1, 2, 3) \) with direction ratios \( (a_1, b_1, c_1) = (2, 3, 4) \).
Substituting these values:
\( \left| \begin{matrix} 1 - 1 & 2 - 1 & 3 - 1 \\ a & b & c \\ 2 & 3 & 4 \end{matrix} \right| = 0 \)
\( \left| \begin{matrix} 0 & 1 & 2 \\ a & b & c \\ 2 & 3 & 4 \end{matrix} \right| = 0 \)
Expanding along the first row:
\( 0 - 1(4a - 2c) + 2(3a - 2b) = 0 \)
\( -4a + 2c + 6a - 4b = 0 \)
\( 2a - 4b + 2c = 0 \)

\(\implies a - 2b + c = 0 \quad \text{--- (2)} \)
The required line also intersects the second given line:
\( L_2: \frac{x+2}{1} = \frac{y-3}{2} = \frac{z+1}{4} \)
Here, \( (x_3, y_3, z_3) = (-2, 3, -1) \) with direction ratios \( (a_2, b_2, c_2) = (1, 2, 4) \).
Using the coplanarity condition with \( L_2 \):
\( \left| \begin{matrix} -2 - 1 & 3 - 1 & -1 - 1 \\ a & b & c \\ 1 & 2 & 4 \end{matrix} \right| = 0 \)
\( \left| \begin{matrix} -3 & 2 & -2 \\ a & b & c \\ 1 & 2 & 4 \end{matrix} \right| = 0 \)
Expanding along the first row:
\( -3(4b - 2c) - 2(4a - c) - 2(2a - b) = 0 \)
\( -12b + 6c - 8a + 2c - 4a + 2b = 0 \)
\( -12a - 10b + 8c = 0 \)

\(\implies 6a + 5b - 4c = 0 \quad \text{--- (3)} \)
Solving equations (2) and (3) for \( a, b, c \) using the cross-multiplication method:
\( \frac{a}{(-2)(-4) - (1)(5)} = \frac{-b}{(1)(-4) - (1)(6)} = \frac{c}{(1)(5) - (-2)(6)} \)
\( \frac{a}{8 - 5} = \frac{-b}{-4 - 6} = \frac{c}{5 + 12} \)
\( \frac{a}{3} = \frac{b}{10} = \frac{c}{17} \)
Thus, the direction ratios of the line are proportional to \( (3, 10, 17) \).
Substituting these into the line equation (1), we get:
\( \frac{x-1}{3} = \frac{y-1}{10} = \frac{z-1}{17} \)
In simple words: Write a general equation for a line through the given point with unknown direction ratios. Set up two determinants to represent the fact that this line crosses both of the other lines. Solving these two equations gives the ratios as \( (3, 10, 17) \), giving us the final equation.

Exam Tip: Set up the coplanarity condition between the required line and the given lines. This is a very powerful technique for intersection problems in three dimensions.

Question 4. Show that the four points (0, –1, –1), (4, 5, 1), (3, 9, 4) and (– 4, 4, 4) are coplanar and find the equation of the common plane.
Answer:
Let the points be \( A(0, -1, -1) \), \( B(4, 5, 1) \), \( C(3, 9, 4) \), and \( D(-4, 4, 4) \).
We can form three vectors starting from point \( A \):
\( \vec{AB} = (4 - 0)\hat{i} + (5 - (-1))\hat{j} + (1 - (-1))\hat{k} = 4\hat{i} + 6\hat{j} + 2\hat{k} \)
\( \vec{AC} = (3 - 0)\hat{i} + (9 - (-1))\hat{j} + (4 - (-1))\hat{k} = 3\hat{i} + 10\hat{j} + 5\hat{k} \)
\( \vec{AD} = (-4 - 0)\hat{i} + (4 - (-1))\hat{j} + (4 - (-1))\hat{k} = -4\hat{i} + 5\hat{j} + 5\hat{k} \)
The four points are coplanar if the scalar triple product of these vectors is zero:
\( [ \vec{AB}, \vec{AC}, \vec{AD} ] = \left| \begin{matrix} 4 & 6 & 2 \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{matrix} \right| \)
Expanding the determinant along the first row:
\( = 4((10)(5) - (5)(5)) - 6((3)(5) - (5)(-4)) + 2((3)(5) - (10)(-4)) \)
\( = 4(50 - 25) - 6(15 + 20) + 2(15 + 40) \)
\( = 4(25) - 6(35) + 2(55) \)
\( = 100 - 210 + 110 = 0 \)
Since the scalar triple product is \( 0 \), the vectors are coplanar, meaning the four points are coplanar.
The equation of the common plane is given by:
\( \left| \begin{matrix} x - 0 & y + 1 & z + 1 \\ 4 & 6 & 2 \\ 3 & 10 & 5 \end{matrix} \right| = 0 \)
\( x(30 - 20) - (y + 1)(20 - 6) + (z + 1)(40 - 18) = 0 \)
\( 10x - 14(y + 1) + 22(z + 1) = 0 \)
\( 10x - 14y - 14 + 22z + 22 = 0 \)
\( 10x - 14y + 22z + 8 = 0 \)
Dividing the entire equation by \( 2 \):
\( 5x - 7y + 11z + 4 = 0 \)
Therefore, the equation of the common plane is \( 5x - 7y + 11z + 4 = 0 \).
In simple words: Construct three vectors starting from the first point to the other three. Since their determinant is zero, the points all lie on a single flat plane. We then find the plane's equation by taking the determinant with a generic variable point.

Exam Tip: Using the vector scalar triple product is often preferred by board examiners as it is direct, elegant, and reduces the chance of algebraic errors.

(iv) Shortest Distance Between Two Lines

Level-II

Question 1. Find the shortest distance between the lines \( l_1 \) and \( l_2 \) given by the following:
(a) \( l_1 : \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{1} \) and \( l_2 : \frac{x-2}{2} = \frac{y+1}{1} = \frac{z+1}{2} \)
(b) \( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k}) \) and \( \vec{r} = (4 + 2\mu)\hat{i} + (5 + 3\mu)\hat{j} + (6 + \mu)\hat{k} \)
Answer:
(a) Comparing the given Cartesian equations with standard symmetric forms:
For \( l_1 \): It passes through \( A(1, 2, 1) \), so \( \vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k} \) and has direction vector \( \vec{b}_1 = \hat{i} - \hat{j} + \hat{k} \).
For \( l_2 \): It passes through \( B(2, -1, -1) \), so \( \vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k} \) and has direction vector \( \vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k} \).
Now, we find the cross product of the direction vectors:
\( \vec{b}_1 \times \vec{b}_2 = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{matrix} \right| = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - (-2)) = -3\hat{i} + 3\hat{k} \)
The magnitude of the cross product is:
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \)
The difference between the position vectors of the passing points is:
\( \vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k} \)
Taking the dot product:
\( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 3\hat{k}) = (1)(-3) + (-3)(0) + (-2)(3) = -9 \)
The shortest distance is:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{|-9|}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \text{ units} \)

(b) The vector equations are:
\( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k}) \)
\( \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k}) \)
Here, we have:
\( \vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k} \)
\( \vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}, \quad \vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k} \)
Finding the cross product of the direction vectors:
\( \vec{b}_1 \times \vec{b}_2 = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{matrix} \right| = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k} \)
Its magnitude is:
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19} \)
The vector between the two starting points is:
\( \vec{a}_2 - \vec{a}_1 = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} \)
Taking the dot product:
\( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = 3(-9) + 3(3) + 3(9) = 9 \)
The shortest distance is:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{|9|}{3\sqrt{19}} = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19} \text{ units} \)
In simple words: The shortest distance is found by taking the vector between any point on the first line and any point on the second line, and projecting it onto the common perpendicular of both lines (which is found using the cross product of their directions).

Exam Tip: Be careful to simplify terms like \( \frac{3}{\sqrt{2}} \) to \( \frac{3\sqrt{2}}{2} \) by rationalizing the denominator, as textbook answers and board schemes usually prefer rationalized values.

Question 2. Show that the lines \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \) and \( \frac{x - 4}{5} = \frac{y - 1}{2} = z \) intersect. Find their point of intersection.
Answer:
Let any point on the first line be written in terms of parameter \( \lambda \):
\( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda \)

\(\implies x = 2\lambda + 1, \quad y = 3\lambda + 2, \quad z = 4\lambda + 3 \)
Let any point on the second line be written in terms of parameter \( \mu \) (writing \( z \) as \( \frac{z-0}{1} \)):
\( \frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1} = \mu \)

\(\implies x = 5\mu + 4, \quad y = 2\mu + 1, \quad z = \mu \)
If the two lines intersect, their coordinates must match:
1) \( 2\lambda + 1 = 5\mu + 4 \implies 2\lambda - 5\mu = 3 \)
2) \( 3\lambda + 2 = 2\mu + 1 \implies 3\lambda - 2\mu = -1 \)
3) \( 4\lambda + 3 = \mu \)
Substituting equation (3) into equation (2):
\( 3\lambda - 2(4\lambda + 3) = -1 \)
\( 3\lambda - 8\lambda - 6 = -1 \)
\( -5\lambda = 5 \)

\(\implies \lambda = -1 \)
Using this value in equation (3):
\( \mu = 4(-1) + 3 = -1 \)
Let us verify if \( \lambda = -1 \) and \( \mu = -1 \) satisfy equation (1):
L.H.S. = \( 2(-1) - 5(-1) = -2 + 5 = 3 \)
R.H.S. = \( 3 \)
Since L.H.S. = R.H.S., the lines intersect.
Substituting \( \lambda = -1 \) into the coordinate equations for the first line:
\( x = 2(-1) + 1 = -1 \)
\( y = 3(-1) + 2 = -1 \)
\( z = 4(-1) + 3 = -1 \)
Thus, the point of intersection is \( (-1, -1, -1) \).
In simple words: Represent points on both lines using parameters, then solve the system of equations. Since we get consistent values for all coordinate axes, the lines cross at the point \( (-1, -1, -1) \).

Exam Tip: Setting up the coordinates parametric-wise is the standard and most reliable method for finding the point of intersection of two lines.

Question 3. Find the shortest distance between the lines \( \vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k}) \), and \( \vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(4\hat{i} - 2\hat{j} + 2\hat{k}) \)
Answer:
Let us compare the given equations with the standard vector equations of lines:
For the first line: \( \vec{a}_1 = \hat{i} + \hat{j} \) and \( \vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k} \).
For the second line: \( \vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k} \) and \( \vec{b}_2 = 4\hat{i} - 2\hat{j} + 2\hat{k} \).
Observe that:
\( \vec{b}_2 = 2(2\hat{i} - \hat{j} + \hat{k}) = 2\vec{b}_1 \)
Since the direction vectors are parallel, the lines are parallel. We can take the common direction vector as \( \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \).
The distance between two parallel lines is given by:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|} \)
First, find \( \vec{a}_2 - \vec{a}_1 \):
\( \vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} = \hat{i} - \hat{k} \)
Now, find the cross product \( (\vec{a}_2 - \vec{a}_1) \times \vec{b} \):
\( (\vec{a}_2 - \vec{a}_1) \times \vec{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 2 & -1 & 1 \end{matrix} \right| = \hat{i}(0 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - 0) = -\hat{i} - 3\hat{j} - \hat{k} \)
Now, calculate the magnitudes:
\( |(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{(-1)^2 + (-3)^2 + (-1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11} \)
\( |\vec{b}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \)
So, the shortest distance is:
\( d = \frac{\sqrt{11}}{\sqrt{6}} = \sqrt{\frac{11}{6}} \text{ units} \).
In simple words: Since the direction vectors are multiples of each other, these lines are parallel. We use a specific formula involving cross product and magnitude to find the gap between them.

Exam Tip: Always check if the lines are parallel before applying the shortest distance formula. Parallel lines have a different formula than skew lines.

Question 4. Find the shortest distance between the lines \( \vec{r} = (1 - t)\hat{i} + (t - 2)\hat{j} + (3 - t)\hat{k} \) and \( \vec{r} = (s + 1)\hat{i} + (2s - 1)\hat{j} + (2s + 1)\hat{k} \) [CBSE 2011]
Answer:
Let us rewrite the given equations of lines in the standard form \( \vec{r} = \vec{a} + \lambda \vec{b} \):
First line:
\( \vec{r} = (1 - t)\hat{i} + (t - 2)\hat{j} + (3 - t)\hat{k} = (\hat{i} - 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} - \hat{k}) \)
Here, \( \vec{a}_1 = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{b}_1 = -\hat{i} + \hat{j} - \hat{k} \).
Second line:
\( \vec{r} = (s + 1)\hat{i} + (2s - 1)\hat{j} + (2s + 1)\hat{k} = (\hat{i} - \hat{j} + \hat{k}) + s(\hat{i} + 2\hat{j} + 2\hat{k}) \)
Here, \( \vec{a}_2 = \hat{i} - \hat{j} + \hat{k} \) and \( \vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k} \).
The shortest distance is given by:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} \)
Let us find the cross product \( \vec{b}_1 \times \vec{b}_2 \):
\( \vec{b}_1 \times \vec{b}_2 = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -1 \\ 1 & 2 & 2 \end{matrix} \right| = \hat{i}(2 - (-2)) - \hat{j}(-2 - (-1)) + \hat{k}(-2 - 1) = 4\hat{i} + \hat{j} - 3\hat{k} \)
The magnitude is:
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{4^2 + 1^2 + (-3)^2} = \sqrt{16 + 1 + 9} = \sqrt{26} \)
The difference vector is:
\( \vec{a}_2 - \vec{a}_1 = (1 - 1)\hat{i} + (-1 - (-2))\hat{j} + (1 - 3)\hat{k} = 0\hat{i} + \hat{j} - 2\hat{k} \)
Now, take the dot product:
\( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (0\hat{i} + \hat{j} - 2\hat{k}) \cdot (4\hat{i} + \hat{j} - 3\hat{k}) = 0(4) + 1(1) + (-2)(-3) = 7 \)
Thus, the shortest distance is:
\( d = \frac{7}{\sqrt{26}} \text{ units} \).
In simple words: First rewrite both lines so they show their points and direction vectors clearly. Use cross product to find a perpendicular direction, and project the vector connecting the lines onto it.

Exam Tip: Separating the parameter terms from the constant terms to convert the equation into standard vector form is a critical first step that must be shown clearly.

Question 5. Find the distance between the parallel planes x + y – z = -4 and 2x + 2y – 2z + 10 = 0.
Answer:
The given planes are:
Plane 1: \( x + y - z + 4 = 0 \)
Plane 2: \( 2x + 2y - 2z + 10 = 0 \)
To find the distance, we must first make the coefficients of \( x, y, z \) identical. Dividing the equation of Plane 2 by \( 2 \):
\( x + y - z + 5 = 0 \)
Now comparing both equations with the standard parallel planes \( ax+by+cz+d_1=0 \) and \( ax+by+cz+d_2=0 \):
Here, \( a = 1, \quad b = 1, \quad c = -1, \quad d_1 = 4, \quad d_2 = 5 \)
The formula for the distance between parallel planes is:
\( d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} \)
\( d = \frac{|4 - 5|}{\sqrt{1^2 + 1^2 + (-1)^2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \text{ units} \).
In simple words: Divide the second plane's equation by 2 so that the variables look identical to the first. Then, take the difference of the constant numbers on the end and divide by the length of the coefficient vector.

Exam Tip: Always check that the coefficients of \( x, y, \) and \( z \) are the same in both plane equations before using the parallel plane distance formula.

Question 6. Find the vector equation of the line parallel to the line \( \frac{x-1}{5} = \frac{3-y}{2} = \frac{z+1}{4} \) and passing through (3,0,-4). Also, find the distance between these two lines.
Answer:
First, let us rewrite the given line in standard form:
\( \frac{x-1}{5} = \frac{y-3}{-2} = \frac{z+1}{4} \)
The direction ratios of this line are \( (5, -2, 4) \). Therefore, the direction vector is:
\( \vec{b} = 5\hat{i} - 2\hat{j} + 4\hat{k} \)
The parallel line passes through the point \( A(3, 0, -4) \), so its position vector is \( \vec{a}_2 = 3\hat{i} - 4\hat{k} \).
The vector equation of this line is:
\( \vec{r} = (3\hat{i} - 4\hat{k}) + \lambda(5\hat{i} - 2\hat{j} + 4\hat{k}) \)
Next, we calculate the distance between these two parallel lines using the formula:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|} \)
Here, the starting point of the first line is \( \vec{a}_1 = \hat{i} + 3\hat{j} - \hat{k} \).
\( \vec{a}_2 - \vec{a}_1 = (3-1)\hat{i} + (0-3)\hat{j} + (-4 - (-1))\hat{k} = 2\hat{i} - 3\hat{j} - 3\hat{k} \)
Now find the cross product \( (\vec{a}_2 - \vec{a}_1) \times \vec{b} \):
\( (\vec{a}_2 - \vec{a}_1) \times \vec{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -3 \\ 5 & -2 & 4 \end{matrix} \right| = \hat{i}(-12 - 6) - \hat{j}(8 - (-15)) + \hat{k}(-4 - (-15)) = -18\hat{i} - 23\hat{j} + 11\hat{k} \)
Let us find the magnitudes:
\( |(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{(-18)^2 + (-23)^2 + 11^2} = \sqrt{324 + 529 + 121} = \sqrt{974} \)
\( |\vec{b}| = \sqrt{5^2 + (-2)^2 + 4^2} = \sqrt{25 + 4 + 16} = \sqrt{45} = 3\sqrt{5} \)
So, the distance is:
\( d = \frac{\sqrt{974}}{3\sqrt{5}} = \frac{\sqrt{4870}}{15} \text{ units} \).
In simple words: Find the direction of the line by rewriting its middle term with a positive \( y \). Use this direction and the new point to get the vector equation. The gap between them is found by using the parallel lines vector distance formula.

Exam Tip: Changing \( \frac{3-y}{2} \) to \( \frac{y-3}{-2} \) is essential for finding the correct direction ratios of the line.

(v) Cartesian and Vector Equation of a Plane in Space & Conversion of One into Another Form

Level-I

Question 1. Find the equation of a plane passing through the origin and perpendicular to x-axis
Answer:
A plane passing through the origin \( (0,0,0) \) and perpendicular to the x-axis has its normal vector along the x-axis.
The direction ratios of the normal vector are \( (1, 0, 0) \).
Using the standard equation of a plane \( a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 \):
\( 1(x - 0) + 0(y - 0) + 0(z - 0) = 0 \)
\( x = 0 \)
Thus, the equation of the plane is \( x = 0 \).
In simple words: A flat surface that goes through the origin and stands straight perpendicular to the x-axis is simply the \( yz \)-plane, which has the equation \( x = 0 \).

Exam Tip: Recall that the equations of coordinate planes are \( x = 0 \) (\( yz \)-plane), \( y = 0 \) (\( xz \)-plane), and \( z = 0 \) (\( xy \)-plane).

Question 2. Find the equation of plane with intercepts 2, 3, 4 on the x ,y, z –axis respectively.
Answer:
The intercept form of the equation of a plane is:
\( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \)
Here, the intercepts are \( a = 2 \), \( b = 3 \), and \( c = 4 \).
Substituting these values:
\( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1 \)
To convert this to general form, multiply the entire equation by the LCM of 2, 3, and 4, which is 12:
\( 12 \left( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} \right) = 12(1) \)
\( 6x + 4y + 3z = 12 \)
Thus, the equation of the plane is \( 6x + 4y + 3z - 12 = 0 \).
In simple words: Use the intercept formula where you put the axis lengths under each variable, then multiply everything by 12 to clear the fractions.

Exam Tip: Be sure to write the final plane equation in general form \( ax+by+cz+d=0 \), as it is preferred for marks.

Question 3. Find the direction cosines of the unit vector perpendicular to the plane \( \vec{r} \cdot (6\hat{i} - 3\hat{j} - 2\hat{k}) + 1 = 0 \) passing through the origin.
Answer:
The equation of the plane is:
\( \vec{r} \cdot (6\hat{i} - 3\hat{j} - 2\hat{k}) + 1 = 0 \)
First, we rewrite this so that the constant term on the right-hand side is positive (representing positive distance from origin):
\( \vec{r} \cdot (6\hat{i} - 3\hat{j} - 2\hat{k}) = -1 \)
Multiplying both sides by \( -1 \):
\( \vec{r} \cdot (-6\hat{i} + 3\hat{j} + 2\hat{k}) = 1 \)
Now, the normal vector is \( \vec{n} = -6\hat{i} + 3\hat{j} + 2\hat{k} \).
Its magnitude is:
\( |\vec{n}| = \sqrt{(-6)^2 + 3^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \)
Dividing by this magnitude gives the unit normal vector:
\( \hat{n} = \frac{-6\hat{i} + 3\hat{j} + 2\hat{k}}{7} = -\frac{6}{7}\hat{i} + \frac{3}{7}\hat{j} + \frac{2}{7}\hat{k} \)
The direction cosines are the components of this unit vector:
\( l = -\frac{6}{7}, \quad m = \frac{3}{7}, \quad n = \frac{2}{7} \)
Thus, the direction cosines of the perpendicular unit vector are \( \left(-\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\right) \).
In simple words: Rewrite the equation so the number on the right is positive. Then, divide the direction coefficients by their total vector length (which is 7) to get the unit values.

Exam Tip: Don't forget to multiply the entire equation by \( -1 \) if the constant is negative. The normal form requires the distance \( p \) on the right side to be positive.

Question 4. Find the Cartesian equation of the following planes: (a) \( \vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 2 \) (b) \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1 \)
Answer:
To find the Cartesian equations, we substitute the general position vector \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) into the vector equations:

(a) \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 2 \)
Taking the dot product:
\( x(1) + y(1) + z(-1) = 2 \)
\( x + y - z = 2 \)
Thus, the Cartesian equation is \( x + y - z = 2 \).

(b) \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1 \)
Taking the dot product:
\( x(2) + y(3) + z(-4) = 1 \)
\( 2x + 3y - 4z = 1 \)
Thus, the Cartesian equation is \( 2x + 3y - 4z = 1 \).
In simple words: Replace \( \vec{r} \) with \( x, y, z \) and multiply them by the corresponding numbers inside the parentheses.

Exam Tip: Substituting \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) is the standard, foolproof way to convert any vector equation in 3D geometry to its Cartesian equivalent.

Level II

Question 1. Find the vector and cartesian equations of the plane which passes through the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1.
Answer:
The plane passes through the point \( A(5, 2, -4) \). Its position vector is:
\( \vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k} \)
Since the plane is perpendicular to a line with direction ratios \( (2, 3, -1) \), this line is normal to the plane. Thus, the normal vector \( \vec{n} \) is:
\( \vec{n} = 2\hat{i} + 3\hat{j} - \hat{k} \)
The vector equation of a plane is:
\( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \)
Substituting the vectors:
\( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = (5\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) \)
\( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = (5)(2) + (2)(3) + (-4)(-1) = 10 + 6 + 4 = 20 \)
Thus, the vector equation is \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 20 \).
The Cartesian equation is found by substituting \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \):
\( 2x + 3y - z = 20 \).
In simple words: The direction of the perpendicular line serves as the normal vector. Multiply this normal vector by a variable point vector, and set it equal to the dot product of the normal with the given point.

Exam Tip: Remember that a line perpendicular to a plane has direction ratios that are proportional to the direction ratios of the normal to that plane.

Question 2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3 iˆ + 5 ˆj - 6 kˆ.
Answer:
Let the given normal vector be \( \vec{n} = 3\hat{i} + 5\hat{j} - 6\hat{k} \).
First, we find its unit vector \( \hat{n} \) by dividing it by its magnitude:
\( |\vec{n}| = \sqrt{3^2 + 5^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70} \)
The unit normal vector is:
\( \hat{n} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}} \)
The equation of the plane in normal form is \( \vec{r} \cdot \hat{n} = d \), where \( d \) is the distance from the origin (7 units):
\( \vec{r} \cdot \left( \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}} \right) = 7 \)
Multiplying both sides by \( \sqrt{70} \):
\( \vec{r} \cdot (3\hat{i} + 5\hat{j} - 6\hat{k}) = 7\sqrt{70} \).
In simple words: Find the length of the normal vector, divide the vector by this length to make it a unit vector, and set up the standard normal form equation with the distance.

Exam Tip: Be sure to divide the normal vector by its magnitude to obtain the unit normal vector \( \hat{n} \) before setting up the equation.

Question 3. Find the vector and cartesian equations of the planes that passes through the point (1, 0, – 2) and the normal to the plane is iˆ + ˆj - kˆ.
Answer:
The plane passes through point \( A(1, 0, -2) \), so the position vector is \( \vec{a} = \hat{i} - 2\hat{k} \).
The normal vector is \( \vec{n} = \hat{i} + \hat{j} - \hat{k} \).
Using the plane equation formula:
\( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \)
\( \vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = (\hat{i} - 2\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) \)
\( \vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = (1)(1) + (0)(1) + (-2)(-1) = 3 \)
Thus, the vector equation is:
\( \vec{r} \cdot (\hat{i} + \hat{j} - \hat{k}) = 3 \).
Substituting \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) gives the Cartesian equation:
\( x + y - z = 3 \).
In simple words: Find the dot product of the normal direction and the point coordinates, which is 3. The plane equation is then the coordinates \( x, y, -z \) added up to equal 3.

Exam Tip: This is a fundamental type of question; writing down the formula \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \) secures partial marks immediately.

(vi) Angle Between (i) Two lines (ii) Two planes (iii) Line & plane

Level-I

Question 1. Find the angle between the lines whose direction ratios are (1, 1, 2) and (√3 – 1, -√3 – 1, 4).
Answer:
Let the direction ratios of the two lines be \( (a_1, b_1, c_1) = (1, 1, 2) \) and \( (a_2, b_2, c_2) = (\sqrt{3}-1, -\sqrt{3}-1, 4) \).
The angle \( \theta \) between the lines is given by:
\( \cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \)
Let us calculate the numerator term:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6 \)
Now, calculate the magnitudes in the denominator:
\( \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6} \)
\( \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 4^2} \)
\( = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{8 + 16} = \sqrt{24} = 2\sqrt{6} \)
Substituting these values back into the cosine equation:
\( \cos \theta = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2} \)
Since \( \cos \theta = \frac{1}{2} \), we have:
\( \theta = 60^\circ \text{ or } \frac{\pi}{3} \text{ radians} \).
In simple words: Plug the ratios into the angle formula. The complicated square root terms simplify nicely, leaving you with a cosine value of a half, which corresponds to 60 degrees.

Exam Tip: Don't be intimidated by terms containing square roots; they often cancel out or simplify to neat values during intermediate steps.

Question 2. Find the angle between line \( \frac{x-2}{3} = \frac{y+1}{-1} = \frac{z-3}{2} \) and the plane 3x + 4y + z + 5 = 0.
Answer:
The direction ratios of the line are \( \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \).
The normal vector of the plane is \( \vec{n} = 3\hat{i} + 4\hat{j} + \hat{k} \).
The angle \( \theta \) between a line and a plane is given by:
\( \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \)
First, find the dot product:
\( \vec{b} \cdot \vec{n} = (3)(3) + (-1)(4) + (2)(1) = 9 - 4 + 2 = 7 \)
Next, find the magnitudes:
\( |\vec{b}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \)
\( |\vec{n}| = \sqrt{3^2 + 4^2 + 1^2} = \sqrt{9 + 16 + 1} = \sqrt{26} \)
Substituting into the sine equation:
\( \sin \theta = \frac{7}{\sqrt{14} \cdot \sqrt{26}} = \frac{7}{\sqrt{364}} = \frac{7}{2\sqrt{91}} \)
Thus, the angle is:
\( \theta = \sin^{-1} \left( \frac{7}{2\sqrt{91}} \right) \).
In simple words: The angle between a line and a plane is calculated using sine instead of cosine. Find the dot product of the line's direction and the plane's normal, then divide by the product of their lengths.

Exam Tip: Remember that the angle between a line and a plane uses \( \sin \theta \) in the formula, whereas the angle between two lines or two planes uses \( \cos \theta \).

Question 3. Find the value of \(\lambda\) such that the line \( \frac{x-2}{9} = \frac{y-1}{\lambda} = \frac{z+3}{-6} \) is perpendicular to the plane 3x – y – 2z = 7.
Answer:
If a line is perpendicular to a plane, its direction ratios must be proportional to the direction ratios of the normal to the plane.
The direction ratios of the line are \( (9, \lambda, -6) \).
The direction ratios of the plane's normal are \( (3, -1, -2) \).
For them to be parallel, we set up the proportionality ratios:
\( \frac{9}{3} = \frac{\lambda}{-1} = \frac{-6}{-2} \)
\( 3 = -\lambda = 3 \)

\(\implies \lambda = -3 \).
Thus, the required value of \( \lambda \) is \( -3 \).
In simple words: If a line is perpendicular to a plane, it points in the exact same direction as the plane's normal. Set up fractions comparing their components, which tells us that \( \lambda \) must be \( -3 \).

Exam Tip: A line being perpendicular to a plane means its direction vector is parallel to the normal vector of the plane, so coordinate-wise components are proportional.

Question 4. Find the angle between the planes whose vector equations are \( \vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) = 5 \) and \( \vec{r} \cdot (3\hat{i} - 3\hat{j} + 5\hat{k}) = 3 \).
Answer:
The normal vectors of the two planes are:
\( \vec{n}_1 = 2\hat{i} + 2\hat{j} - 3\hat{k} \)
\( \vec{n}_2 = 3\hat{i} - 3\hat{j} + 5\hat{k} \)
The angle \( \theta \) between the planes is equal to the angle between their normal vectors:
\( \cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|} \)
Let us calculate the dot product:
\( \vec{n}_1 \cdot \vec{n}_2 = (2)(3) + (2)(-3) + (-3)(5) = 6 - 6 - 15 = -15 \)
Now, calculate the magnitudes:
\( |\vec{n}_1| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \)
\( |\vec{n}_2| = \sqrt{3^2 + (-3)^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43} \)
Substituting these into the cosine equation:
\( \cos \theta = \frac{|-15|}{\sqrt{17} \cdot \sqrt{43}} = \frac{15}{\sqrt{731}} \)
Thus, the angle is:
\( \theta = \cos^{-1} \left( \frac{15}{\sqrt{731}} \right) \).
In simple words: The angle between two planes is the angle between their normal vectors. Take the dot product and divide by the product of their vector lengths to get the cosine of the angle.

Exam Tip: Make sure to apply the absolute value on the numerator of the angle formula so that you obtain an acute angle between the planes.

Question 5. Find the angle between the line \( \frac{x+1}{2} = \frac{y}{3} = \frac{z-3}{6} \) and the plane 10x + 2y – 11z = 3.
Answer:
The direction ratios of the line are \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \).
The normal vector of the plane is \( \vec{n} = 10\hat{i} + 2\hat{j} - 11\hat{k} \).
The angle \( \theta \) between the line and the plane is given by:
\( \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \)
First, find the dot product:
\( \vec{b} \cdot \vec{n} = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = -40 \)
Next, find the magnitudes:
\( |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \)
\( |\vec{n}| = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15 \)
Substituting these values:
\( \sin \theta = \frac{|-40|}{7 \cdot 15} = \frac{40}{105} = \frac{8}{21} \)
Thus, the angle is:
\( \theta = \sin^{-1} \left( \frac{8}{21} \right) \).
In simple words: Find the dot product of the line's direction and the plane's normal, then divide by their lengths multiplied together. This gives you the sine of the angle, which simplifies to 8/21.

Exam Tip: Since both magnitudes simplify to integers (7 and 15), this is a common and high-scoring type of board exam question. Double-check your arithmetic to get full marks.

Level-II

Question 1. Find the value of p, such that the lines \( \frac{x}{1} = \frac{y}{3} = \frac{z}{2p} \) and \( \frac{x}{-3} = \frac{y}{5} = \frac{z}{2} \) are perpendicular to each other.
Answer:
The direction ratios of the two lines are:
\( (a_1, b_1, c_1) = (1, 3, 2p) \)
\( (a_2, b_2, c_2) = (-3, 5, 2) \)
Since the two lines are perpendicular, their dot product must be zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
Substituting the values:
\( (1)(-3) + (3)(5) + (2p)(2) = 0 \)
\( -3 + 15 + 4p = 0 \)
\( 12 + 4p = 0 \)

\(\implies 4p = -12 \)

\(\implies p = -3 \).
Thus, the required value of \( p \) is \( -3 \).
In simple words: When two lines are at right angles, multiplying their direction numbers coordinate-by-coordinate and adding them up must equal zero. This gives us a simple equation to solve for \( p \).

Exam Tip: The condition for perpendicularity \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \) is one of the most frequently tested concepts in 3D geometry exams.

Question 2. A line makes angles \(\alpha, \beta, \gamma, \delta\) with the four diagonals of a cube, Prove that \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma + \cos^2\delta = \frac{4}{3} \).
Answer:
Let us place a cube of side \( a \) with one corner at the origin and edges along the coordinate axes \( OX, OY, OZ \).
The four diagonals are:
1. \( D_1 \) joining \( O(0,0,0) \) and \( P(a,a,a) \) with direction cosines: \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \)
2. \( D_2 \) joining \( A(a,0,0) \) and \( B(0,a,a) \) with direction cosines: \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \)
3. \( D_3 \) joining \( C(0,a,0) \) and \( D(a,0,a) \) with direction cosines: \( \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \)
4. \( D_4 \) joining \( E(0,0,a) \) and \( F(a,a,0) \) with direction cosines: \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) \)
Let the given line have direction cosines \( (l, m, n) \), where \( l^2 + m^2 + n^2 = 1 \).
Using the angle formula between two lines, the cosines of the angles with the four diagonals are:
\( \cos \alpha = \frac{l + m + n}{\sqrt{3}} \)
\( \cos \beta = \frac{-l + m + n}{\sqrt{3}} \)
\( \cos \gamma = \frac{l - m + n}{\sqrt{3}} \)
\( \cos \delta = \frac{l + m - n}{\sqrt{3}} \)
Squaring and adding these equations:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{1}{3} \left[ (l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2 \right] \)
Using algebraic expansion:
\( (l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2 = 4(l^2 + m^2 + n^2) \)
Since \( l^2 + m^2 + n^2 = 1 \):
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{1}{3} \left[ 4(1) \right] = \frac{4}{3} \). (Hence proved).
In simple words: Represent the four cube diagonals using coordinates. Using the direction cosines of a general line, calculate the cosine of the angle for each diagonal, square them, and add them up. Since the sum of squared direction cosines is 1, the result simplifies to exactly 4/3.

Exam Tip: This is a standard theoretical proof. Carefully write out the coordinates of the vertices of the cube to show how the direction cosines of the diagonals are derived.

(vii) Distance of a Point from a Plane

Level-I

Question 1. Write the distance of plane 2x– y + 2z + 1 = 0 from the origins.
Answer:
The given plane is \( 2x - y + 2z + 1 = 0 \).
The perpendicular distance from the origin \( (0,0,0) \) to the plane is given by:
\( d = \frac{|d|}{\sqrt{a^2 + b^2 + c^2}} \)
Substituting \( a = 2, \quad b = -1, \quad c = 2, \quad d = 1 \):
\( d = \frac{|1|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{1}{\sqrt{4 + 1 + 4}} = \frac{1}{\sqrt{9}} = \frac{1}{3} \text{ units} \).
In simple words: Plug the point \( (0,0,0) \) into the plane equation, which leaves just the constant 1 on top, and divide it by the length of the normal vector (which is 3).

Exam Tip: When calculating distance, always keep the result positive by using absolute values in the numerator.

Question 2. Find the point through which the line 2x = 3y = 4z passes.
Answer:
The given equation is:
\( 2x = 3y = 4z \)
To convert this to standard symmetric form, divide by the LCM of 2, 3, and 4, which is 12:
\( \frac{2x}{12} = \frac{3y}{12} = \frac{4z}{12} \)
\( \frac{x}{6} = \frac{y}{4} = \frac{z}{3} \)
This can be written as:
\( \frac{x - 0}{6} = \frac{y - 0}{4} = \frac{z - 0}{3} \)
Comparing this with \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), the line passes through \( (x_1, y_1, z_1) = (0, 0, 0) \).
Thus, the line passes through the origin \( (0, 0, 0) \).
In simple words: Dividing the equations to put them in fraction form shows that there are no numbers subtracted from \( x, y, z \). This means the line starts exactly at the origin.

Exam Tip: Converting to symmetric form is the standard way to find the direction ratios as well as a passing point on the line.

Question 3. Find the distance of a point (2, 5, – 3) from the plane \( \vec{r} \cdot (6\hat{i} - 3\hat{j} + 2\hat{k}) = 4 \)
Answer:
First, let us express the equation of the plane in Cartesian form:
\( 6x - 3y + 2z - 4 = 0 \)
The perpendicular distance from a point \( (x_1, y_1, z_1) = (2, 5, -3) \) to the plane is:
\( d = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \)
Substituting the coordinates and coefficients:
\( d = \frac{|6(2) - 3(5) + 2(-3) - 4|}{\sqrt{6^2 + (-3)^2 + 2^2}} \)
\( d = \frac{|12 - 15 - 6 - 4|}{\sqrt{36 + 9 + 4}} = \frac{|-13|}{\sqrt{49}} = \frac{13}{7} \text{ units} \).
In simple words: Write the plane in Cartesian coordinates, substitute the point's numbers into it, and divide the absolute result by the length of the coefficient vector (which is 7).

Exam Tip: Be sure to move the constant term to the left-hand side (making it \( -4 \)) before using the distance formula to avoid sign confusion.

Question 4. Find the distance of the following plane from origin: 2x – y + 2z + 1 = 0
Answer:
The given plane is \( 2x - y + 2z + 1 = 0 \).
The perpendicular distance from the origin \( (0,0,0) \) is given by:
\( d = \frac{|d|}{\sqrt{a^2 + b^2 + c^2}} \)
Substituting \( a = 2, \quad b = -1, \quad c = 2, \quad d = 1 \):
\( d = \frac{|1|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{1}{\sqrt{4 + 1 + 4}} = \frac{1}{\sqrt{9}} = \frac{1}{3} \text{ units} \).
In simple words: Substitute \( (0,0,0) \) into the equation, which leaves 1 on top, and divide it by the length of the normal vector, which is 3.

Exam Tip: The distance from the origin is a special case of the general point-to-plane formula where the variable coordinates are set to zero.

Question 5. Find the distance of the point (a,b,c) from x-axis
Answer:
Let \( P(a, b, c) \) be the given point. The projection of \( P \) on the x-axis is \( A(a, 0, 0) \).
The distance of point \( P \) from the x-axis is the distance \( PA \):
\( d = \sqrt{(a - a)^2 + (b - 0)^2 + (c - 0)^2} = \sqrt{b^2 + c^2} \).
Thus, the distance is \( \sqrt{b^2 + c^2} \).
In simple words: To find the distance from the x-axis, ignore the x-coordinate and find the square root of the sum of the squares of the y and z coordinates.

Exam Tip: Similarly, the distance of a point \( (a,b,c) \) from the y-axis is \( \sqrt{a^2+c^2} \), and from the z-axis is \( \sqrt{a^2+b^2} \).

Level II

Question 1. Find the points on the line \( \frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2} \) at a distance of 5 units from the point P(1,3,3).
Answer:
Any point \( Q \) on the given line can be written in terms of parameter \( \lambda \):
\( \frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2} = \lambda \)
\( x = 3\lambda - 2, \quad y = 2\lambda - 1, \quad z = 2\lambda + 3 \)
So, \( Q = (3\lambda - 2, 2\lambda - 1, 2\lambda + 3) \).
The distance between \( P(1,3,3) \) and \( Q \) is given as 5 units:
\( PQ = 5 \implies PQ^2 = 25 \)
\( (3\lambda - 2 - 1)^2 + (2\lambda - 1 - 3)^2 + (2\lambda + 3 - 3)^2 = 25 \)
\( (3\lambda - 3)^2 + (2\lambda - 4)^2 + (2\lambda)^2 = 25 \)
\( (9\lambda^2 - 18\lambda + 9) + (4\lambda^2 - 16\lambda + 16) + 4\lambda^2 = 25 \)
\( 17\lambda^2 - 34\lambda + 25 = 25 \)
\( 17\lambda^2 - 34\lambda = 0 \)
\( 17\lambda(\lambda - 2) = 0 \)

\(\implies \lambda = 0 \text{ or } \lambda = 2 \).
For \( \lambda = 0 \): the point is \( (-2, -1, 3) \).
For \( \lambda = 2 \): the point is \( (3(2)-2, 2(2)-1, 2(2)+3) = (4, 3, 7) \).
Therefore, the required points on the line are \( (-2, -1, 3) \) and \( (4, 3, 7) \).
In simple words: Represent a general point on the line using a helper variable \( \lambda \). Write the distance formula between this general point and the given point, set it equal to 5, and solve the quadratic equation to get two values for \( \lambda \).

Exam Tip: Quadratic equations in such distance problems typically yield two values, representing two distinct points on the line that are at the specified distance from \( P \).

Question 2. Find the distance of the point (3,4,5) from the plane x + y + z = 2 measured parallel to the line 2x = y = z.
Answer:
The given point is \( P(3, 4, 5) \).
The line along which distance is measured is \( 2x = y = z \). We rewrite this in standard form by dividing by 2:
\( \frac{x}{1} = \frac{y}{2} = \frac{z}{2} \)
The direction ratios of this line are \( (1, 2, 2) \).
The equation of the line passing through \( P(3, 4, 5) \) and parallel to this direction is:
\( \frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = \lambda \)
Any point \( Q \) on this line is:
\( Q = (\lambda + 3, 2\lambda + 4, 2\lambda + 5) \)
Since \( Q \) lies on the plane \( x + y + z = 2 \), its coordinates must satisfy the plane's equation:
\( (\lambda + 3) + (2\lambda + 4) + (2\lambda + 5) = 2 \)
\( 5\lambda + 12 = 2 \)
\( 5\lambda = -10 \)

\(\implies \lambda = -2 \).
Substituting \( \lambda = -2 \) back into the coordinates of \( Q \):
\( Q = (-2 + 3, 2(-2) + 4, 2(-2) + 5) = (1, 0, 1) \).
The required distance is the distance between \( P(3, 4, 5) \) and \( Q(1, 0, 1) \):
\( PQ = \sqrt{(3-1)^2 + (4-0)^2 + (5-1)^2} = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \text{ units} \).
In simple words: Write the equation of a line through \( (3,4,5) \) pointing in the same direction as the given line. Find where this line crosses the plane, and then calculate the distance between the starting point and this crossing point.

Exam Tip: "Distance measured parallel to a line" means finding the intersection point of a parallel line through the given point with the plane, and then finding the distance between these two points.

Question 3. Find the distance between the point P(6, 5, 9) and the plane determined by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6).
Answer:
The equation of a plane determined by three points \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) is:
\( \left| \begin{matrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{matrix} \right| = 0 \)
Substituting the coordinates of \( A(3, -1, 2) \), \( B(5, 2, 4) \), and \( C(-1, -1, 6) \):
\( \left| \begin{matrix} x - 3 & y + 1 & z - 2 \\ 5 - 3 & 2 - (-1) & 4 - 2 \\ -1 - 3 & -1 - (-1) & 6 - 2 \end{matrix} \right| = 0 \)
\( \left| \begin{matrix} x - 3 & y + 1 & z - 2 \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{matrix} \right| = 0 \)
Expanding this determinant along the first row:
\( (x - 3)(12 - 0) - (y + 1)(8 - (-8)) + (z - 2)(0 - (-12)) = 0 \)
\( 12(x - 3) - 16(y + 1) + 12(z - 2) = 0 \)
Dividing the entire equation by 4:
\( 3(x - 3) - 4(y + 1) + 3(z - 2) = 0 \)
\( 3x - 9 - 4y - 4 + 3z - 6 = 0 \)
\( 3x - 4y + 3z - 19 = 0 \)
Now, the perpendicular distance from \( P(6, 5, 9) \) to this plane is:
\( d = \frac{|3(6) - 4(5) + 3(9) - 19|}{\sqrt{3^2 + (-4)^2 + 3^2}} \)
\( d = \frac{|18 - 20 + 27 - 19|}{\sqrt{9 + 16 + 9}} = \frac{6}{\sqrt{34}} \text{ units} \).
In simple words: First write down the equation of the plane that passes through the three given points by using a determinant. Then, use the point-to-plane distance formula to find how far \( (6, 5, 9) \) is from this plane.

Exam Tip: Be sure to write the plane's equation in its simplest form before applying the distance formula to avoid large numbers and calculation errors.

Question 4. Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line \( \vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k}) \) and the plane \( \vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 \) 
Answer:
First, let us find the point of intersection of the line and the plane.
The parametric equation of the line is:
\( \vec{r} = (3\lambda + 2)\hat{i} + (4\lambda - 1)\hat{j} + (2\lambda + 2)\hat{k} \)
Substituting this vector into the plane's equation:
\( \left[ (3\lambda + 2)\hat{i} + (4\lambda - 1)\hat{j} + (2\lambda + 2)\hat{k} \right] \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 \)
Taking the dot product:
\( (3\lambda + 2)(1) + (4\lambda - 1)(-1) + (2\lambda + 2)(1) = 5 \)
\( 3\lambda + 2 - 4\lambda + 1 + 2\lambda + 2 = 5 \)
\( \lambda + 5 = 5 \)

\(\implies \lambda = 0 \).
Substituting \( \lambda = 0 \) back into the line equation gives the intersection point \( Q \):
\( Q = (2, -1, 2) \).
Now, we calculate the distance between the given point \( P(-1, -5, -10) \) and the intersection point \( Q(2, -1, 2) \):
\( PQ = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2} \)
\( PQ = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \text{ units} \).
In simple words: Put the line's coordinates into the plane's equation to find the matching parameter value, which is 0. This gives the crossing point as \( (2, -1, 2) \). Then, find the distance between this crossing point and the target point.

Exam Tip: Substituting the parametric vector form of a line into the plane equation is the most efficient way to find their point of intersection.

Level III

Question 1. Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point (1,3,4) from the plane 2x – y + z + 3 = 0. Find also, the image of the point in the plane.
Answer:
Let the given point be \( P(1, 3, 4) \).
The direction ratios of the normal to the plane \( 2x - y + z + 3 = 0 \) are \( (2, -1, 1) \).
The equation of the line passing through \( P(1, 3, 4) \) and perpendicular to the plane is:
\( \frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = \lambda \)
Any point \( F \) on this line is:
\( F = (2\lambda + 1, -\lambda + 3, \lambda + 4) \)
If \( F \) is the foot of the perpendicular, it must lie on the plane:
\( 2(2\lambda + 1) - (-\lambda + 3) + (\lambda + 4) + 3 = 0 \)
\( 4\lambda + 2 + \lambda - 3 + \lambda + 4 + 3 = 0 \)
\( 6\lambda + 6 = 0 \)

\(\implies \lambda = -1 \).
Substituting \( \lambda = -1 \) back into the coordinates of \( F \):
\( F = (2(-1) + 1, -(-1) + 3, -1 + 4) = (-1, 4, 3) \).
Thus, the foot of the perpendicular is \( (-1, 4, 3) \).
The perpendicular distance is the distance \( PF \):
\( PF = \sqrt{(-1 - 1)^2 + (4 - 3)^2 + (3 - 4)^2} = \sqrt{(-2)^2 + 1^2 + (-1)^2} = \sqrt{6} \text{ units} \).
Let the image of point \( P \) in the plane be \( P'(x', y', z') \). Since \( F \) is the midpoint of \( PP' \):
\( \frac{1 + x'}{2} = -1 \implies x' = -3 \)
\( \frac{3 + y'}{2} = 4 \implies y' = 5 \)
\( \frac{4 + z'}{2} = 3 \implies z' = 2 \)
Thus, the coordinates of the image point are \( (-3, 5, 2) \).
In simple words: Write a line equation through the point along the normal direction. Find where it hits the plane, which gives the foot as \( (-1, 4, 3) \). Calculate the distance as \( \sqrt{6} \). Since the image is the same distance on the opposite side, the foot is the midpoint, allowing us to find the image coordinates as \( (-3, 5, 2) \).

Exam Tip: This combined question (foot, distance, and image) is extremely common in 6-mark sections. Be neat with each part as they build on one another.

(ix) Foot of Perpendicular and Image with Respect to a Line and Plane

Level-III

Question 2. Find the image of the point (1, –2, 1) in the line \( \frac{x-2}{3} = \frac{y+1}{-1} = \frac{z+3}{2} \)
Answer:
Let the given point be \( P(1, -2, 1) \).
Any point \( F \) on the given line can be written in terms of parameter \( \lambda \):
\( \frac{x-2}{3} = \frac{y+1}{-1} = \frac{z+3}{2} = \lambda \)
\( F = (3\lambda + 2, -\lambda - 1, 2\lambda - 3) \)
The direction ratios of the segment \( PF \) are:
\( (3\lambda + 2 - 1, -\lambda - 1 - (-2), 2\lambda - 3 - 1) = (3\lambda + 1, -\lambda + 1, 2\lambda - 4) \).
Since the line segment \( PF \) is perpendicular to the given line (which has direction ratios \( 3, -1, 2 \)), their dot product is zero:
\( 3(3\lambda + 1) - 1(-\lambda + 1) + 2(2\lambda - 4) = 0 \)
\( 9\lambda + 3 + \lambda - 1 + 4\lambda - 8 = 0 \)
\( 14\lambda - 6 = 0 \)

\(\implies \lambda = \frac{3}{7} \).
Substituting \( \lambda = \frac{3}{7} \) back into the coordinates of \( F \):
\( x_f = 3 \left( \frac{3}{7} \right) + 2 = \frac{23}{7} \)
\( y_f = -\left( \frac{3}{7} \right) - 1 = -\frac{10}{7} \)
\( z_f = 2 \left( \frac{3}{7} \right) - 3 = -\frac{15}{7} \)
So, the foot of the perpendicular is \( F\left(\frac{23}{7}, -\frac{10}{7}, -\frac{15}{7}\right) \).
Let the image of the point \( P \) in the line be \( P'(x', y', z') \). Since \( F \) is the midpoint of \( PP' \):
\( \frac{1 + x'}{2} = \frac{23}{7} \implies x' = \frac{46}{7} - 1 = \frac{39}{7} \)
\( \frac{-2 + y'}{2} = -\frac{10}{7} \implies y' = -\frac{20}{7} + 2 = -\frac{6}{7} \)
\( \frac{1 + z'}{2} = -\frac{15}{7} \implies z' = -\frac{30}{7} - 1 = -\frac{37}{7} \)
Thus, the image point is \( \left(\frac{39}{7}, -\frac{6}{7}, -\frac{37}{7}\right) \).
In simple words: Find a general point \( F \) on the line. Make the vector from \( P \) to \( F \) perpendicular to the line's direction to find \( \lambda = 3/7 \). This gives the midpoint, and we double it to find the reflection point on the opposite side.

Exam Tip: Be extra careful with fractional coordinates. Double check your multiplication and subtraction steps, as simple arithmetic errors are common here.

(ix) Foot of Perpendicular and Image with Respect to a Line and Plane

LEVEL II

Question 1. Find the coordinates of the point where the line through (3,-4,-5) and (2,-3,1) crosses the plane determined by points A(1,2,3) , B(2,2,1) and C(-1,3,6).
Answer:
First, we find the equation of the line passing through the points \( (3, -4, -5) \) and \( (2, -3, 1) \).
The direction ratios of the line are:
\( a = 2 - 3 = -1 \)
\( b = -3 - (-4) = 1 \)
\( c = 1 - (-5) = 6 \)
The symmetric equation of this line is:
\( \frac{x - 3}{-1} = \frac{y + 4}{1} = \frac{z + 5}{6} = \lambda \)
Any general point on this line can be written as:
\( T(-\lambda + 3, \lambda - 4, 6\lambda - 5) \)
Now, let us find the equation of the plane containing the points \( A(1, 2, 3) \), \( B(2, 2, 1) \), and \( C(-1, 3, 6) \):
\( \left| \begin{matrix} x - 1 & y - 2 & z - 3 \\ 2 - 1 & 2 - 2 & 1 - 3 \\ -1 - 1 & 3 - 2 & 6 - 3 \end{matrix} \right| = 0 \)
\( \left| \begin{matrix} x - 1 & y - 2 & z - 3 \\ 1 & 0 & -2 \\ -2 & 1 & 3 \end{matrix} \right| = 0 \)
Expanding this determinant along the first row:
\( (x - 1)(0 - (-2)) - (y - 2)(3 - 4) + (z - 3)(1 - 0) = 0 \)
\( 2(x - 1) + (y - 2) + (z - 3) = 0 \)
\( 2x - 2 + y - 2 + z - 3 = 0 \)
\( 2x + y + z - 7 = 0 \)
If the general point \( T \) lies on this plane, its coordinates must satisfy the plane equation:
\( 2(-\lambda + 3) + (\lambda - 4) + (6\lambda - 5) - 7 = 0 \)
\( -2\lambda + 6 + \lambda - 4 + 6\lambda - 5 - 7 = 0 \)
\( 5\lambda - 10 = 0 \)

\(\implies \lambda = 2 \).
Substituting \( \lambda = 2 \) into the coordinates of \( T \):
\( x = -2 + 3 = 1 \)
\( y = 2 - 4 = -2 \)
\( z = 6(2) - 5 = 7 \)
Thus, the point of intersection is \( (1, -2, 7) \).
In simple words: Write a general parametric equation for the line. Find the equation of the plane passing through the three points. Substitute the line's general coordinate expressions into the plane's equation to find the value of the parameter, which gives you the intersection point.

Exam Tip: First find the plane's equation using a determinant of coordinates. This prevents algebra mistakes during the line-intersection step.

Question 2. Find the foot of the perpendicular from P(1,2,3) on the line \( \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} \). Also, obtain the equation of the plane containing the line and the point (1,2,3).
Answer:
Let \( F \) be the foot of the perpendicular on the given line. Any point on this line can be written in terms of parameter \( \lambda \):
\( \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} = \lambda \)
\( F = (3\lambda + 6, \quad 2\lambda + 7, \quad -2\lambda + 7) \)
The direction ratios of the line segment \( PF \) are:
\( (3\lambda + 6 - 1, \quad 2\lambda + 7 - 2, \quad -2\lambda + 7 - 3) = (3\lambda + 5, \quad 2\lambda + 5, \quad -2\lambda + 4) \)
Since \( PF \) is perpendicular to the given line (whose direction ratios are \( 3, 2, -2 \)):
\( 3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0 \)
\( 9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \)
\( 17\lambda + 17 = 0 \)

\(\implies \lambda = -1 \).
Substituting \( \lambda = -1 \) back into the coordinates of \( F \):
\( x = 3(-1) + 6 = 3 \)
\( y = 2(-1) + 7 = 5 \)
\( z = -2(-1) + 7 = 9 \)
Thus, the foot of the perpendicular is \( (3, 5, 9) \).
Next, we find the plane containing this line and point \( P(1, 2, 3) \).
Let a point on the line be \( Q(6, 7, 7) \).
The vector \( \vec{PQ} = (6 - 1)\hat{i} + (7 - 2)\hat{j} + (7 - 3)\hat{k} = 5\hat{i} + 5\hat{j} + 4\hat{k} \).
The line's direction vector is \( \vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k} \).
The normal to the required plane is perpendicular to both \( \vec{PQ} \) and \( \vec{b} \):
\( \vec{n} = \vec{PQ} \times \vec{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 5 & 4 \\ 3 & 2 & -2 \end{matrix} \right| \)
\( \vec{n} = \hat{i}(-10 - 8) - \hat{j}(-10 - 12) + \hat{k}(10 - 15) = -18\hat{i} + 22\hat{j} - 5\hat{k} \)
The equation of the plane passing through \( P(1, 2, 3) \) is:
\( -18(x - 1) + 22(y - 2) - 5(z - 3) = 0 \)
\( -18x + 18 + 22y - 44 - 5z + 15 = 0 \)
\( -18x + 22y - 5z - 11 = 0 \)
Multiplying by \( -1 \), we get:
\( 18x - 22y + 5z + 11 = 0 \).
In simple words: Express the coordinates of the line using \( \lambda \). Make the vector from \( P \) to this point perpendicular to the line's direction to find \( \lambda = -1 \). To get the plane, take the cross product of the line's direction and the vector from \( P \) to a point on the line, giving you the plane's normal.

Exam Tip: Finding the cross product of the line's direction and a vector connecting the external point to the line is the standard technique to find the normal of a plane containing a line and a point.

Question 3. Prove that the image of the point (3,-2,1) in the plane 3x – y + 4z =2 lies on the plane, x + y + z + 4 = 0.
Answer:
Let the given point be \( A(3, -2, 1) \) and the plane be \( 3x - y + 4z - 2 = 0 \).
Let \( A'(x', y', z') \) be the image of \( A \) in this plane. The formula for the image is:
\( \frac{x' - x_1}{a} = \frac{y' - y_1}{b} = \frac{z' - z_1}{c} = -2\frac{a x_1 + b y_1 + c z_1 + d}{a^2 + b^2 + c^2} \)
Here, \( x_1 = 3, \quad y_1 = -2, \quad z_1 = 1 \) and \( a = 3, \quad b = -1, \quad c = 4, \quad d = -2 \).
Let us calculate the numerator term:
\( a x_1 + b y_1 + c z_1 + d = 3(3) - 1(-2) + 4(1) - 2 = 9 + 2 + 4 - 2 = 13 \)
Calculate the denominator:
\( a^2 + b^2 + c^2 = 3^2 + (-1)^2 + 4^2 = 9 + 1 + 16 = 26 \)
Substituting these into the formula:
\( \frac{x' - 3}{3} = \frac{y' + 2}{-1} = \frac{z' - 1}{4} = -2 \left( \frac{13}{26} \right) = -1 \)
Solving for each coordinate:
\( \frac{x' - 3}{3} = -1 \implies x' - 3 = -3 \implies x' = 0 \)
\( \frac{y' + 2}{-1} = -1 \implies y' + 2 = 1 \implies y' = -1 \)
\( \frac{z' - 1}{4} = -1 \implies z' - 1 = -4 \implies z' = -3 \)
So, the image point is \( A'(0, -1, -3) \).
Now, let us verify if this image point lies on the plane \( x + y + z + 4 = 0 \):
Substituting \( A'(0, -1, -3) \) into the L.H.S.:
\( \text{L.H.S.} = 0 + (-1) + (-3) + 4 = -4 + 4 = 0 = \text{R.H.S.} \)
Since the coordinates satisfy the equation, the image point lies on the plane. (Hence proved).
In simple words: Find the reflection of the given point across the first plane using the standard image formula, which yields \( (0, -1, -3) \). Then, plug these coordinates into the second plane equation. Since the sum equals zero, the point lies on it.

Exam Tip: Using the direct coordinate formula for the image of a point with respect to a plane is much faster and reduces intermediate step calculation errors compared to finding the foot first.

LEVEL-III

Question 1. Find the foot of perpendicular drawn from the point A(1, 0, 3) to the joint of the points B(4, 7, 1) and C(3, 5, 3).
Answer:
The line passes through \( B(4, 7, 1) \) and \( C(3, 5, 3) \).
The direction ratios of the line \( BC \) are:
\( a = 3 - 4 = -1 \)
\( b = 5 - 7 = -2 \)
\( c = 3 - 1 = 2 \)
The symmetric equation of the line \( BC \) is:
\( \frac{x - 4}{-1} = \frac{y - 7}{-2} = \frac{z - 1}{2} = \lambda \)
Any general point \( F \) on this line can be written as:
\( F(-\lambda + 4, \quad -2\lambda + 7, \quad 2\lambda + 1) \)
The direction ratios of the line segment \( AF \), where \( A(1, 0, 3) \), are:
\( (-\lambda + 4 - 1, \quad -2\lambda + 7 - 0, \quad 2\lambda + 1 - 3) = (-\lambda + 3, \quad -2\lambda + 7, \quad 2\lambda - 2) \)
Since \( AF \) is perpendicular to the line \( BC \), the dot product of their direction ratios is zero:
\( -1(-\lambda + 3) - 2(-2\lambda + 7) + 2(2\lambda - 2) = 0 \)
\( \lambda - 3 + 4\lambda - 14 + 4\lambda - 4 = 0 \)
\( 9\lambda - 21 = 0 \)

\(\implies \lambda = \frac{21}{9} = \frac{7}{3} \).
Substituting \( \lambda = \frac{7}{3} \) into the coordinates of \( F \):
\( x = -\left( \frac{7}{3} \right) + 4 = \frac{5}{3} \)
\( y = -2\left( \frac{7}{3} \right) + 7 = \frac{7}{3} \)
\( z = 2\left( \frac{7}{3} \right) + 1 = \frac{17}{3} \)
Thus, the foot of the perpendicular is \( \left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right) \).
In simple words: Find the line joining \( B \) and \( C \) in terms of a parameter \( \lambda \). Set up the direction vector from \( A \) to this line and make it perpendicular to the line's direction. Solving this gives \( \lambda = 7/3 \), which determines the foot coordinates.

Exam Tip: Always make sure to write the direction ratios of \( BC \) clearly before writing the symmetric form of the line equation.

Question 2. Find the image of the point (1, –2, 1) in the line \( \frac{x-2}{3} = \frac{y+1}{-1} = \frac{z+3}{2} \).
Answer:
Let \( P(1, -2, 1) \) be the given point.
Any point \( F \) on the given line can be written in terms of parameter \( \lambda \) as:
\( F = (3\lambda + 2, \quad -\lambda - 1, \quad 2\lambda - 3) \)
The direction ratios of the line segment \( PF \) are:
\( (3\lambda + 2 - 1, \quad -\lambda - 1 - (-2), \quad 2\lambda - 3 - 1) = (3\lambda + 1, \quad -\lambda + 1, \quad 2\lambda - 4) \)
Since \( PF \) is perpendicular to the given line (whose direction ratios are \( 3, -1, 2 \)):
\( 3(3\lambda + 1) - 1(-\lambda + 1) + 2(2\lambda - 4) = 0 \)
\( 9\lambda + 3 + \lambda - 1 + 4\lambda - 8 = 0 \)
\( 14\lambda - 6 = 0 \)

\(\implies \lambda = \frac{6}{14} = \frac{3}{7} \).
Substituting \( \lambda = \frac{3}{7} \) back into the coordinates of \( F \):
\( x_f = 3 \left( \frac{3}{7} \right) + 2 = \frac{23}{7} \)
\( y_f = -\left( \frac{3}{7} \right) - 1 = -\frac{10}{7} \)
\( z_f = 2 \left( \frac{3}{7} \right) - 3 = -\frac{15}{7} \)
Thus, the foot of the perpendicular is \( F\left(\frac{23}{7}, -\frac{10}{7}, -\frac{15}{7}\right) \).
Let the image of the point \( P \) in the line be \( P'(x', y', z') \). Since \( F \) is the midpoint of \( PP' \):
\( \frac{1 + x'}{2} = \frac{23}{7} \implies x' = \frac{46}{7} - 1 = \frac{39}{7} \)
\( \frac{-2 + y'}{2} = -\frac{10}{7} \implies y' = -\frac{20}{7} + 2 = -\frac{6}{7} \)
\( \frac{1 + z'}{2} = -\frac{15}{7} \implies z' = -\frac{30}{7} - 1 = -\frac{37}{7} \)
Therefore, the image point is \( \left(\frac{39}{7}, -\frac{6}{7}, -\frac{37}{7}\right) \).
In simple words: Represent a general point on the line using \( \lambda \). Use the perpendicular dot product to find \( \lambda = 3/7 \), which gives the foot of the perpendicular. Use this foot as the midpoint to find the image coordinates.

Exam Tip: After finding \( \lambda \), carefully verify the midpoint relations to avoid simple calculation slips with the coordinates of the image.

Question 3. The foot of the perpendicular from the origin to the plane is (12, – 4, 3). Find the equation of the plane
Answer:
Let \( P(12, -4, 3) \) be the foot of the perpendicular from the origin \( O(0, 0, 0) \) to the plane.
The vector \( \vec{OP} \) is normal to the plane. Thus, the normal vector \( \vec{n} \) is:
\( \vec{n} = 12\hat{i} - 4\hat{j} + 3\hat{k} \)
The plane passes through the point \( P(12, -4, 3) \).
Using the equation of a plane \( a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 \):
\( 12(x - 12) - 4(y - (-4)) + 3(z - 3) = 0 \)
\( 12(x - 12) - 4(y + 4) + 3(z - 3) = 0 \)
\( 12x - 144 - 4y - 16 + 3z - 9 = 0 \)
\( 12x - 4y + 3z - 169 = 0 \)
Thus, the equation of the plane is \( 12x - 4y + 3z - 169 = 0 \).
In simple words: The vector from the origin to the foot of the perpendicular is the normal of the plane. Use this normal vector and the foot coordinates to write down the standard Cartesian plane equation.

Exam Tip: The direction ratios of the normal of the plane are simply the coordinates of the foot of the perpendicular when it is drawn from the origin.

Question 4. Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3,2,1) from the plane 2x – y+z+1=0. Find also, the image of the point in the plane.
Answer:
Let the given point be \( P(3, 2, 1) \).
The direction ratios of the normal to the plane \( 2x - y + z + 1 = 0 \) are \( (2, -1, 1) \).
The equation of the line passing through \( P(3, 2, 1) \) and perpendicular to the plane is:
\( \frac{x-3}{2} = \frac{y-2}{-1} = \frac{z-1}{1} = \lambda \)
Any general point \( F \) on this line can be written as:
\( F = (2\lambda + 3, \quad -\lambda + 2, \quad \lambda + 1) \)
If \( F \) is the foot of the perpendicular, it must lie on the plane:
\( 2(2\lambda + 3) - (-\lambda + 2) + (\lambda + 1) + 1 = 0 \)
\( 4\lambda + 6 + \lambda - 2 + \lambda + 1 + 1 = 0 \)
\( 6\lambda + 6 = 0 \)

\(\implies \lambda = -1 \).
Substituting \( \lambda = -1 \) back into the coordinates of \( F \):
\( x = 2(-1) + 3 = 1 \)
\( y = -(-1) + 2 = 3 \)
\( z = -1 + 1 = 0 \)
Thus, the foot of the perpendicular is \( F(1, 3, 0) \).
The perpendicular distance is the distance \( PF \):
\( PF = \sqrt{(1-3)^2 + (3-2)^2 + (0-1)^2} = \sqrt{(-2)^2 + 1^2 + (-1)^2} = \sqrt{6} \text{ units} \).
Let the image of the point \( P \) in the plane be \( P'(x', y', z') \). Since \( F \) is the midpoint of \( PP' \):
\( \frac{3 + x'}{2} = 1 \implies x' = -1 \)
\( \frac{2 + y'}{2} = 3 \implies y' = 4 \)
\( \frac{1 + z'}{2} = 0 \implies z' = -1 \)
Thus, the coordinates of the image point are \( P'(-1, 4, -1) \).
In simple words: Setup a line through \( P \) along the plane's normal direction. Find its intersection with the plane, which is \( F(1,3,0) \). Calculate the distance using the distance formula. Finally, find the image point using the midpoint formula.

Exam Tip: Always state each of the three results (foot, distance, image) clearly at the end of your solution, as each carries separate step-wise marks.

Questions for self evaluation

Question 1. Find the equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes x + 2y + 3z – 7 = 0 and 2x – 3y + 4z = 0.
Answer:
Let the normal vector of the required plane be \( \vec{n} = a\hat{i} + b\hat{j} + c\hat{k} \).
Since the plane is perpendicular to the plane \( x + 2y + 3z - 7 = 0 \) (normal \( \vec{n}_1 = \hat{i} + 2\hat{j} + 3\hat{k} \)):
\( a + 2b + 3c = 0 \quad \text{--- (1)} \)
Since it is also perpendicular to the plane \( 2x - 3y + 4z = 0 \) (normal \( \vec{n}_2 = 2\hat{i} - 3\hat{j} + 4\hat{k} \)):
\( 2a - 3b + 4c = 0 \quad \text{--- (2)} \)
Solving equations (1) and (2) using the cross-multiplication method:
\( \frac{a}{(2)(4) - (3)(-3)} = \frac{-b}{(1)(4) - (3)(2)} = \frac{c}{(1)(-3) - (2)(2)} \)
\( \frac{a}{8 + 9} = \frac{-b}{4 - 6} = \frac{c}{-3 - 4} \)
\( \frac{a}{17} = \frac{b}{2} = \frac{c}{-7} = k \)
Thus, the direction ratios of the normal are proportional to \( (17, 2, -7) \).
The equation of the plane passing through \( (1, 1, 1) \) is:
\( 17(x - 1) + 2(y - 1) - 7(z - 1) = 0 \)
\( 17x - 17 + 2y - 2 - 7z + 7 = 0 \)
\( 17x + 2y - 7z - 12 = 0 \).
In simple words: Since the required plane is perpendicular to two other planes, its normal vector is perpendicular to their normals. Find this vector using cross-multiplication, and use the point to write the final plane equation.

Exam Tip: Using cross-multiplication on the normal coefficients of the two perpendicular planes is the most direct way to find the normal vector of the required plane.

Question 2. Find the vector equation of a line joining the points with position vectors i ˆ – 2 j ˆ – 3 k ˆ and parallel to the line joining the points with position vectors i ˆ – j ˆ + 4 k ˆ , and 2 i ˆ + j ˆ + 2 k ˆ . Also find the cartesian equivalent of this equation.
Answer:
The line passes through the point with position vector:
\( \vec{a} = \hat{i} - 2\hat{j} - 3\hat{k} \)
The line is parallel to the line segment connecting \( \vec{c} = \hat{i} - \hat{j} + 4\hat{k} \) and \( \vec{d} = 2\hat{i} + \hat{j} + 2\hat{k} \).
The direction vector \( \vec{b} \) of this parallel line is:
\( \vec{b} = \vec{d} - \vec{c} = (2 - 1)\hat{i} + (1 - (-1))\hat{j} + (2 - 4)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k} \)
The vector equation of the line is:
\( \vec{r} = \vec{a} + \lambda \vec{b} \)
\( \vec{r} = (\hat{i} - 2\hat{j} - 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} - 2\hat{k}) \)
The Cartesian equivalent is written using the point coordinates and the direction components:
\( \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z + 3}{-2} \).
In simple words: Subtract the two points' vectors to find the line's direction. Combine this direction vector with the given passing point's vector to write down both the vector and Cartesian equations.

Exam Tip: Be sure to provide both the vector and Cartesian forms when asked, as many students only write one form and lose partial marks.

Question 3. Find the foot of perpendicular drawn from the point A(1, 0, 3) to the joint of the points B(4, 7, 1) and C(3, 5, 3).
Answer:
Let us represent the line joining \( B(4, 7, 1) \) and \( C(3, 5, 3) \).
The direction ratios of the line \( BC \) are:
\( a = 3 - 4 = -1, \quad b = 5 - 7 = -2, \quad c = 3 - 1 = 2 \)
The symmetric equation of the line \( BC \) is:
\( \frac{x - 4}{-1} = \frac{y - 7}{-2} = \frac{z - 1}{2} = \lambda \)
Any point \( F \) on this line is of the form:
\( F(-\lambda + 4, \quad -2\lambda + 7, \quad 2\lambda + 1) \)
The direction ratios of \( AF \), where \( A(1, 0, 3) \), are:
\( (-\lambda + 4 - 1, \quad -2\lambda + 7 - 0, \quad 2\lambda + 1 - 3) = (-\lambda + 3, \quad -2\lambda + 7, \quad 2\lambda - 2) \)
Since \( AF \) is perpendicular to the direction vector of the line \( BC \):
\( -1(-\lambda + 3) - 2(-2\lambda + 7) + 2(2\lambda - 2) = 0 \)
\( \lambda - 3 + 4\lambda - 14 + 4\lambda - 4 = 0 \)
\( 9\lambda - 21 = 0 \)

\(\implies \lambda = \frac{21}{9} = \frac{7}{3} \).
Substituting \( \lambda = \frac{7}{3} \) back into the coordinates of \( F \):
\( x = -\left( \frac{7}{3} \right) + 4 = \frac{5}{3} \)
\( y = -2\left( \frac{7}{3} \right) + 7 = \frac{7}{3} \)
\( z = 2\left( \frac{7}{3} \right) + 1 = \frac{17}{3} \)
Thus, the coordinates of the foot of the perpendicular are \( \left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right) \).
In simple words: Express a general point on the line using \( \lambda \). Set up the direction ratios of the perpendicular line segment and use the dot product to find \( \lambda = 7/3 \), giving you the foot of the perpendicular.

Exam Tip: Be neat when working with fractions to ensure that you do not make simple arithmetic errors when finding the coordinates.

Question 4. Find the shortest distance between the lines \( \vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k}) \), and \( \vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(4\hat{i} - 2\hat{j} + 2\hat{k}) \)
Answer:
Comparing the two vector equations of lines:
First line: \( \vec{a}_1 = \hat{i} + \hat{j} \) and \( \vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k} \).
Second line: \( \vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k} \) and \( \vec{b}_2 = 4\hat{i} - 2\hat{j} + 2\hat{k} \).
Since \( \vec{b}_2 = 2\vec{b}_1 \), the lines are parallel. We take the common direction vector as \( \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \).
The distance between two parallel lines is given by:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|} \)
Calculate \( \vec{a}_2 - \vec{a}_1 \):
\( \vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} = \hat{i} - \hat{k} \)
Calculate the cross product \( (\vec{a}_2 - \vec{a}_1) \times \vec{b} \):
\( (\vec{a}_2 - \vec{a}_1) \times \vec{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 2 & -1 & 1 \end{matrix} \right| = \hat{i}(0 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - 0) = -\hat{i} - 3\hat{j} - \hat{k} \)
Calculate the magnitudes:
\( |(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{(-1)^2 + (-3)^2 + (-1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11} \)
\( |\vec{b}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \)
Thus, the shortest distance is:
\( d = \sqrt{\frac{11}{6}} \text{ units} \).
In simple words: Since the direction vectors are multiples, the lines are parallel. Calculate the cross product of the difference in starting points and the direction vector, then divide its length by the direction vector's length.

Exam Tip: Be sure to use the specific parallel-lines distance formula instead of the general skew-lines formula, which would yield zero in the denominator.

Question 5. Find the image of the point (1, –2, 1) in the line \( \frac{x-2}{3} = \frac{y+1}{-1} = \frac{z+3}{2} \).
Answer:
Let \( P(1, -2, 1) \) be the given point.
Let \( F \) be a general point on the given line:
\( F = (3\lambda + 2, \quad -\lambda - 1, \quad 2\lambda - 3) \)
The direction ratios of the line segment \( PF \) are:
\( (3\lambda + 2 - 1, \quad -\lambda - 1 - (-2), \quad 2\lambda - 3 - 1) = (3\lambda + 1, \quad -\lambda + 1, \quad 2\lambda - 4) \).
Since \( PF \) is perpendicular to the given line (whose direction ratios are \( 3, -1, 2 \)):
\( 3(3\lambda + 1) - 1(-\lambda + 1) + 2(2\lambda - 4) = 0 \)
\( 9\lambda + 3 + \lambda - 1 + 4\lambda - 8 = 0 \)
\( 14\lambda - 6 = 0 \)

\(\implies \lambda = \frac{3}{7} \).
Substituting \( \lambda = \frac{3}{7} \) back into the coordinates of \( F \):
\( F = \left(\frac{23}{7}, \quad -\frac{10}{7}, \quad -\frac{15}{7}\right) \).
Let the image of the point \( P \) in the line be \( P'(x', y', z') \). Since \( F \) is the midpoint of \( PP' \):
\( \frac{1 + x'}{2} = \frac{23}{7} \implies x' = \frac{39}{7} \)
\( \frac{-2 + y'}{2} = -\frac{10}{7} \implies y' = -\frac{6}{7} \)
\( \frac{1 + z'}{2} = -\frac{15}{7} \implies z' = -\frac{37}{7} \)
Thus, the image point is \( \left(\frac{39}{7}, -\frac{6}{7}, -\frac{37}{7}\right) \).
In simple words: Represent a point on the line with \( \lambda \). Set up the perpendicularity condition to find \( \lambda = 3/7 \), which gives the foot of the perpendicular. Reflect the point across this foot to find the image coordinates.

Exam Tip: Be careful when solving the linear equations with fractions to ensure that you get the correct coordinates for the image.

Question 6. Show that the four points (0, –1, –1), (4, 5, 1), (3, 9, 4) and (– 4, 4, 4) are coplanar and find the equation of the common plane.
Answer:
Let the points be \( A(0, -1, -1) \), \( B(4, 5, 1) \), \( C(3, 9, 4) \), and \( D(-4, 4, 4) \).
The equation of any plane passing through \( A(0, -1, -1) \) is:
\( a(x - 0) + b(y + 1) + c(z + 1) = 0 \)
\( ax + b(y + 1) + c(z + 1) = 0 \quad \text{--- (1)} \)
If this plane also passes through \( B(4, 5, 1) \) and \( C(3, 9, 4) \):
For \( B(4, 5, 1) \):
\( 4a + 6b + 2c = 0 \implies 2a + 3b + c = 0 \quad \text{--- (2)} \)
For \( C(3, 9, 4) \):
\( 3a + 10b + 5c = 0 \quad \text{--- (3)} \)
Solving equations (2) and (3) using the cross-multiplication method:
\( \frac{a}{(3)(5) - (1)(10)} = \frac{-b}{(2)(5) - (1)(3)} = \frac{c}{(2)(10) - (3)(3)} \)
\( \frac{a}{5} = \frac{b}{-7} = \frac{c}{11} \)
Substituting \( a = 5 \), \( b = -7 \), and \( c = 11 \) into equation (1):
\( 5x - 7(y + 1) + 11(z + 1) = 0 \)
\( 5x - 7y + 11z + 4 = 0 \quad \text{--- (4)} \)
Now, let us verify if the fourth point \( D(-4, 4, 4) \) lies on this plane by substituting its coordinates:
\( \text{L.H.S.} = 5(-4) - 7(4) + 11(4) + 4 = -20 - 28 + 44 + 4 = 0 = \text{R.H.S.} \)
Since the fourth point satisfies the equation, the four points are coplanar.
The equation of the common plane is \( 5x - 7y + 11z + 4 = 0 \).
In simple words: Find the equation of the plane passing through any three points. Substitute the fourth point into this equation. Since it satisfies the equation, the points are coplanar and the plane is \( 5x - 7y + 11z + 4 = 0 \).

Exam Tip: Substituting the fourth point into the plane equation of the first three points is the easiest way to prove coplanarity.

Question 7. The foot of the perpendicular from the origin to the plane is (12, – 4, 3). Find the equation of the plane.
Answer:
Let \( P(12, -4, 3) \) be the foot of the perpendicular from the origin \( O(0, 0, 0) \) to the plane.
The vector \( \vec{OP} \) is normal to the plane:
\( \vec{n} = 12\hat{i} - 4\hat{j} + 3\hat{k} \)
The plane passes through the point \( P(12, -4, 3) \).
Using the standard equation of a plane:
\( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \)
\( 12(x - 12) - 4(y + 4) + 3(z - 3) = 0 \)
\( 12x - 144 - 4y - 16 + 3z - 9 = 0 \)
\( 12x - 4y + 3z - 169 = 0 \).
In simple words: The vector from the origin to the foot of the perpendicular is normal to the plane. Use this normal vector and the point's coordinates to write down the plane equation.

Exam Tip: The coordinates of the foot of the perpendicular from the origin directly serve as the coefficients \( a, b, c \) in the plane equation.

Question 8. Show that the lines \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \) and \( \frac{x - 4}{5} = \frac{y - 1}{2} = z \) intersect. Find their point of intersection.
Answer:
Let any point on the first line be represented by parameter \( \lambda \):
\( x = 2\lambda + 1, \quad y = 3\lambda + 2, \quad z = 4\lambda + 3 \)
Let any point on the second line be represented by parameter \( \mu \):
\( x = 5\mu + 4, \quad y = 2\mu + 1, \quad z = \mu \)
If the two lines intersect:
1) \( 2\lambda + 1 = 5\mu + 4 \implies 2\lambda - 5\mu = 3 \)
2) \( 3\lambda + 2 = 2\mu + 1 \implies 3\lambda - 2\mu = -1 \)
3) \( 4\lambda + 3 = \mu \)
Substituting equation (3) into equation (2):
\( 3\lambda - 2(4\lambda + 3) = -1 \)
\( 3\lambda - 8\lambda - 6 = -1 \)
\( -5\lambda = 5 \)

\(\implies \lambda = -1 \).
Using this in equation (3):
\( \mu = 4(-1) + 3 = -1 \)
Let us verify if \( \lambda = -1 \) and \( \mu = -1 \) satisfy equation (1):
L.H.S. = \( 2(-1) - 5(-1) = -2 + 5 = 3 \)
R.H.S. = \( 3 \)
Since L.H.S. = R.H.S., the lines intersect.
Substituting \( \lambda = -1 \) into the coordinates of the first line:
\( x = 2(-1) + 1 = -1 \)
\( y = 3(-1) + 2 = -1 \)
\( z = 4(-1) + 3 = -1 \)
Thus, the point of intersection is \( (-1, -1, -1) \).
In simple words: Represent points on both lines using parameters. Solve the system of coordinates to find \( \lambda = -1 \) and \( \mu = -1 \). Since the values are consistent, the lines cross at the point \( (-1, -1, -1) \).

Exam Tip: Be sure to show the verification of the third equation after solving for the parameters from the first two.

Question 9. A line makes angles α, ȕ, Ȗ, δ with the four diagonals of a cube, Prove that \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma + \cos^2\delta = \frac{4}{3} \)
Answer:
Let us place a cube of side \( a \) with one corner at the origin and edges along the coordinate axes \( OX, OY, OZ \).
The four diagonals are:
1. \( D_1 \) joining \( O(0,0,0) \) and \( P(a,a,a) \) with direction cosines: \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \)
2. \( D_2 \) joining \( A(a,0,0) \) and \( B(0,a,a) \) with direction cosines: \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \)
3. \( D_3 \) joining \( C(0,a,0) \) and \( D(a,0,a) \) with direction cosines: \( \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \)
4. \( D_4 \) joining \( E(0,0,a) \) and \( F(a,a,0) \) with direction cosines: \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) \)
Let the given line have direction cosines \( (l, m, n) \), where \( l^2 + m^2 + n^2 = 1 \).
Using the angle formula between two lines, the cosines of the angles with the four diagonals are:
\( \cos \alpha = \frac{l + m + n}{\sqrt{3}} \)
\( \cos \beta = \frac{-l + m + n}{\sqrt{3}} \)
\( \cos \gamma = \frac{l - m + n}{\sqrt{3}} \)
\( \cos \delta = \frac{l + m - n}{\sqrt{3}} \)
Squaring and adding these terms:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{1}{3} \left[ (l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2 \right] \)
Expanding the terms inside the bracket:
\( (l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2 = 4(l^2 + m^2 + n^2) \)
Since \( l^2 + m^2 + n^2 = 1 \):
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{4}{3}(1) = \frac{4}{3} \). (Hence proved).
In simple words: Represent the four cube diagonals using coordinates. Using the direction cosines of a general line, calculate the cosine of the angle for each diagonal, square them, and add them up. Since the sum of squared direction cosines is 1, the result simplifies to exactly 4/3.

Exam Tip: Be precise when expanding the algebraic terms containing \( l, m, n \) to show how all cross-multiplication terms cancel out.

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