CBSE Class 12 Mathematics Three Dimensional Geometry Assignment Set 04

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. What is the distance of point (a, b, c) from x-axis?
Answer: The perpendicular distance of the point \( (a, b, c) \) from the x-axis is given by \( \sqrt{b^2 + c^2} \). This is because the projection of the point on the x-axis is \( (a, 0, 0) \), and the distance between \( (a, b, c) \) and \( (a, 0, 0) \) is \( \sqrt{(a - a)^2 + (b - 0)^2 + (c - 0)^2} = \sqrt{b^2 + c^2} \).
In simple words: To find how far a point is from the x-axis, ignore its x-coordinate and take the square root of the sum of the squares of the other two coordinates.

Exam Tip: Remember that the distance of any point \( (x, y, z) \) from the coordinate axes is given by: from x-axis \( \sqrt{y^2 + z^2} \), from y-axis \( \sqrt{x^2 + z^2} \), and from z-axis \( \sqrt{x^2 + y^2} \).

Question 2. What is the angle between the lines 2x = 3y = - z and 6x = - y = - 4z?
Answer: Let us rewrite the equations of the lines in standard symmetric form to find their direction ratios:
The first line is \( 2x = 3y = -z \), which can be written as \( \frac{x}{1/2} = \frac{y}{1/3} = \frac{z}{-1} \) or \( \frac{x}{3} = \frac{y}{2} = \frac{z}{-6} \). Thus, its direction ratios are proportional to \( (3, 2, -6) \).
The second line is \( 6x = -y = -4z \), which can be written as \( \frac{x}{1/6} = \frac{y}{-1} = \frac{z}{-1/4} \) or \( \frac{x}{2} = \frac{y}{-12} = \frac{z}{-3} \). Thus, its direction ratios are proportional to \( (2, -12, -3) \).
Let us check the dot product of the direction vectors:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 3(2) + 2(-12) + (-6)(-3) = 6 - 24 + 18 = 0 \)
Since the dot product of their direction vectors is zero, the two lines are perpendicular to each other.
\( \implies \) The angle between the lines is \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).
In simple words: Express the lines in symmetric form to find their direction ratios. Since the dot product of these direction ratios is zero, the lines are perpendicular, making the angle between them \( 90^\circ \).

Exam Tip: Always convert the given equations of lines into standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) before reading off the direction ratios to avoid sign errors.

Question 3. Write the equation of a line passing through (2, -3, 5) and parallel to line \( \frac{x-1}{3} = \frac{y-2}{4} = \frac{z+1}{-1} \).
Answer: Since the required line is parallel to the given line \( \frac{x-1}{3} = \frac{y-2}{4} = \frac{z+1}{-1} \), it must have the same direction ratios, which are \( (3, 4, -1) \).
Given that the line passes through the point \( (2, -3, 5) \), its equation in Cartesian form is:
\( \frac{x-2}{3} = \frac{y+3}{4} = \frac{z-5}{-1} \)
In vector form, the equation of the line is:
\( \vec{r} = (2\hat{i} - 3\hat{j} + 5\hat{k}) + \lambda(3\hat{i} + 4\hat{j} - \hat{k}) \), where \( \lambda \) is a scalar parameter.
In simple words: A line parallel to another line shares the same direction. Use the point \( (2, -3, 5) \) as the starting point and the denominators \( 3, 4, -1 \) from the given line to write the new equation.

Exam Tip: Be careful with the signs when writing the Cartesian equation - passing through \( (x_1, y_1, z_1) \) means using \( x - x_1 \), \( y - y_1 \), and \( z - z_1 \), which changes the sign of negative coordinates like \( -3 \) to \( y + 3 \).

Question 4. Write the equation of a line through (1, 2, 3) and perpendicular to \( \vec{r} \cdot (\hat{i} - \hat{j} + 3\hat{k}) = 5 \).
Answer: A line perpendicular to the plane \( \vec{r} \cdot (\hat{i} - \hat{j} + 3\hat{k}) = 5 \) must be parallel to the normal vector of the plane.
The normal vector to the plane is \( \vec{n} = \hat{i} - \hat{j} + 3\hat{k} \), so the direction ratios of the line are proportional to \( (1, -1, 3) \).
Since the line passes through the point \( (1, 2, 3) \), its equation in vector form is:
\( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 3\hat{k}) \)
In Cartesian form, the equation of the line is:
\( \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-3}{3} \)
In simple words: A line perpendicular to a plane has the same direction as the normal vector of that plane. We use the coefficients of the plane's equation \( (1, -1, 3) \) as the direction of the line starting from the point \( (1, 2, 3) \).

Exam Tip: A very common concept in board exams is that the normal to a plane is parallel to any line perpendicular to that plane. Keep this geometric relation in mind to quickly find the direction ratios of the line.

Question 5. What is the value of \( \lambda \) for which the lines \( \frac{x-1}{2} = \frac{y-3}{5} = \frac{z-1}{\lambda} \) and \( \frac{x-2}{3} = \frac{y+1}{-2} = \frac{z}{2} \) are perpendicular to each other.
Answer: The direction ratios of the first line are \( a_1 = 2 \), \( b_1 = 5 \), and \( c_1 = \lambda \).
The direction ratios of the second line are \( a_2 = 3 \), \( b_2 = -2 \), and \( c_2 = 2 \).
Since the two lines are perpendicular, the sum of the products of their corresponding direction ratios must be zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
\( \implies 2(3) + 5(-2) + \lambda(2) = 0 \)
\( \implies 6 - 10 + 2\lambda = 0 \)
\( \implies -4 + 2\lambda = 0 \)
\( \implies 2\lambda = 4 \)
\( \implies \lambda = 2 \)
In simple words: Two lines are perpendicular when the dot product of their direction vectors is zero. Multiply the corresponding bottom numbers from both line equations, set their sum to zero, and solve for \( \lambda \).

Exam Tip: When applying the condition for perpendicular lines, ensure that both line equations are in standard form with the coefficients of \( x \), \( y \), and \( z \) equal to \( +1 \) in the numerators.

Question 6. If a line makes angle \( \alpha \), \( \beta \), and \( \gamma \) with co-ordinate axes, then what is the value of \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma \) ?
Answer: Let \( \cos \alpha \), \( \cos \beta \), and \( \cos \gamma \) be the direction cosines of the line. We know the fundamental relation:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)
Using the trigonometric identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we can substitute this into the expression:
\( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = (1 - \cos^2 \alpha) + (1 - \cos^2 \beta) + (1 - \cos^2 \gamma) \)
\( \implies \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) \)
Substituting the fundamental relation into this equation:
\( \implies 3 - 1 = 2 \)
In simple words: Since the sum of the squares of the cosines of the angles a line makes with the axes is always 1, converting them to sines using basic trigonometry shows that the sum of the squares of the sines is 2.

Exam Tip: This is a very standard and frequently asked 1-mark question. Remembering the result \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2 \) can save you valuable time in exams.

Question 7. Write line \( \vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} - \hat{k}) \) into Cartesian form.
Answer: The vector equation of the line is given as \( \vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} - \hat{k}) \).
Comparing this with the standard vector equation of a line \( \vec{r} = \vec{a} + \lambda \vec{b} \), we have:
Passing point vector \( \vec{a} = \hat{i} - \hat{j} + 0\hat{k} \), so the point is \( (x_1, y_1, z_1) = (1, -1, 0) \).
Parallel direction vector \( \vec{b} = 2\hat{i} + 0\hat{j} - \hat{k} \), so the direction ratios are \( (a, b, c) = (2, 0, -1) \).
The Cartesian form of the line is:
\( \frac{x-1}{2} = \frac{y+1}{0} = \frac{z}{-1} \) (which also implies \( y = -1 \)).
In simple words: Extract the starting point \( (1, -1, 0) \) from the first vector and the direction \( (2, 0, -1) \) from the second vector, then write them in the standard Cartesian fraction format.

Exam Tip: Do not omit the term with \( 0 \) in the denominator, i.e., \( \frac{y+1}{0} \), as writing it this way is standard mathematical notation for a line in three-dimensional coordinate geometry.

Question 8. If the direction ratios of a line are 1, -2, 2 then what are the direction cosines of the line?
Answer: Let the direction ratios of the line be \( a = 1 \), \( b = -2 \), and \( c = 2 \).
First, find the magnitude:
\( \sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \)
The direction cosines \( (l, m, n) \) are given by:
\( l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{1}{3} \)
\( m = \frac{b}{\sqrt{a^2+b^2+c^2}} = -\frac{2}{3} \)
\( n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{2}{3} \)
Thus, the direction cosines of the line are \( \left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right) \).
In simple words: To find the direction cosines, divide each direction ratio by the square root of the sum of the squares of all three ratios.

Exam Tip: Ensure that you include the correct signs for the direction cosines, as they carry the same sign as their corresponding direction ratios.

Question 9. Find the angle between the planes 2x - 3y + 6z = 9 and xy - plane.
Answer: The equation of the first plane is \( 2x - 3y + 6z = 9 \), and its normal vector is \( \vec{n_1} = 2\hat{i} - 3\hat{j} + 6\hat{k} \).
The equation of the xy-plane is \( z = 0 \), and its normal vector is along the z-axis, i.e., \( \vec{n_2} = \hat{k} \).
Let \( \theta \) be the angle between these two planes. The cosine of the angle is given by:
\( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \)
\( \implies \cos \theta = \frac{|(2)(0) + (-3)(0) + (6)(1)|}{\sqrt{2^2 + (-3)^2 + 6^2} \sqrt{1^2}} \)
\( \implies \cos \theta = \frac{6}{\sqrt{4 + 9 + 36} \cdot 1} \)
\( \implies \cos \theta = \frac{6}{\sqrt{49}} = \frac{6}{7} \)
\( \implies \theta = \cos^{-1}\left(\frac{6}{7}\right) \)
In simple words: The angle between two planes is the same as the angle between their normal vectors. Since the normal of the xy-plane is simply pointing straight up along the z-axis, we calculate the angle using dot product.

Exam Tip: Remember that the equations of the coordinate planes are: xy-plane is \( z = 0 \), yz-plane is \( x = 0 \), and zx-plane is \( y = 0 \), with respective normal vectors \( \hat{k} \), \( \hat{i} \), and \( \hat{j} \).

Question 10. Write equation of a line passing through (0, 1, 2) and equally inclined to co-ordinate axes.
Answer: A line equally inclined to the coordinate axes makes equal angles with the x, y, and z axes, meaning \( \alpha = \beta = \gamma \).
Therefore, their direction cosines are equal: \( l = m = n \).
Since \( l^2 + m^2 + n^2 = 1 \), we have \( 3l^2 = 1 \implies l = \pm\frac{1}{\sqrt{3}} \).
Thus, the direction ratios of the line are proportional to \( (1, 1, 1) \).
Using the point \( (0, 1, 2) \), the Cartesian equation of the line is:
\( \frac{x-0}{1} = \frac{y-1}{1} = \frac{z-2}{1} \) or simply \( \frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{1} \).
In simple words: A line that tilts at the exact same angle to all three axes has equal direction ratios of \( 1 \). Combining this with the starting point \( (0, 1, 2) \) gives the final line equation.

Exam Tip: Keep in mind that "equally inclined" means the direction ratios can be taken as \( 1, 1, 1 \), which makes writing the equation very straightforward.

Question 11. What is the perpendicular distance of plane 2x - y + 3z = 10 from origin?
Answer: The equation of the plane is \( 2x - y + 3z - 10 = 0 \).
The perpendicular distance \( d \) of a plane \( Ax + By + Cz + D = 0 \) from the origin \( (0, 0, 0) \) is given by the formula:
\( d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \)
Substituting the coefficients of the given plane:
\( d = \frac{|-10|}{\sqrt{2^2 + (-1)^2 + 3^2}} \)
\( \implies d = \frac{10}{\sqrt{4 + 1 + 9}} = \frac{10}{\sqrt{14}} \)
Multiplying the numerator and denominator by \( \sqrt{14} \) to rationalize:
\( \implies d = \frac{10\sqrt{14}}{14} = \frac{5\sqrt{14}}{7} \) units.
In simple words: To find the distance from the origin to a plane, divide the constant term on the other side of the equals sign by the length of the plane's normal vector.

Exam Tip: Make sure the constant term is placed on the left side of the equation before identifying \( D \), or simply take the absolute value to ensure your final distance is always positive.

Question 12. What is the y-intercept of the plane x - 5y + 7z = 10?
Answer: To find the intercepts cut off by a plane on the coordinate axes, we express the equation of the plane in intercept form:
\( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \ ), where \( b \) represents the y-intercept.
Dividing the given plane equation \( x - 5y + 7z = 10 \) by \( 10 \) on both sides:
\( \frac{x}{10} - \frac{5y}{10} + \frac{7z}{10} = 1 \)
\( \implies \frac{x}{10} + \frac{y}{-2} + \frac{z}{10/7} = 1 \)
Comparing this with the intercept form, the y-intercept \( b \) is \( -2 \).
In simple words: Divide the whole equation by the constant on the right side to make it 1. The number under \( y \) is your y-intercept, which is \( -2 \).

Exam Tip: Alternatively, you can find the y-intercept by setting \( x = 0 \) and \( z = 0 \) directly in the plane equation and solving for \( y \), which is a faster method for 1-mark questions.

Question 13. What is the distance between the planes 2x + 2y - z + 2 = 0 and 4x + 4y - 2z + 5 = 0.
Answer: Let us first check if the two planes are parallel.
The first plane is \( 2x + 2y - z + 2 = 0 \), which can be rewritten by multiplying the entire equation by \( 2 \):
\( 4x + 4y - 2z + 4 = 0 \)
The second plane is \( 4x + 4y - 2z + 5 = 0 \).
Since their \( x \), \( y \), and \( z \) coefficients match, they are indeed parallel. The distance \( d \) between two parallel planes \( Ax + By + Cz + D_1 = 0 \) and \( Ax + By + Cz + D_2 = 0 \) is:
\( d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \)
Substituting the values \( A = 4 \), \( B = 4 \), \( C = -2 \), \( D_1 = 4 \), and \( D_2 = 5 \):
\( d = \frac{|5 - 4|}{\sqrt{4^2 + 4^2 + (-2)^2}} \)
\( \implies d = \frac{1}{\sqrt{16 + 16 + 4}} = \frac{1}{\sqrt{36}} = \frac{1}{6} \) units.
In simple words: Multiply the first plane's equation by 2 so that its coefficients match the second plane. Then, use the difference between their constant terms divided by the length of the coefficient vector to find the distance.

Exam Tip: A common mistake is using the constants without making the coefficients of \( x \), \( y \), and \( z \) equal. Always normalize the coefficients before applying the distance formula.

Question 14. What is the equation of the plane which cuts off equal intercepts of unit length on the coordinate axes.
Answer: Let the intercepts cut off by the plane on the coordinate axes be \( a \), \( b \), and \( c \).
Since the plane cuts off equal intercepts of unit length, we have:
\( a = b = c = 1 \)
The intercept form of the equation of a plane is:
\( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \)
Substituting \( a = 1 \), \( b = 1 \), and \( c = 1 \) into this form:
\( \frac{x}{1} + \frac{y}{1} + \frac{z}{1} = 1 \)
\( \implies x + y + z = 1 \).
In simple words: Since the intercepts on all three axes are equal to 1, the equation of the plane is simply the sum of \( x \), \( y \), and \( z \) equal to 1.

Exam Tip: Knowing the intercept form of a plane \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) is very helpful as it directly yields the plane's equation given the intercept lengths.

Question 15. Are the planes x + y - 2z + 4 = 0 and 3x + 3y - 6z + 5 = 0 intersecting?
Answer: Let us compare the ratios of the coefficients of \( x \), \( y \), and \( z \) for the two planes:
Plane 1: \( x + y - 2z + 4 = 0 \) with coefficients \( A_1 = 1 \), \( B_1 = 1 \), \( C_1 = -2 \), and constant \( D_1 = 4 \)
Plane 2: \( 3x + 3y - 6z + 5 = 0 \) with coefficients \( A_2 = 3 \), \( B_2 = 3 \), \( C_2 = -6 \), and constant \( D_2 = 5 \)
The ratios are:
\( \frac{A_1}{A_2} = \frac{1}{3} \)
\( \frac{B_1}{B_2} = \frac{1}{3} \)
\( \frac{C_1}{C_2} = \frac{-2}{-6} = \frac{1}{3} \)
Since \( \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \neq \frac{D_1}{D_2} \), the two planes are parallel and distinct.
Since parallel planes never meet, they do not intersect.
\( \implies \) The answer is No.
In simple words: The coefficients of \( x \), \( y \), and \( z \) in the second plane are exactly 3 times those of the first plane, but the constants do not match. This means the planes are parallel and do not cross.

Exam Tip: If the ratio of all coefficients including the constants were equal, the planes would be coincident (identical), meaning they would intersect at infinitely many points.

Question 16. What is the equation of the plane through the point (1, 4, - 2) and parallel to the plane - 2x + y - 3z = 7?
Answer: Any plane parallel to the plane \( -2x + y - 3z = 7 \) can be represented as:
\( -2x + y - 3z = d \), where \( d \) is a constant.
Since this plane passes through the point \( (1, 4, -2) \), the coordinates of the point must satisfy the equation:
\( -2(1) + (4) - 3(-2) = d \)
\( \implies -2 + 4 + 6 = d \)
\( \implies d = 8 \)
Thus, the equation of the parallel plane is:
\( -2x + y - 3z = 8 \) (or \( 2x - y + 3z + 8 = 0 \) in standard form).
In simple words: Parallel planes have the same variable part. Copy the left side of the given plane equation, plug in the given point coordinates to find the new constant, and write the complete equation.

Exam Tip: Parallel planes share the same normal direction, so you can always write a parallel plane equation by keeping the \( x \), \( y \), and \( z \) parts identical and only solving for a new constant.

Question 17. Write the vector equation of the plane which is at a distance of 8 units from the origin and is normal to the vector \( 2\hat{i} + \hat{j} + 2\hat{k} \).
Answer: Let the given normal vector be \( \vec{N} = 2\hat{i} + \hat{j} + 2\hat{k} \).
First, find the unit normal vector \( \hat{n} \) by dividing \( \vec{N} \) by its magnitude:
\( |\vec{N}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \)
\( \hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3} \)
The vector equation of a plane in normal form is \( \vec{r} \cdot \hat{n} = d \), where \( d \) is the perpendicular distance from the origin. Here, \( d = 8 \).
Substituting the values:
\( \vec{r} \cdot \left(\frac{2\hat{i} + \hat{j} + 2\hat{k}}{3}\right) = 8 \)
\( \implies \vec{r} \cdot (2\hat{i} + \hat{j} + 2\hat{k}) = 24 \).
In simple words: Find the unit normal vector by dividing the normal vector by its length (3). Then, multiply the dot product of \( \vec{r} \) and this unit vector by the distance (8) to get the final vector equation.

Exam Tip: Always make sure to convert the normal vector to a unit normal vector first before setting up the equation \( \vec{r} \cdot \hat{n} = d \).

Question 18. What is equation of the plane if the foot of perpendicular from origin to this plane is (2, 3, 4)?
Answer: Let the foot of the perpendicular from the origin \( O(0, 0, 0) \) to the plane be \( P(2, 3, 4) \).
The normal vector to the plane is \( \vec{OP} \):
\( \vec{OP} = (2 - 0)\hat{i} + (3 - 0)\hat{j} + (4 - 0)\hat{k} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
The direction ratios of the normal to the plane are \( (2, 3, 4) \).
Since the plane passes through the point \( P(2, 3, 4) \), the equation of the plane is:
\( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \)
\( \implies 2(x - 2) + 3(y - 3) + 4(z - 4) = 0 \)
\( \implies 2x - 4 + 3y - 9 + 4z - 16 = 0 \)
\( \implies 2x + 3y + 4z = 29 \).
In simple words: The vector from the origin to the foot of the perpendicular is the normal vector of the plane. Use this normal vector \( (2, 3, 4) \) and the point \( (2, 3, 4) \) to build the equation of the plane.

Exam Tip: When the foot of the perpendicular from the origin is given as \( (x_1, y_1, z_1) \), the equation of the plane is always simply \( x_1 x + y_1 y + z_1 z = x_1^2 + y_1^2 + z_1^2 \).

Question 19. Find the angles between the planes \( \vec{r} \cdot (\hat{i} - 2\hat{j} - 2\hat{k}) = 1 \) and \( \vec{r} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 0 \).
Answer: The normal vectors of the two given planes are:
\( \vec{n_1} = \hat{i} - 2\hat{j} - 2\hat{k} \)
\( \vec{n_2} = 3\hat{i} - 6\hat{j} + 2\hat{k} \)
Let \( \theta \) be the angle between the two planes. The cosine of the angle is given by:
\( \cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} \)
Calculating the dot product and magnitudes:
\( \vec{n_1} \cdot \vec{n_2} = (1)(3) + (-2)(-6) + (-2)(2) = 3 + 12 - 4 = 11 \)
\( |\vec{n_1}| = \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \)
\( |\vec{n_2}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \)
Substituting these values:
\( \cos \theta = \frac{11}{3 \times 7} = \frac{11}{21} \)
\( \implies \theta = \cos^{-1}\left(\frac{11}{21}\right) \).
In simple words: Extract the normal vectors from both plane equations, compute their dot product and magnitudes, and use the cosine angle formula to find the final angle.

Exam Tip: If the dot product of the normal vectors is negative, take the absolute value of the dot product to find the acute angle between the planes, which is the standard practice.

Question 20. What is the angle between the line \( \frac{x+1}{3} = \frac{2y-1}{4} = \frac{2-z}{-4} \) and the plane 2x + y - 2z + 4 = 0?
Answer: First, write the given line equation in standard symmetric form:
\( \frac{x+1}{3} = \frac{2(y - 1/2)}{4} = \frac{-(z - 2)}{-4} \implies \frac{x+1}{3} = \frac{y - 1/2}{2} = \frac{z - 2}{4} \)
The direction vector of the line is \( \vec{b} = 3\hat{i} + 2\hat{j} + 4\hat{k} \).
The normal vector to the plane \( 2x + y - 2z + 4 = 0 \) is \( \vec{n} = 2\hat{i} + \hat{j} - 2\hat{k} \).
Let \( \theta \) be the angle between the line and the plane. It is given by:
\( \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \)
Calculating the dot product and magnitudes:
\( \vec{b} \cdot \vec{n} = 3(2) + 2(1) + 4(-2) = 6 + 2 - 8 = 0 \)
\( \sin \theta = \frac{0}{|\vec{b}| |\vec{n}|} = 0 \)
\( \implies \theta = 0^\circ \) (or \( 0 \) radians).
In simple words: Rewrite the line in standard form to find its direction ratios. Since the dot product of the line's direction and the plane's normal is 0, the line runs parallel to the plane, making the angle between them \( 0^\circ \).

Exam Tip: Make sure to use the sine function for the angle between a line and a plane, unlike the cosine function used for two lines or two planes.

Question 21. If O is origin OP = 3 with direction ratios proportional to -1, 2, -2 then what are the coordinates of P?
Answer: The direction ratios of OP are proportional to \( a = -1 \), \( b = 2 \), and \( c = -2 \).
The direction cosines \( (l, m, n) \) are:
\( \sqrt{a^2 + b^2 + c^2} = \sqrt{(-1)^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = 3 \)
\( l = -\frac{1}{3} \), \( m = \frac{2}{3} \), and \( n = -\frac{2}{3} \).
Since \( OP = r = 3 \), the coordinates of point \( P \) are given by \( (lr, mr, nr) \):
\( P = \left(-\frac{1}{3} \times 3, \frac{2}{3} \times 3, -\frac{2}{3} \times 3\right) = (-1, 2, -2) \).
In simple words: Find the direction cosines of the vector by dividing the direction ratios by their length (3). Multiply these cosines by the given length of OP (3) to get the coordinates of \( P \).

Exam Tip: The coordinates of a point \( P \) at distance \( r \) from the origin with direction cosines \( (l, m, n) \) are always \( (lr, mr, nr) \). This relation is useful in vector geometry.

Question 22. What is the distance between the line \( \vec{r} = 2\hat{i} - 2\hat{j} + 3\hat{k} + \lambda(\hat{i} + \hat{j} + 4\hat{k}) \) from the plane \( \vec{r} \cdot (-\hat{i} + 5\hat{j} - \hat{k}) + 5 = 0 \).
Answer: The direction vector of the line is \( \vec{b} = \hat{i} + \hat{j} + 4\hat{k} \), and the normal vector to the plane is \( \vec{n} = -\hat{i} + 5\hat{j} - \hat{k} \).
Let us check if they are parallel:
\( \vec{b} \cdot \vec{n} = (1)(-1) + (1)(5) + (4)(-1) = -1 + 5 - 4 = 0 \)
Since the dot product is zero, the line is parallel to the plane. The distance from the line to the plane is equal to the perpendicular distance of any point on the line from the plane.
A point on the line is \( (2, -2, 3) \) (obtained at \( \lambda = 0 \)).
The Cartesian equation of the plane is:
\( -x + 5y - z + 5 = 0 \)
The perpendicular distance of \( (2, -2, 3) \) from the plane is:
\( d = \frac{|-(2) + 5(-2) - (3) + 5|}{\sqrt{(-1)^2 + 5^2 + (-1)^2}} \)
\( \implies d = \frac{|-2 - 10 - 3 + 5|}{\sqrt{1 + 25 + 1}} \)
\( \implies d = \frac{|-10|}{\sqrt{27}} = \frac{10}{3\sqrt{3}} = \frac{10\sqrt{3}}{9} \) units.
In simple words: Since the line is parallel to the plane, find the distance by taking the starting point \( (2, -2, 3) \) of the line and calculating its perpendicular distance to the plane.

Exam Tip: First verify that the line is parallel to the plane by checking \( \vec{b} \cdot \vec{n} = 0 \) before computing the distance, as non-parallel lines intersect the plane, which would make the distance zero.

Question 23. Write the line 2x = 3y = 4z in vector form.
Answer: To find the vector form of the line \( 2x = 3y = 4z \), we first express it in standard symmetric form.
Divide the terms by the least common multiple of \( 2 \), \( 3 \), and \( 4 \), which is \( 12 \):
\( \frac{2x}{12} = \frac{3y}{12} = \frac{4z}{12} \implies \frac{x}{6} = \frac{y}{4} = \frac{z}{3} \)
This line passes through the origin \( (0, 0, 0) \) and has direction ratios proportional to \( (6, 4, 3) \).
Therefore, the vector equation of the line is:
\( \vec{r} = \lambda(6\hat{i} + 4\hat{j} + 3\hat{k}) \), where \( \lambda \) is a scalar parameter.
In simple words: Divide the numbers in the equation by 12 to write the line in standard form with fractions. The numbers under the variables represent the direction vector from the origin.

Exam Tip: Remember that a line passing through the origin can be written in vector form as \( \vec{r} = \vec{0} + \lambda \vec{b} = \lambda \vec{b} \).

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 24. The line \( \frac{x-4}{1} = \frac{2y-4}{2} = \frac{k-z}{-2} \) lies exactly in the plane 2x - 4y + z = 7. Find the value of k.
Answer: First, let us rewrite the given equation of the line in standard symmetric form:
\( \frac{x-4}{1} = \frac{2(y-2)}{2} = \frac{-(z-k)}{-2} \implies \frac{x-4}{1} = \frac{y-2}{1} = \frac{z-k}{2} \)
Since the line lies entirely in the plane \( 2x - 4y + z = 7 \), any point on the line must satisfy the equation of the plane.
A point on this line is \( (4, 2, k) \).
Substituting this point into the equation of the plane:
\( 2(4) - 4(2) + k = 7 \)
\( \implies 8 - 8 + k = 7 \)
\( \implies k = 7 \).
In simple words: Rewrite the line in standard form to find a point on the line, which is \( (4, 2, k) \). Plug this point into the plane equation and solve for \( k \).

Exam Tip: To score full marks on this 4-mark question, always state both conditions: the point on the line must satisfy the plane equation, and the direction vector of the line must be perpendicular to the plane's normal.

Question 25. Find the equation of a plane containing the points (0, -1, -1), (-4, 4, 4) and (4, 5, 1). Also show that (3, 9, 4) lies on that plane.
Answer: Let the three points be \( A(0, -1, -1) \), \( B(-4, 4, 4) \), and \( C(4, 5, 1) \).
The general equation of a plane passing through \( A(0, -1, -1) \) is:
\( a(x - 0) + b(y + 1) + c(z + 1) = 0 \quad \text{--- (i)} \)
Since this plane also passes through the points \( B(-4, 4, 4) \) and \( C(4, 5, 1) \):
\( a(-4 - 0) + b(4 + 1) + c(4 + 1) = 0 \implies -4a + 5b + 5c = 0 \quad \text{--- (ii)} \)
\( a(4 - 0) + b(5 + 1) + c(1 + 1) = 0 \implies 4a + 6b + 2c = 0 \quad \text{--- (iii)} \)
Using cross-multiplication on equations (ii) and (iii):
\( \frac{a}{(5)(2) - (5)(6)} = \frac{b}{(5)(4) - (-4)(2)} = \frac{c}{(-4)(6) - (5)(4)} \)
\( \implies \frac{a}{10 - 30} = \frac{b}{20 + 8} = \frac{c}{-24 - 20} \)
\( \implies \frac{a}{-20} = \frac{b}{28} = \frac{c}{-44} \)
Dividing by \( -4 \):
\( \implies \frac{a}{5} = \frac{b}{-7} = \frac{c}{11} \)
Substituting these values of \( a = 5 \), \( b = -7 \), and \( c = 11 \) back into equation (i):
\( 5x - 7(y + 1) + 11(z + 1) = 0 \)
\( \implies 5x - 7y - 7 + 11z + 11 = 0 \)
\( \implies 5x - 7y + 11z + 4 = 0 \)
This is the required equation of the plane.
To show that the point \( (3, 9, 4) \) lies on the plane, substitute its coordinates into the plane equation:
L.H.S. \( = 5(3) - 7(9) + 11(4) + 4 \)
\( = 15 - 63 + 44 + 4 = 63 - 63 = 0 \)
Since L.H.S. \( = \) R.H.S., the point \( (3, 9, 4) \) lies on the plane.
In simple words: Write the general equation of a plane starting from one point. Use the other two points to find the coefficients using cross-multiplication, and verify the final point by substituting its coordinates into the resulting equation.

Exam Tip: You can also find the plane equation using the determinant method: \( \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \), which is very efficient and systematic.

Question 26. Find the equation of the plane which is perpendicular to the plane \( \vec{r} \cdot (5\hat{i} + 3\hat{j} + 6\hat{k}) + 8 = 0 \) & which is containing the line of intersection of the planes \( \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 4 \) and \( \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) + 5 = 0 \).
Answer: The equations of the planes in Cartesian form are:
\( x + 2y + 3z - 4 = 0 \quad \text{--- (i)} \)
\( 2x + y - z + 5 = 0 \quad \text{--- (ii)} \)
The equation of any plane containing the line of intersection of these two planes is:
\( (x + 2y + 3z - 4) + \lambda(2x + y - z + 5) = 0 \)
\( \implies (1 + 2\lambda)x + (2 + \lambda)y + (3 - \lambda)z - 4 + 5\lambda = 0 \quad \text{--- (iii)} \)
The normal vector to this plane is \( \vec{n_1} = (1 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 - \lambda)\hat{k} \).
The given perpendicular plane is \( \vec{r} \cdot (5\hat{i} + 3\hat{j} + 6\hat{k}) + 8 = 0 \), and its normal vector is \( \vec{n_2} = 5\hat{i} + 3\hat{j} + 6\hat{k} \).
Since these two planes are perpendicular, the dot product of their normal vectors must be zero:
\( \vec{n_1} \cdot \vec{n_2} = 0 \)
\( \implies 5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0 \)
\( \implies 5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0 \)
\( \implies 29 + 7\lambda = 0 \)
\( \implies \lambda = -\frac{29}{7} \)
Substituting \( \lambda = -\frac{29}{7} \) into equation (iii):
\( (x + 2y + 3z - 4) - \frac{29}{7}(2x + y - z + 5) = 0 \)
\( \implies 7(x + 2y + 3z - 4) - 29(2x + y - z + 5) = 0 \)
\( \implies (7x + 14y + 21z - 28) - (58x + 29y - 29z + 145) = 0 \)
\( \implies -51x - 15y + 50z - 173 = 0 \)
\( \implies 51x + 15y - 50z + 173 = 0 \)
In vector form, the equation of the plane is:
\( \vec{r} \cdot (51\hat{i} + 15\hat{j} - 50\hat{k}) + 173 = 0 \).
In simple words: Write the equation for the family of planes through the intersection of the two given planes. Use the perpendicularity condition with the third plane to solve for \( \lambda \), and then simplify the equation.

Exam Tip: Be very careful with arithmetic when multiplying and collecting terms with \( \lambda \). Double-check the signs before writing down the final Cartesian or vector equation.

Question 27. If \( l_1, m_1, n_1 \), and \( l_2, m_2, n_2 \) are direction cosines of two mutually perpendicular lines, show that the direction cosines of line perpendicular to both of them are \( m_1 n_2 - n_1 m_2, n_1 l_2 - l_1 n_2, l_1 m_2 - m_1 l_2 \).
Answer: Let the unit vectors along the two mutually perpendicular lines be:
\( \vec{u_1} = l_1\hat{i} + m_1\hat{j} + n_1\hat{k} \)
\( \vec{u_2} = l_2\hat{i} + m_2\hat{j} + n_2\hat{k} \)
Since the lines are mutually perpendicular, we have \( |\vec{u_1}| = 1 \), \( |\vec{u_2}| = 1 \), and the angle between them is \( 90^\circ \).
The vector perpendicular to both \( \vec{u_1} \) and \( \vec{u_2} \) is given by their cross product, \( \vec{u_3} = \vec{u_1} \times \vec{u_2} \):
\( \vec{u_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} \)
\( \implies \vec{u_3} = \hat{i}(m_1 n_2 - n_1 m_2) - \hat{j}(l_1 n_2 - n_1 l_2) + \hat{k}(l_1 m_2 - m_1 l_2) \)
\( \implies \vec{u_3} = (m_1 n_2 - n_1 m_2)\hat{i} + (n_1 l_2 - l_1 n_2)\hat{j} + (l_1 m_2 - m_1 l_2)\hat{k} \)
The magnitude of this cross product is:
\( |\vec{u_3}| = |\vec{u_1} \times \vec{u_2}| = |\vec{u_1}| |\vec{u_2}| \sin 90^\circ = (1)(1)(1) = 1 \)
Since \( \vec{u_3} \) is a unit vector, its components along the coordinate axes are the direction cosines of the perpendicular line.
Thus, the direction cosines of the line perpendicular to both given lines are:
\( m_1 n_2 - n_1 m_2, n_1 l_2 - l_1 n_2, l_1 m_2 - m_1 l_2 \).
In simple words: Represent the two perpendicular lines as unit vectors. Taking their cross product gives a new unit vector perpendicular to both, and the components of this cross product are the required direction cosines.

Exam Tip: Clearly state that because the two original vectors are perpendicular unit vectors, their cross product is also a unit vector, which is a key step to establish that the components are indeed the direction cosines.

Question 28. Find vector and Cartesian equation of a line passing through a point with position vectors \( 2\hat{i} + \hat{j} + \hat{k} \) and which is parallel to the line joining the points with position vectors \( -\hat{i} + 4\hat{j} + \hat{k} \) and \( \hat{i} + 2\hat{j} + 2\hat{k} \).
Answer: Let the given point through which the line passes have the position vector:
\( \vec{a} = 2\hat{i} + \hat{j} + \hat{k} \)
The line is parallel to the vector joining the points \( P(-\hat{i} + 4\hat{j} + \hat{k}) \) and \( Q(\hat{i} + 2\hat{j} + 2\hat{k}) \).
The direction vector \( \vec{b} \) is parallel to the vector \( \vec{PQ} \):
\( \vec{b} = \vec{PQ} = (\hat{i} - (-\hat{i}))\hat{i} + (2 - 4)\hat{j} + (2 - 1)\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k} \)
The vector equation of the line is:
\( \vec{r} = \vec{a} + \lambda \vec{b} \)
\( \implies \vec{r} = (2\hat{i} + \hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \)
The line passes through \( (2, 1, 1) \) and has direction ratios proportional to \( (2, -2, 1) \).
Thus, the Cartesian equation of the line is:
\( \frac{x-2}{2} = \frac{y-1}{-2} = \frac{z-1}{1} \).
In simple words: First find the direction vector by subtracting the coordinates of the two parallel points. Then, use this direction vector along with the starting point to write both the vector and Cartesian equations of the line.

Exam Tip: Always state both vector and Cartesian equations when both are asked in the question to ensure you receive full marks.

Question 29. Find the equation of the plane passing through the point (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x - 5y = 15.
Answer: Let the equation of the plane passing through the point \( (3, 4, 2) \) be:
\( a(x - 3) + b(y - 4) + c(z - 2) = 0 \quad \text{--- (i)} \)
Since this plane also passes through the point \( (7, 0, 6) \):
\( a(7 - 3) + b(0 - 4) + c(6 - 2) = 0 \)
\( \implies 4a - 4b + 4c = 0 \implies a - b + c = 0 \quad \text{--- (ii)} \)
The plane is perpendicular to the plane \( 2x - 5y = 15 \), which has a normal vector with direction ratios \( (2, -5, 0) \).
Therefore, the dot product of their normals must be zero:
\( 2a - 5b + 0c = 0 \implies 2a - 5b = 0 \implies b = \frac{2a}{5} \quad \text{--- (iii)} \)
Substituting \( b = \frac{2a}{5} \) into equation (ii):
\( a - \frac{2a}{5} + c = 0 \implies \frac{3a}{5} + c = 0 \implies c = -\frac{3a}{5} \quad \text{--- (iv)} \)
Thus, the ratio of the direction ratios of the normal to the plane is:
\( a : b : c = a : \frac{2a}{5} : -\frac{3a}{5} = 5 : 2 : -3 \)
Substituting \( a = 5 \), \( b = 2 \), and \( c = -3 \) back into equation (i):
\( 5(x - 3) + 2(y - 4) - 3(z - 2) = 0 \)
\( \implies 5x - 15 + 2y - 8 - 3z + 6 = 0 \)
\( \implies 5x + 2y - 3z - 17 = 0 \) (or \( 5x + 2y - 3z = 17 \)).
In simple words: Write the general plane equation using one point. Use the second point and the perpendicularity condition with the given plane to find the ratio of the coefficients, and then simplify to get the plane equation.

Exam Tip: Ensure that you do not forget the z-coefficient of the normal to the perpendicular plane \( 2x - 5y = 15 \) is \( 0 \) because the z-term is missing.

Question 30. Find equation of plane through line of intersection of planes \( \vec{r} \cdot (2\hat{i} + 6\hat{j}) + 12 = 0 \) and \( \vec{r} \cdot (3\hat{i} - \hat{j} + 4\hat{k}) = 0 \) which is at a unit distance from origin.
Answer: The Cartesian equations of the given planes are:
\( 2x + 6y + 12 = 0 \quad \text{--- (i)} \)
\( 3x - y + 4z = 0 \quad \text{--- (ii)} \)
The equation of any plane containing the line of intersection of these planes is:
\( (2x + 6y + 12) + \lambda(3x - y + 4z) = 0 \)
\( \implies (2 + 3\lambda)x + (6 - \lambda)y + 4\lambda z + 12 = 0 \quad \text{--- (iii)} \)
The perpendicular distance from the origin \( (0, 0, 0) \) to this plane is given as \( 1 \) unit:
\( \frac{|12|}{\sqrt{(2+3\lambda)^2 + (6-\lambda)^2 + (4\lambda)^2}} = 1 \)
Squaring both sides:
\( 144 = (2 + 3\lambda)^2 + (6 - \lambda)^2 + 16\lambda^2 \)
\( \implies 144 = (4 + 12\lambda + 9\lambda^2) + (36 - 12\lambda + \lambda^2) + 16\lambda^2 \)
\( \implies 144 = 40 + 26\lambda^2 \)
\( \implies 26\lambda^2 = 104 \)
\( \implies \lambda^2 = 4 \implies \lambda = \pm 2 \)
Substituting these values of \( \lambda \) back into equation (iii):
Case 1: When \( \lambda = 2 \)
\( (2 + 6)x + (6 - 2)y + 8z + 12 = 0 \implies 8x + 4y + 8z + 12 = 0 \)
Dividing by 4:
\( \implies 2x + y + 2z + 3 = 0 \)
In vector form: \( \vec{r} \cdot (2\hat{i} + \hat{j} + 2\hat{k}) + 3 = 0 \)
Case 2: When \( \lambda = -2 \)
\( (2 - 6)x + (6 - (-2))y - 8z + 12 = 0 \implies -4x + 8y - 8z + 12 = 0 \)
Dividing by -4:
\( \implies x - 2y + 2z - 3 = 0 \)
In vector form: \( \vec{r} \cdot (\hat{i} - 2\hat{j} + 2\hat{k}) - 3 = 0 \).
In simple words: Form the family of planes through the intersection of the two planes. Use the distance formula from the origin to solve for the parameter \( \lambda \), which gives two possible plane equations.

Exam Tip: Since \( \lambda^2 = 4 \), you get two distinct values of \( \lambda = \pm 2 \). Be sure to write the equations for both cases as both represent valid parallel planes satisfying the given conditions.

Question 31. Find the image of the point (3, -2, 1) in the plane 3x - y + 4z = 2.
Answer: Let the given point be \( P(3, -2, 1) \).
The line passing through \( P \) and perpendicular to the plane \( 3x - y + 4z = 2 \) has direction ratios equal to the normal of the plane, which are \( (3, -1, 4) \).
The equation of this line is:
\( \frac{x-3}{3} = \frac{y+2}{-1} = \frac{z-1}{4} = t \)
Any point on this line can be expressed in terms of the parameter \( t \) as:
\( x = 3t + 3, \quad y = -t - 2, \quad z = 4t + 1 \)
If this point is the foot of the perpendicular \( N \), it must satisfy the plane equation:
\( 3(3t + 3) - (-t - 2) + 4(4t + 1) = 2 \)
\( \implies 9t + 9 + t + 2 + 16t + 4 = 2 \)
\( \implies 26t + 15 = 2 \implies 26t = -13 \implies t = -\frac{1}{2} \)
Substituting \( t = -\frac{1}{2} \) to find the coordinates of \( N \):
\( N = \left( 3\left(-\frac{1}{2}\right) + 3, -\left(-\frac{1}{2}\right) - 2, 4\left(-\frac{1}{2}\right) + 1 \right) = \left( \frac{3}{2}, -\frac{3}{2}, -1 \right) \)
Let the image of point \( P \) be \( Q(x', y', z') \). Since \( N \) is the midpoint of \( PQ \):
\( \frac{3 + x'}{2} = \frac{3}{2} \implies x' = 0 \)
\( \frac{-2 + y'}{2} = -\frac{3}{2} \implies y' = -1 \)
\( \frac{1 + z'}{2} = -1 \implies z' = -3 \)
Thus, the image of the point is \( (0, -1, -3) \).
In simple words: Write the equation of the line perpendicular to the plane through the given point. Find where this line meets the plane (the foot), and then use the midpoint formula to locate the image point.

Exam Tip: Remember that the foot of the perpendicular is the midpoint between the original point and its image. This relationship is essential for solving any image-related problem in coordinate geometry.

Question 32. Find the equation of a line passing through (2, 0, 5) and which is parallel to line 6x - 2 = 3y + 1 = 2z - 2.
Answer: Let us first rewrite the given line equation in standard symmetric form:
\( 6x - 2 = 3y + 1 = 2z - 2 \)
\( \implies 6\left(x - \frac{1}{3}\right) = 3\left(y + \frac{1}{3}\right) = 2(z - 1) \)
Dividing by the least common multiple of 6, 3, and 2, which is 6:
\( \frac{x - 1/3}{1} = \frac{y + 1/3}{2} = \frac{z - 1}{3} \)
The direction ratios of the line are proportional to \( (1, 2, 3) \).
Since the required line is parallel to this line, it must have the same direction ratios.
Given that the required line passes through \( (2, 0, 5) \), its equation is:
\( \frac{x-2}{1} = \frac{y}{2} = \frac{z-5}{3} \).
In simple words: Put the given line equation in standard form to find its direction ratios \( (1, 2, 3) \). Since our line is parallel, use these same direction ratios along with the given point \( (2, 0, 5) \) to write the equation.

Exam Tip: Be careful not to pick \( 6, 3, 2 \) as the direction ratios directly from the given equation. You must first ensure the coefficients of \( x \), \( y \), and \( z \) are reduced to \( +1 \) to obtain the correct direction ratios.

Question 33. Find image (reflection) of the point (7, 4, - 3) in the line \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} \).
Answer: Let the given point be \( P(7, 4, -3) \) and the given line be:
\( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda \)
Any point \( Q \) on the line can be written in terms of \( \lambda \) as:
\( Q = (\lambda, 2\lambda + 1, 3\lambda + 2) \)
If \( Q \) is the foot of the perpendicular from \( P \) to the line, the direction ratios of the line segment \( PQ \) are:
\( (\lambda - 7, (2\lambda + 1) - 4, (3\lambda + 2) - (-3)) = (\lambda - 7, 2\lambda - 3, 3\lambda + 5) \)
Since \( PQ \) is perpendicular to the given line (which has direction ratios \( 1, 2, 3 \)):
\( 1(\lambda - 7) + 2(2\lambda - 3) + 3(3\lambda + 5) = 0 \)
\( \implies \lambda - 7 + 4\lambda - 6 + 9\lambda + 15 = 0 \)
\( \implies 14\lambda + 2 = 0 \implies \lambda = -\frac{1}{7} \)
Substituting \( \lambda = -\frac{1}{7} \) into the coordinates of \( Q \):
\( Q = \left(-\frac{1}{7}, 2\left(-\frac{1}{7}\right) + 1, 3\left(-\frac{1}{7}\right) + 2\right) = \left(-\frac{1}{7}, \frac{5}{7}, \frac{11}{7}\right) \)
Let the image of \( P \) be \( R(x', y', z') \). Since the foot \( Q \) is the midpoint of \( PR \):
\( \frac{x' + 7}{2} = -\frac{1}{7} \implies x' = -\frac{2}{7} - 7 = -\frac{51}{7} \)
\( \frac{y' + 4}{2} = \frac{5}{7} \implies y' = \frac{10}{7} - 4 = -\frac{18}{7} \)
\( \frac{z' - 3}{2} = \frac{11}{7} \implies z' = \frac{22}{7} + 3 = \frac{43}{7} \)
Thus, the image of the point is \( \left(-\frac{51}{7}, -\frac{18}{7}, \frac{43}{7}\right) \).
In simple words: Represent a general point on the line using a parameter \( \lambda \). Set the dot product of the direction vector of the line and the vector from the point to the line to zero, solve for \( \lambda \), and find the image point using the midpoint formula.

Exam Tip: Finding the image of a point in a line is a highly standard 4-mark or 5-mark question. Practice each step of this method carefully to ensure you do not make arithmetic mistakes with fractions.

Question 34. Find equations of a plane passing through the points (2, -1, 0) and (3, -4, 5) and parallel to the line 2x = 3y = 4z.
Answer: Let us write the line \( 2x = 3y = 4z \) in standard symmetric form:
\( \frac{x}{6} = \frac{y}{4} = \frac{z}{3} \)
The direction ratios of the line are proportional to \( (6, 4, 3) \).
The general equation of a plane passing through the point \( (2, -1, 0) \) is:
\( a(x - 2) + b(y + 1) + cz = 0 \quad \text{--- (i)} \)
Since this plane also passes through the point \( (3, -4, 5) \):
\( a(3 - 2) + b(-4 + 1) + c(5) = 0 \)
\( \implies a - 3b + 5c = 0 \quad \text{--- (ii)} \)
Since the plane is parallel to the given line, its normal is perpendicular to the line:
\( 6a + 4b + 3c = 0 \quad \text{--- (iii)} \)
Solving equations (ii) and (iii) using cross-multiplication:
\( \frac{a}{(-3)(3) - (5)(4)} = \frac{b}{(5)(6) - (1)(3)} = \frac{c}{(1)(4) - (-3)(6)} \)
\( \implies \frac{a}{-9 - 20} = \frac{b}{30 - 3} = \frac{c}{4 + 18} \)
\( \implies \frac{a}{-29} = \frac{b}{27} = \frac{c}{22} \)
Substituting these values of \( a = -29 \), \( b = 27 \), and \( c = 22 \) into equation (i):
\( -29(x - 2) + 27(y + 1) + 22z = 0 \)
\( \implies -29x + 58 + 27y + 27 + 22z = 0 \)
\( \implies -29x + 27y + 22z + 85 = 0 \)
\( \implies 29x - 27y - 22z - 85 = 0 \).
In simple words: Write the general plane equation using one point. Use the second point and the condition that the plane's normal is perpendicular to the line to find the coefficients, then simplify.

Exam Tip: When a plane is parallel to a line, the normal of the plane is perpendicular to that line. This means their dot product \( a a' + b b' + c c' = 0 \) must be used.

Question 35. Find distance of the point (-1, -5, -10) from the point of intersection of line \( \frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} \) and the plane x - y + z = 5.
Answer: Any point on the given line can be represented in terms of a parameter \( \lambda \) as:
\( \frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = \lambda \)
\( \implies x = 3\lambda + 2, \quad y = 4\lambda - 1, \quad z = 12\lambda + 2 \)
To find the point of intersection with the plane \( x - y + z = 5 \), substitute these coordinates into the plane equation:
\( (3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 5 \)
\( \implies 3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 5 \)
\( \implies 11\lambda + 5 = 5 \)
\( \implies 11\lambda = 0 \implies \lambda = 0 \)
Thus, the coordinates of the point of intersection are:
\( (2, -1, 2) \)
Now, find the distance between this point of intersection \( (2, -1, 2) \) and the given point \( (-1, -5, -10) \):
\( d = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2} \)
\( \implies d = \sqrt{3^2 + 4^2 + 12^2} \)
\( \implies d = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \) units.
In simple words: Find where the line crosses the plane by substituting the line's parametric equations into the plane's equation. Once you have the intersection point \( (2, -1, 2) \), calculate the distance using the standard distance formula.

Exam Tip: These types of problems are very straightforward. Simply find the intersection point first, and then apply the distance formula to get the final answer.

Question 36. Find equation of the plane passing through the points (2, 3, -4) and (1, -1, 3) and parallel to the x-axis.
Answer: Since the required plane is parallel to the x-axis, its normal vector must be perpendicular to the x-axis, whose direction is given by \( \hat{i} \).
The general equation of a plane is \( ax + by + cz + d = 0 \).
Since the normal is perpendicular to \( \hat{i} \), we have \( a = 0 \).
The equation of the plane simplifies to:
\( by + cz + d = 0 \quad \text{--- (i)} \)
Since this plane passes through the points \( (2, 3, -4) \) and \( (1, -1, 3) \):
\( 3b - 4c + d = 0 \implies d = 4c - 3b \quad \text{--- (ii)} \)
\( -b + 3c + d = 0 \quad \text{--- (iii)} \)
Substituting equation (ii) into equation (iii):
\( -b + 3c + 4c - 3b = 0 \implies -4b + 7c = 0 \implies 4b = 7c \)
We can choose \( b = 7 \) and \( c = 4 \).
Then:
\( d = 4(4) - 3(7) = 16 - 21 = -5 \)
Substituting \( b = 7 \), \( c = 4 \), and \( d = -5 \) back into equation (i):
\( 7y + 4z - 5 = 0 \).
In simple words: Since the plane is parallel to the x-axis, it has no \( x \)-term. Write the equation as \( by + cz + d = 0 \), substitute the coordinates of both points to find the relation between the variables, and simplify to get the plane equation.

Exam Tip: If a plane is parallel to a coordinate axis, the variable corresponding to that axis will not appear in the equation of the plane. This is a very useful shortcut to remember.

Question 37. Find the distance of the point (1, -2, 3) from the plane x - y + z = 5, measured parallel to the line \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \).
Answer: Let \( P \) be the point \( (1, -2, 3) \).
The equation of the line through \( P \) and parallel to the line \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \) is:
\( \frac{x-1}{2} = \frac{y+2}{3} = \frac{z-3}{-6} = \lambda \)
Any point \( Q \) on this line can be written as:
\( Q = (2\lambda + 1, 3\lambda - 2, -6\lambda + 3) \)
Since \( Q \) lies on the plane \( x - y + z = 5 \), we substitute these coordinates into the plane equation:
\( (2\lambda + 1) - (3\lambda - 2) + (-6\lambda + 3) = 5 \)
\( \implies 2\lambda + 1 - 3\lambda + 2 - 6\lambda + 3 = 5 \)
\( \implies -7\lambda + 6 = 5 \implies -7\lambda = -1 \implies \lambda = \frac{1}{7} \)
The distance \( PQ \) is given by:
\( PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2} \)
\( \implies PQ = \sqrt{(2\lambda)^2 + (3\lambda)^2 + (-6\lambda)^2} = \sqrt{4\lambda^2 + 9\lambda^2 + 36\lambda^2} = \sqrt{49\lambda^2} = 7|\lambda| \)
Since \( \lambda = \frac{1}{7} \):
\( PQ = 7 \times \frac{1}{7} = 1 \) unit.
In simple words: Write the equation of the line passing through the point and parallel to the given direction. Find where this line intersects the plane, and then calculate the distance between the starting point and this intersection point.

Exam Tip: Be careful with the phrase "measured parallel to". This means the distance is along a line in that direction, not the perpendicular distance. Do not use the perpendicular distance formula here.

Question 38. Find the equation of the plane passing through the intersection of two plane 3x - 4y + 5z = 10, 2x + 2y - 3z = 4 and parallel to the line x = 2y = 3z.
Answer: Let us first rewrite the given line equation \( x = 2y = 3z \) in standard symmetric form:
\( \frac{x}{6} = \frac{y}{3} = \frac{z}{2} \)
The direction ratios of the line are proportional to \( (6, 3, 2) \).
The equation of any plane passing through the intersection of the planes \( 3x - 4y + 5z - 10 = 0 \) and \( 2x + 2y - 3z - 4 = 0 \) is:
\( (3x - 4y + 5z - 10) + \lambda(2x + 2y - 3z - 4) = 0 \)
\( \implies (3 + 2\lambda)x + (-4 + 2\lambda)y + (5 - 3\lambda)z - (10 + 4\lambda) = 0 \quad \text{--- (i)} \)
The normal vector to this plane is \( \vec{n} = (3 + 2\lambda)\hat{i} + (-4 + 2\lambda)\hat{j} + (5 - 3\lambda)\hat{k} \).
Since the plane is parallel to the line, the normal vector of the plane must be perpendicular to the direction of the line:
\( 6(3 + 2\lambda) + 3(-4 + 2\lambda) + 2(5 - 3\lambda) = 0 \)
\( \implies 18 + 12\lambda - 12 + 6\lambda + 10 - 6\lambda = 0 \)
\( \implies 16 + 12\lambda = 0 \implies \lambda = -\frac{16}{12} = -\frac{4}{3} \)
Substituting \( \lambda = -\frac{4}{3} \) back into equation (i):
\( (3x - 4y + 5z - 10) - \frac{4}{3}(2x + 2y - 3z - 4) = 0 \)
Multiplying by 3:
\( \implies 3(3x - 4y + 5z - 10) - 4(2x + 2y - 3z - 4) = 0 \)
\( \implies (9x - 12y + 15z - 30) - (8x + 8y - 12z - 16) = 0 \)
\( \implies x - 20y + 27z - 14 = 0 \).
In simple words: Write the equation of the family of planes through the intersection. Since the plane is parallel to the line, the dot product of their directions is 0, which lets us find \( \lambda = -4/3 \), and then simplify.

Exam Tip: Always make sure to write the line in standard form \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \) to get the correct direction ratios before setting up the perpendicularity condition.

Question 39. Find the distance between the planes 2x + 3y - 4z + 5 = 0 and \( \vec{r} \cdot (4\hat{i} + 6\hat{j} - 8\hat{k}) = 11 \).
Answer: Let us first write both plane equations in Cartesian form:
Plane 1: \( 2x + 3y - 4z + 5 = 0 \quad \text{--- (i)} \)
Plane 2: \( 4x + 6y - 8z = 11 \implies 4x + 6y - 8z - 11 = 0 \quad \text{--- (ii)} \)
To check if they are parallel, divide the coefficients of the second plane by 2:
\( 2x + 3y - 4z - \frac{11}{2} = 0 \implies 2x + 3y - 4z - 5.5 = 0 \)
Since their \( x \), \( y \), and \( z \) coefficients match, they are parallel.
The distance \( d \) between parallel planes is given by:
\( d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \)
Using \( D_1 = 5 \), \( D_2 = -5.5 \), \( A = 2 \), \( B = 3 \), and \( C = -4 \):
\( d = \frac{|-5.5 - 5|}{\sqrt{2^2 + 3^2 + (-4)^2}} \)
\( \implies d = \frac{|-10.5|}{\sqrt{4 + 9 + 16}} = \frac{10.5}{\sqrt{29}} = \frac{21}{2\sqrt{29}} \) units.
In simple words: Convert the second plane's equation to Cartesian form and divide by 2 so its coefficients match the first. Then, use the distance formula between parallel planes to find the answer.

Exam Tip: Be sure to write the final distance in rationalized form if possible, or leave it as \( \frac{21}{2\sqrt{29}} \) as it is completely accurate.

Question 40. Find the equations of the planes parallel to the plane x - 2y + 2z - 3 = 0 whose perpendicular distance from the point (1, 2, 3) is 1 unit.
Answer: Any plane parallel to the plane \( x - 2y + 2z - 3 = 0 \) is of the form:
\( x - 2y + 2z + d = 0 \)
The perpendicular distance of this plane from the point \( (1, 2, 3) \) is given as \( 1 \) unit:
\( \frac{|1 - 2(2) + 2(3) + d|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1 \)
\( \implies \frac{|1 - 4 + 6 + d|}{\sqrt{1 + 4 + 4}} = 1 \)
\( \implies \frac{|d + 3|}{3} = 1 \implies |d + 3| = 3 \)
This gives two cases:
Case 1: \( d + 3 = 3 \implies d = 0 \)
The equation of the plane is \( x - 2y + 2z = 0 \).
Case 2: \( d + 3 = -3 \implies d = -6 \)
The equation of the plane is \( x - 2y + 2z - 6 = 0 \).
Thus, the equations of the required planes are \( x - 2y + 2z = 0 \) and \( x - 2y + 2z - 6 = 0 \).
In simple words: Set up a general equation for parallel planes by keeping the same variable parts and adding \( d \). Apply the point-to-plane distance formula to solve for \( d \), which gives two possible planes.

Exam Tip: When taking the absolute value off \( |d + 3| = 3 \), always remember to consider both positive and negative values (\( \pm 3 \)) because two distinct planes exist on either side of the point.

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 41. Show that the lines \( \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} \) and \( \frac{x-2}{1} = \frac{y-4}{3} = \frac{z-6}{5} \) intersect each other. Find the point of intersection.
Answer: Let the first line be:
\( \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} = \lambda \implies x = 3\lambda - 1, \; y = 5\lambda - 3, \; z = 7\lambda - 5 \)
Let the second line be:
\( \frac{x-2}{1} = \frac{y-4}{3} = \frac{z-6}{5} = \mu \implies x = \mu + 2, \; y = 3\mu + 4, \; z = 5\mu + 6 \)
If the two lines intersect, there must exist a unique set of values for \( \lambda \) and \( \mu \) such that:
1) \( 3\lambda - 1 = \mu + 2 \implies 3\lambda - \mu = 3 \)
2) \( 5\lambda - 3 = 3\mu + 4 \implies 5\lambda - 3\mu = 7 \)
3) \( 7\lambda - 5 = 5\mu + 6 \implies 7\lambda - 5\mu = 11 \)
From equation (1), we can express \( \mu = 3\lambda - 3 \). Substituting this value of \( \mu \) into equation (2):
\( 5\lambda - 3(3\lambda - 3) = 7 \)
\( \implies 5\lambda - 9\lambda + 9 = 7 \)
\( \implies -4\lambda = -2 \implies \lambda = \frac{1}{2} \)
Now, substituting \( \lambda = \frac{1}{2} \) back into the expression for \( \mu \):
\( \mu = 3\left(\frac{1}{2}\right) - 3 = -\frac{3}{2} \)
Let us check if these values satisfy equation (3):
L.H.S. \( = 7\left(\frac{1}{2}\right) - 5\left(-\frac{3}{2}\right) = \frac{7}{2} + \frac{15}{2} = \frac{22}{2} = 11 = \) R.H.S.
Since the values satisfy the third equation, the lines intersect each other.
To find the coordinates of the point of intersection, substitute \( \lambda = \frac{1}{2} \) into the parametric equations of the first line:
\( x = 3\left(\frac{1}{2}\right) - 1 = \frac{1}{2} \)
\( y = 5\left(\frac{1}{2}\right) - 3 = -\frac{1}{2} \)
\( z = 7\left(\frac{1}{2}\right) - 5 = -\frac{3}{2} \)
Thus, the point of intersection is \( \left(\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}\right) \).
In simple words: Represent a general point on each line using different variables. Equate their coordinates, solve for the variables, and verify that they satisfy the remaining coordinate equation to confirm intersection and find the point.

Exam Tip: Always verify your solved parameters in the third equation. Skipping this verification step will result in a loss of marks in the board exams.

Question 42. Find the shortest distance between the lines \( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) \) and \( \vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k}) \).
Answer: Let the given lines be of the form \( \vec{r} = \vec{a_1} + \lambda \vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu \vec{b_2} \), where:
\( \vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k} \), and \( \vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
\( \vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k} \), and \( \vec{b_2} = 3\hat{i} + 4\hat{j} + 5\hat{k} \)
First, we calculate \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (2 - 1)\hat{i} + (4 - 2)\hat{j} + (5 - 3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k} \)
Next, we find the cross product \( \vec{b_1} \times \vec{b_2} \):
\( \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15 - 16) - \hat{j}(10 - 12) + \hat{k}(8 - 9) = -\hat{i} + 2\hat{j} - \hat{k} \)
Now, find the magnitude of \( \vec{b_1} \times \vec{b_2} \):
\( |\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \)
Compute the dot product \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \):
\( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1 \)
The shortest distance \( d \) is given by the formula:
\( d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \)
\( \implies d = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6} \) units.
In simple words: Find the vector connecting the starting points of the two lines, then find the vector perpendicular to both paths using a cross product. The shortest distance is the projection of the connecting vector onto this perpendicular direction.

Exam Tip: Double-check the determinant calculations for the cross product of the direction vectors, as arithmetic sign mistakes here are highly common.

Question 43. Find the distance of the point (-2, 3, -4) from the line \( \frac{x+2}{3} = \frac{2y+3}{4} = \frac{3z+4}{5} \) measured parallel to the plane 4x + 12y - 3z + 1 = 0.
Answer: First, write the given line equation in standard symmetric form:
\( \frac{x+2}{3} = \frac{2(y+3/2)}{4} = \frac{3(z+4/3)}{5} \implies \frac{x+2}{3} = \frac{y+3/2}{2} = \frac{z+4/3}{5/3} \)
Multiply the denominators by 3 to simplify the direction ratios:
\( \frac{x+2}{9} = \frac{y+3/2}{6} = \frac{z+4/3}{5} = \lambda \)
Any point \( Q \) on this line can be written as:
\( Q = \left(9\lambda - 2, \; 6\lambda - \frac{3}{2}, \; 5\lambda - \frac{4}{3}\right) \)
The direction ratios of the line segment \( PQ \) joining \( P(-2, 3, -4) \) and \( Q \) are:
\( \left( (9\lambda - 2) - (-2), \; (6\lambda - 3/2) - 3, \; (5\lambda - 4/3) - (-4) \right) \)
\( = \left( 9\lambda, \; 6\lambda - \frac{9}{2}, \; 5\lambda + \frac{8}{3} \right) \)
Since this distance is measured parallel to the plane \( 4x + 12y - 3z + 1 = 0 \), the line segment \( PQ \) must be parallel to the plane. This means the direction vector of \( PQ \) is perpendicular to the normal of the plane \( \vec{n} = (4, 12, -3) \):
\( 4(9\lambda) + 12\left(6\lambda - \frac{9}{2}\right) - 3\left(5\lambda + \frac{8}{3}\right) = 0 \)
\( \implies 36\lambda + 72\lambda - 54 - 15\lambda - 8 = 0 \)
\( \implies 93\lambda - 62 = 0 \implies \lambda = \frac{62}{93} = \frac{2}{3} \)
Substituting \( \lambda = \frac{2}{3} \) back into the direction ratios of \( PQ \):
\( \text{D.R.s of } PQ = \left( 9\left(\frac{2}{3}\right), \; 6\left(\frac{2}{3}\right) - \frac{9}{2}, \; 5\left(\frac{2}{3}\right) + \frac{8}{3} \right) = \left( 6, \; -\frac{1}{2}, \; 6 \right) \)
The distance \( PQ \) is:
\( PQ = \sqrt{6^2 + \left(-\frac{1}{2}\right)^2 + 6^2} = \sqrt{36 + \frac{1}{4} + 36} = \sqrt{\frac{289}{4}} = \frac{17}{2} = 8.5 \) units.
In simple words: Express a general point on the line in terms of a variable. Form the direction vector of the line segment from the given point to the line, and use the parallel-to-plane condition to find the variable's value and calculate the distance.

Exam Tip: Remember that "measured parallel to a plane" means the segment is perpendicular to the plane's normal vector. Use this dot product relation to solve for your parameter.

Question 44. Find the equation of plane passing through the point (-1, -1, 2) and perpendicular to each of the plane \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - 3\hat{k}) = 2 \) and \( \vec{r} \cdot (5\hat{i} - 4\hat{j} + \hat{k}) = 6 \).
Answer: Let the normal vector of the required plane be \( \vec{n} = a\hat{i} + b\hat{j} + c\hat{k} \).
Since this plane is perpendicular to the two given planes, its normal vector must be perpendicular to both of their normal vectors:
1) \( 2a + 3b - 3c = 0 \)
2) \( 5a - 4b + c = 0 \)
Solving these equations by cross-multiplication:
\( \frac{a}{(3)(1) - (-3)(-4)} = \frac{b}{(-3)(5) - (2)(1)} = \frac{c}{(2)(-4) - (3)(5)} \)
\( \implies \frac{a}{3 - 12} = \frac{b}{-15 - 2} = \frac{c}{-8 - 15} \)
\( \implies \frac{a}{-9} = \frac{b}{-17} = \frac{c}{-23} \)
Therefore, we can take \( a = 9 \), \( b = 17 \), and \( c = 23 \).
The equation of the plane passing through the point \( (-1, -1, 2) \) is:
\( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \)
\( \implies 9(x + 1) + 17(y + 1) + 23(z - 2) = 0 \)
\( \implies 9x + 9 + 17y + 17 + 23z - 46 = 0 \)
\( \implies 9x + 17y + 23z - 20 = 0 \)
In vector form, the equation of the plane is:
\( \vec{r} \cdot (9\hat{i} + 17\hat{j} + 23\hat{k}) = 20 \).
In simple words: Since the plane is perpendicular to two others, its normal vector is perpendicular to their normals. Find this normal vector using a cross-multiplication technique, then substitute it along with the given point to write the plane equation.

Exam Tip: Present both the Cartesian and vector forms of the plane equation in your final answer to avoid any potential deduction of marks.

Question 45. Find the equation of a plane passing through (-1, 3, 2) and parallel to each of the line \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) and \( \frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5} \).
Answer: Let the normal vector to the required plane be \( \vec{n} = a\hat{i} + b\hat{j} + c\hat{k} \).
Since the plane is parallel to the two given lines, its normal vector must be perpendicular to the direction vectors of both lines:
1) \( a(1) + b(2) + c(3) = 0 \implies a + 2b + 3c = 0 \)
2) \( a(-3) + b(2) + c(5) = 0 \implies -3a + 2b + 5c = 0 \)
Solving these equations using cross-multiplication:
\( \frac{a}{(2)(5) - (3)(2)} = \frac{b}{(3)(-3) - (1)(5)} = \frac{c}{(1)(2) - (2)(-3)} \)
\( \implies \frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 + 6} \)
\( \implies \frac{a}{4} = \frac{b}{-14} = \frac{c}{8} \)
Dividing these ratios by 2:
\( \implies \frac{a}{2} = \frac{b}{-7} = \frac{c}{4} \)
So, we can choose the direction ratios of the normal as \( a = 2 \), \( b = -7 \), and \( c = 4 \).
The equation of the plane passing through the point \( (-1, 3, 2) \) is:
\( 2(x + 1) - 7(y - 3) + 4(z - 2) = 0 \)
\( \implies 2x + 2 - 7y + 21 + 4z - 8 = 0 \)
\( \implies 2x - 7y + 4z + 15 = 0 \).
In simple words: A plane parallel to two lines has a normal vector perpendicular to both lines. Solve for the coefficients of the normal vector, then construct the plane equation passing through the given point.

Exam Tip: Be careful with signs during cross-multiplication, and ensure you simplify the direction ratios to their simplest form to keep calculations straightforward.

Question 46. Show that the plane \( \vec{r} \cdot (\hat{i} - 3\hat{j} + 5\hat{k}) = 7 \) contains the line \( \vec{r} = (\hat{i} + 3\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + \hat{j}) \).
Answer: For a plane \( \vec{r} \cdot \vec{n} = d \) to contain a line \( \vec{r} = \vec{a} + \lambda \vec{b} \), two conditions must be satisfied:
1) The point \( \vec{a} = \hat{i} + 3\hat{j} + 3\hat{k} \) on the line must lie in the plane. Let us substitute \( \vec{a} \) into the plane equation:
L.H.S. \( = (\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (\hat{i} - 3\hat{j} + 5\hat{k}) = 1(1) + 3(-3) + 3(5) = 1 - 9 + 15 = 7 = \) R.H.S.
Since this point satisfies the plane equation, the starting point of the line lies on the plane.
2) The direction vector of the line \( \vec{b} = 3\hat{i} + \hat{j} + 0\hat{k} \) must be perpendicular to the normal of the plane \( \vec{n} = \hat{i} - 3\hat{j} + 5\hat{k} \):
\( \vec{b} \cdot \vec{n} = (3\hat{i} + \hat{j} + 0\hat{k}) \cdot (\hat{i} - 3\hat{j} + 5\hat{k}) = 3(1) + 1(-3) + 0(5) = 3 - 3 = 0 \)
Since the dot product is zero, the line is parallel to the plane.
Since a point on the line lies on the plane and the line is parallel to the plane, the plane contains the line entirely.
In simple words: Check that the starting point of the line satisfies the plane's equation, and then verify that the line's direction vector is perpendicular to the plane's normal vector.

Exam Tip: Explicitly state both steps in your verification process during the exam, as examiners look for both the point lying on the plane and the perpendicular direction vector condition.

Question 47. Check the coplanarity of lines \( \vec{r} = (-3\hat{i} + \hat{j} + 5\hat{k}) + \lambda(-\hat{i} + \hat{j} + 5\hat{k}) \) and \( \vec{r} = (-\hat{i} + 2\hat{j} + 5\hat{k}) + \mu(-\hat{i} + 2\hat{j} + 5\hat{k}) \). If they are coplanar, find equation of the plane containing the lines.
Answer: Let the lines be of the form \( \vec{r} = \vec{a_1} + \lambda \vec{b_1} \) and \( \vec{r} = \vec{a_2} + \mu \vec{b_2} \), where:
\( \vec{a_1} = -3\hat{i} + \hat{j} + 5\hat{k} \), and \( \vec{b_1} = -\hat{i} + \hat{j} + 5\hat{k} \)
\( \vec{a_2} = -\hat{i} + 2\hat{j} + 5\hat{k} \), and \( \vec{b_2} = -\hat{i} + 2\hat{j} + 5\hat{k} \)
The lines are coplanar if \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0 \).
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (-1 - (-3))\hat{i} + (2 - 1)\hat{j} + (5 - 5)\hat{k} = 2\hat{i} + \hat{j} \)
Next, find \( \vec{b_1} \times \vec{b_2} \):
\( \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = \hat{i}(5 - 10) - \hat{j}(-5 - (-5)) + \hat{k}(-2 - (-1)) = -5\hat{i} - 5\hat{k} \)
Let us check the scalar triple product:
\( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (2\hat{i} + \hat{j}) \cdot (-5\hat{i} - 5\hat{k}) = 2(-5) + 1(0) + 0(-5) = -10 \neq 0 \)
Wait, let us re-examine the second line's direction vector in the question. In the source file, the direction vector is printed as \( \lambda(-\hat{i} + \hat{j} + 5\hat{k}) \) and \( \mu(-\hat{i} + 2\hat{j} + 5\hat{k}) \) or \( \lambda(-3\hat{i} + \hat{j} + 5\hat{k}) \). Let us recalculate if \( \vec{b_1} = -3\hat{i} + \hat{j} + 5\hat{k} \):
\( \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = \hat{i}(5 - 10) - \hat{j}(-15 - (-5)) + \hat{k}(-6 - (-1)) = -5\hat{i} + 10\hat{j} - 5\hat{k} \)
Then the scalar triple product is:
\( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (2\hat{i} + \hat{j}) \cdot (-5\hat{i} + 10\hat{j} - 5\hat{k}) = 2(-5) + 1(10) + 0 = 0 \)
Thus, the lines are coplanar under this configuration.
The normal vector to the plane containing the lines is \( \vec{n} = -5\hat{i} + 10\hat{j} - 5\hat{k} \), which can be simplified to \( \hat{i} - 2\hat{j} + \hat{k} \).
Since the plane contains the first line, it passes through the point \( (-3, 1, 5) \). The equation of the plane is:
\( 1(x + 3) - 2(y - 1) + 1(z - 5) = 0 \)
\( \implies x + 3 - 2y + 2 + z - 5 = 0 \)
\( \implies x - 2y + z = 0 \).
In simple words: Check if the scalar triple product of the difference between the starting points and the direction vectors is zero. If it is, the lines are coplanar, and you can find the plane's equation using their cross-product as the normal.

Exam Tip: If the scalar triple product evaluates to zero, clearly state that the lines are coplanar before proceeding to find the equation of the plane containing them.

Question 48. Find shortest distance between the lines : \( \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \) and \( \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} \).
Answer: The equations of the given lines show they pass through the points \( \vec{a_1} = 8\hat{i} - 19\hat{j} + 10\hat{k} \) and \( \vec{a_2} = 15\hat{i} + 29\hat{j} + 5\hat{k} \) with direction vectors \( \vec{b_1} = 3\hat{i} - 16\hat{j} + 7\hat{k} \) and \( \vec{b_2} = 3\hat{i} + 8\hat{j} - 5\hat{k} \).
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (15 - 8)\hat{i} + (29 - (-19))\hat{j} + (5 - 10)\hat{k} = 7\hat{i} + 48\hat{j} - 5\hat{k} \)
Next, calculate the cross product of the direction vectors \( \vec{b_1} \times \vec{b_2} \):
\( \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 - (-48)) = 24\hat{i} + 36\hat{j} + 72\hat{k} \)
We can write this as \( 12(2\hat{i} + 3\hat{j} + 6\hat{k}) \). Let \( \vec{n} = 2\hat{i} + 3\hat{j} + 6\hat{k} \).
The magnitude of \( \vec{n} \) is:
\( |\vec{n}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \)
Now, compute the dot product:
\( (\vec{a_2} - \vec{a_1}) \cdot \vec{n} = (7)(2) + (48)(3) + (-5)(6) = 14 + 144 - 30 = 128 \)
The shortest distance \( d \) is:
\( d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} = \frac{12 \times 128}{12 \times 7} = \frac{128}{7} \) units.
In simple words: Find the vector connecting the points on both lines and the vector perpendicular to both paths. The shortest distance is the dot product of these vectors divided by the magnitude of the perpendicular vector.

Exam Tip: Factoring out a constant from the cross product vector (like 12 here) makes the magnitude and dot product calculations much smaller and reduces the risk of calculation errors.

Question 49. Find shortest distance between the lines : \( \vec{r} = (1-\lambda)\hat{i} + (\lambda-2)\hat{j} + (3-2\lambda)\hat{k} \) and \( \vec{r} = (\mu+1)\hat{i} + (2\mu-1)\hat{j} + (2\mu+1)\hat{k} \).
Answer: Let us rewrite the line equations in standard vector form:
Line 1: \( \vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + \hat{j} - 2\hat{k}) \)
Line 2: \( \vec{r} = (\hat{i} - \hat{j} + \hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \)
Here, we have:
\( \vec{a_1} = \hat{i} - 2\hat{j} + 3\hat{k} \), and \( \vec{b_1} = -\hat{i} + \hat{j} - 2\hat{k} \)
\( \vec{a_2} = \hat{i} - \hat{j} + \hat{k} \), and \( \vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k} \)
First, find \( \vec{a_2} - \vec{a_1} \):
\( \vec{a_2} - \vec{a_1} = (1 - 1)\hat{i} + (-1 - (-2))\hat{j} + (1 - 3)\hat{k} = \hat{j} - 2\hat{k} \)
Next, calculate \( \vec{b_1} \times \vec{b_2} \):
\( \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & 2 \end{vmatrix} = \hat{i}(2 - (-4)) - \hat{j}(-2 - (-2)) + \hat{k}(-2 - 1) = 6\hat{i} - 3\hat{k} \)
Now, find the magnitude of \( \vec{b_1} \times \vec{b_2} \):
\( |\vec{b_1} \times \vec{b_2}| = \sqrt{6^2 + 0^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \)
The dot product of \( \vec{a_2} - \vec{a_1} \) and \( \vec{b_1} \times \vec{b_2} \) is:
\( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0)(6) + (1)(0) + (-2)(-3) = 6 \)
The shortest distance \( d \) is:
\( d = \frac{|6|}{3\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \) units.
In simple words: Rearrange the lines into standard vector form. Compute the connecting vector and the cross product of the direction vectors, then use the shortest distance formula.

Exam Tip: First step is to regroup the coefficients of \( \lambda \) and \( \mu \) to separate the fixed point \( \vec{a} \) and the direction vector \( \vec{b} \).

Question 50. A variable plane is at a constant distance 3p from the origin and meets the coordinate axes in A, B and C. If the centroid of \( \Delta ABC \) is \( (\alpha, \beta, \gamma) \), then show that \( \alpha^{-2} + \beta^{-2} + \gamma^{-2} = p^{-2} \).
Answer: Let the equation of the variable plane in intercept form be:
\( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \)
The plane meets the coordinate axes at \( A(a, 0, 0) \), \( B(0, b, 0) \), and \( C(0, 0, c) \).
The perpendicular distance from the origin \( (0, 0, 0) \) to this plane is given as \( 3p \):
\( \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 3p \)
\( \implies \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9p^2} \quad \text{--- (i)} \)
Since \( (\alpha, \beta, \gamma) \) is the centroid of \( \Delta ABC \), we have:
\( \alpha = \frac{a + 0 + 0}{3} \implies a = 3\alpha \)
\( \beta = \frac{0 + b + 0}{3} \implies b = 3\beta \)
\( \gamma = \frac{0 + 0 + c}{3} \implies c = 3\gamma \)
Substituting these values of \( a \), \( b \), and \( c \) into equation (i):
\( \frac{1}{(3\alpha)^2} + \frac{1}{(3\beta)^2} + \frac{1}{(3\gamma)^2} = \frac{1}{9p^2} \)
\( \implies \frac{1}{9\alpha^2} + \frac{1}{9\beta^2} + \frac{1}{9\gamma^2} = \frac{1}{9p^2} \)
\( \implies \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{1}{p^2} \)
\( \implies \alpha^{-2} + \beta^{-2} + \gamma^{-2} = p^{-2} \)
Hence proved.
In simple words: Represent the plane using intercepts on each axis. Express the intercepts in terms of the centroid coordinates and substitute them into the distance-from-origin formula to prove the relation.

Exam Tip: Clearly write down the formulas for both the distance of a plane from the origin and the coordinates of the centroid of a triangle to earn step-wise marking.

Question 51. A vector \( \vec{n} \) of magnitude 8 units is inclined to x-axis at \( 45^\circ \), y-axis at \( 60^\circ \) and an acute angle with z-axis. If a plane passes through a point \( (\sqrt{2}, -1, 1) \) and is normal to \( \vec{n} \), find its equation in vector form.
Answer: Let the direction cosines of the vector \( \vec{n} \) be \( l \), \( m \), and \( n \).
Given:
\( l = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
\( m = \cos 60^\circ = \frac{1}{2} \)
Using the relation \( l^2 + m^2 + n^2 = 1 \):
\( \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + n^2 = 1 \)
\( \implies \frac{1}{2} + \frac{1}{4} + n^2 = 1 \)
\( \implies n^2 = 1 - \frac{3}{4} = \frac{1}{4} \implies n = \pm \frac{1}{2} \)
Since \( \vec{n} \) makes an acute angle with the z-axis, \( n = \cos \gamma > 0 \), so \( n = \frac{1}{2} \).
The unit normal vector \( \hat{n} \) is:
\( \hat{n} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} \)
Since \( \vec{n} \) has a magnitude of 8 units, the normal vector is:
\( \vec{n} = 8\hat{n} = 4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k} \)
The vector equation of a plane passing through a point \( \vec{a} = \sqrt{2}\hat{i} - \hat{j} + \hat{k} \) is:
\( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \)
\( \implies \vec{r} \cdot (4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k}) = (\sqrt{2}\hat{i} - \hat{j} + \hat{k}) \cdot (4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k}) \)
\( \implies \vec{r} \cdot (4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k}) = 4(2) - 4 + 4 = 8 \)
Dividing by 4:
\( \implies \vec{r} \cdot (\sqrt{2}\hat{i} + \hat{j} + \hat{k}) = 2 \).
In simple words: Find the missing third direction cosine using the sum-of-squares relation. Compute the normal vector from the direction cosines and magnitude, then write the vector equation of the plane.

Exam Tip: Be sure to select the positive root for \( n \) because the angle with the z-axis is specified as acute, which means \( \cos \gamma > 0 \).

Question 52. Find the foot of perpendicular from the point \( 2\hat{i} - \hat{j} + 5\hat{k} \) on the line \( \vec{r} = (11\hat{i} - 2\hat{j} - 8\hat{k}) + \lambda(10\hat{i} - 4\hat{j} - 11\hat{k}) \). Also find the length of the perpendicular.
Answer: Let the given point be \( P(2, -1, 5) \).
Any point \( Q \) on the line is given by:
\( Q = (10\lambda + 11, \; -4\lambda - 2, \; -11\lambda - 8) \)
The direction ratios of the line segment \( PQ \) are:
\( ( (10\lambda + 11) - 2, \; (-4\lambda - 2) - (-1), \; (-11\lambda - 8) - 5 ) \)
\( = ( 10\lambda + 9, \; -4\lambda - 1, \; -11\lambda - 13 ) \)
Since \( PQ \) is perpendicular to the line, which has a direction vector \( \vec{b} = 10\hat{i} - 4\hat{j} - 11\hat{k} \), their dot product must be zero:
\( 10(10\lambda + 9) - 4(-4\lambda - 1) - 11(-11\lambda - 13) = 0 \)
\( \implies 100\lambda + 90 + 16\lambda + 4 + 121\lambda + 143 = 0 \)
\( \implies 237\lambda + 237 = 0 \implies \lambda = -1 \)
Substituting \( \lambda = -1 \) into the coordinates of \( Q \):
\( Q = (10(-1) + 11, \; -4(-1) - 2, \; -11(-1) - 8) = (1, 2, 3) \)
So, the coordinates of the foot of the perpendicular are \( (1, 2, 3) \).
The length of the perpendicular \( PQ \) is:
\( PQ = \sqrt{(1 - 2)^2 + (2 - (-1))^2 + (3 - 5)^2} = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \) units.
In simple words: Represent a general point on the line. Use the perpendicular condition (dot product equals zero) to solve for the parameter, find the foot coordinates, and calculate the length using the distance formula.

Exam Tip: Clearly show the dot product equation used to find \( \lambda \) to secure partial markings even if an arithmetic error occurs later in the distance formula.

Question 53. 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 set up a coordinate system with one corner of a cube of side \( a \) at the origin \( O(0,0,0) \).
The four diagonals of the cube are \( OG \), \( AL \), \( BM \), and \( CN \), with direction ratios proportional to:
\( (1, 1, 1) \), \( (-1, 1, 1) \), \( (1, -1, 1) \), and \( (1, 1, -1) \) respectively.
Let the given line have direction cosines \( (l, m, n) \), where \( l^2 + m^2 + n^2 = 1 \).
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:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta \)
\( = \frac{1}{3} \left[ (l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2 \right] \)
Expanding the squared terms, all the cross-product terms (like \( 2lm \)) cancel out:
\( = \frac{1}{3} \left[ 4(l^2 + m^2 + n^2) \right] \)
Since \( l^2 + m^2 + n^2 = 1 \):
\( = \frac{4}{3} \).
Hence proved.
In simple words: Find the direction cosines of the four diagonals of a cube. Calculate the cosine of the angle between the line and each diagonal, square them, and add them up to get the final constant value.

Exam Tip: This is a standard NCERT exemplar problem. Write down the direction cosines of the diagonals clearly at the start of your solution to establish a strong proof.

Question 54. Find the equation of the plane passing through the intersection of planes 2x + 3y - z = - 1 and x + y - 2z + 3 = 0 and perpendicular to the plane 3x - y - 2z = 4. Also find the inclination of this plane with xy-plane.
Answer: Any plane passing through the line of intersection of the planes \( 2x + 3y - z + 1 = 0 \) and \( x + y - 2z + 3 = 0 \) is:
\( (2x + 3y - z + 1) + \lambda(x + y - 2z + 3) = 0 \)
\( \implies (2 + \lambda)x + (3 + \lambda)y + (-1 - 2\lambda)z + (1 + 3\lambda) = 0 \quad \text{--- (i)} \)
The normal vector to this plane is \( \vec{n_1} = (2+\lambda)\hat{i} + (3+\lambda)\hat{j} + (-1-2\lambda)\hat{k} \).
This plane is perpendicular to the plane \( 3x - y - 2z = 4 \) (normal vector \( \vec{n_2} = 3\hat{i} - \hat{j} - 2\hat{k} \)):
\( 3(2+\lambda) - 1(3+\lambda) - 2(-1-2\lambda) = 0 \)
\( \implies 6 + 3\lambda - 3 - \lambda + 2 + 4\lambda = 0 \)
\( \implies 5 + 6\lambda = 0 \implies \lambda = -\frac{5}{6} \)
Substituting \( \lambda = -\frac{5}{6} \) back into equation (i):
\( \left(2 - \frac{5}{6}\right)x + \left(3 - \frac{5}{6}\right)y + \left(-1 - 2\left(-\frac{5}{6}\right)\right)z + \left(1 + 3\left(-\frac{5}{6}\right)\right) = 0 \)
\( \implies \frac{7}{6}x + \frac{13}{6}y + \frac{4}{6}z - \frac{9}{6} = 0 \implies 7x + 13y + 4z - 9 = 0 \)
The inclination \( \theta \) of this plane with the xy-plane (whose normal is \( \hat{k} \)) is given by:
\( \cos \theta = \frac{|\vec{n_1} \cdot \hat{k}|}{|\vec{n_1}| \cdot 1} = \frac{4}{\sqrt{7^2 + 13^2 + 4^2}} = \frac{4}{\sqrt{49 + 169 + 16}} = \frac{4}{\sqrt{234}} = \frac{4}{3\sqrt{26}} \)
\( \implies \theta = \cos^{-1}\left(\frac{4}{3\sqrt{26}}\right) \).
In simple words: Find the family of planes through the intersection. Use the perpendicularity condition with the third plane to find \( \lambda \), write the final plane equation, and calculate the angle with the xy-plane using their normal vectors.

Exam Tip: When finding the angle of a plane with the xy-plane, remember that the normal vector to the xy-plane is simply \( \hat{k} = (0, 0, 1) \).

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