CBSE Class 12 Mathematics Three Dimensional Geometry Assignment Set 05

Very Short Answer Questions

Question. If a line has direction ratios 2, -1, -2, then what are its direction cosines?
Answer: Here direction ratios of line are 2, -1, -2
\( \therefore \) Direction cosines of line are \( \frac{2}{\sqrt{2^2 + (-1)^2 + (-2)^2}}, \frac{-1}{\sqrt{2^2 + (-1)^2 + (-2)^2}}, \frac{-2}{\sqrt{2^2 + (-1)^2 + (-2)^2}} \)
i.e., \( \frac{2}{3}, \frac{-1}{3}, \frac{-2}{3} \)
Note: If \( a, b, c \) are the direction ratios of a line, the direction cosines are \( \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}} \)

Question. Find the co-ordinate of the point, where the line \( \frac{x+2}{1} = \frac{y-5}{3} = \frac{z+1}{5} \) cuts the yz-plane. 
Answer: Let the required point be \( (\alpha, \beta, \gamma) \) where given line cuts yz-plane.
\( \therefore \frac{\alpha+2}{1} = \frac{\beta-5}{3} = \frac{\gamma+1}{5} = k \) (say)
If \( \frac{\alpha+2}{1} = k \)

\( \Rightarrow \) \( \alpha = -2+k \), \( \beta = 5+3k \), \( \gamma = -1+5k \)
Since this point lies in yz-plane.
\( \therefore \alpha = 0 \)

\( \Rightarrow \) \( -2+k = 0 \)

\( \Rightarrow \) \( k = 2 \)
So, \( \alpha = 0, \beta = 11, \gamma = 9 \)
\( \therefore \) Required point is (0, 11, 9) where given line cuts yz-plane.

Question. Write the direction cosine of a line equally inclined to the three coordinate axes. 
Answer: Any line equally inclined to coordinate axes will have direction cosines \( l, l, l \)
\( \therefore l^2 + l^2 + l^2 = 1 \)
\( 3l^2 = 1 \)

\( \Rightarrow \) \( l = \pm \frac{1}{\sqrt{3}} \)
\( \therefore \) Direction cosines are \( +\frac{1}{\sqrt{3}}, +\frac{1}{\sqrt{3}}, +\frac{1}{\sqrt{3}} \) or \( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \)

Question. Write the distance of the following plane from the origin. 
\( 2x - y + 2z + 1 = 0 \)
Answer: We have given plane
\( 2x - y + 2z + 1 = 0 \)
Distance from origin = \( \left| \frac{(2 \times 0) - (1 \times 0) + (2 \times 0) + 1}{\sqrt{(2)^2 + (-1)^2 + (2)^2}} \right| = \left| \frac{1}{\sqrt{4 + 1 + 4}} \right| = \frac{1}{3} \)

Question. Find the acute angle 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: We have, \( \vec{n}_1 = \hat{i} - 2\hat{j} - 2\hat{k} \) and \( \vec{n}_2 = 3\hat{i} - 6\hat{j} + 2\hat{k} \)
Let \( \theta \) be the angle between the normals to the planes drawns from some common point.
We have, \( \cos \theta = \left| \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1| |\vec{n}_2|} \right| = \left| \frac{3 + 12 - 4}{\sqrt{9} \sqrt{49}} \right| = \left| \frac{11}{3 \times 7} \right| = \frac{11}{21} \)
\( \therefore \theta = \cos^{-1} \left( \frac{11}{21} \right) \).

Question. Write the direction cosines of a line parallel to z-axis. 
Answer: The angle made by a line parallel to z-axis with x, y and z-axis are \( 90^\circ, 90^\circ \) and \( 0^\circ \) respectively.
\( \therefore \) The direction cosines of the line are \( \cos 90^\circ, \cos 90^\circ, \cos 0^\circ \) i.e., 0, 0, 1.

Question. Write the cartesian equation of a plane, bisecting the line segment joining the points A(2, 3, 5) and B(4, 5, 7) at right angles. 
Answer: One point of required plane = mid point of given line segment.
\( = \left( \frac{2 + 4}{2}, \frac{3 + 5}{2}, \frac{5 + 7}{2} \right) = (3, 4, 6) \)
Also dr's of normal to the plane = \( (4 - 2), (5 - 3), (7 - 5) = (2, 2, 2) \)
Therefore, required equation of plane is
\( 2(x - 3) + 2(y - 4) + 2(z - 6) = 0 \)
\( 2x + 2y + 2z = 26 \) or \( x + y + z = 13 \)

Question. Write the vector equation of the plane, passing through the point \( (a, b, c) \) and parallel to the plane \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2 \). 
Answer: Since, the required plane is parallel to plane \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2 \).
\( \therefore \) Normal of required plane is normal of given plane.

\( \Rightarrow \) Normal of required plane = \( \hat{i} + \hat{j} + \hat{k} \)
\( \therefore \) Required vector equation of plane
\( \{ \vec{r} - (a\hat{i} + b\hat{j} + c\hat{k}) \} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0 \)

Question. Find the sum of the intercepts cut off by the plane \( 2x + y - z = 5 \), on the coordinate axes. 
Answer: Let \( a, b, c \) be the intercepts cut off by the plane
\( 2x + y - z = 5 \dots (i) \) on x, y and z-axis respectively.

\( \Rightarrow \) \( A(a, 0, 0), B(0, b, 0) \) and \( C(0, 0, c) \) satisfy the equation (i)
Hence, \( 2a + 0 - 0 = 5 \)

\( \Rightarrow \) \( a = \frac{5}{2} \)
and \( 2 \times 0 + b - 0 = 5 \)

\( \Rightarrow \) \( b = 5 \)
and \( 2 \times 0 + 0 - c = 5 \)

\( \Rightarrow \) \( c = -5 \)
\( \therefore a + b + c = \frac{5}{2} + 5 - 5 = \frac{5}{2} \)

Question. Write the coordinates of the point which is the reflection of the point \( (\alpha, \beta, \gamma) \) in the XZ-plane. 
Answer: The reflection of the point \( (\alpha, \beta, \gamma) \) in the XZ plane is \( (\alpha, -\beta, \gamma) \).

Question. Find the distance between the planes \( \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) - 4 = 0 \) and \( \vec{r} \cdot (6\hat{i} - 9\hat{j} + 18\hat{k}) + 30 = 0 \) 
Answer: Given two planes are
\( \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) - 4 = 0 \) and \( \vec{r} \cdot (6\hat{i} - 9\hat{j} + 18\hat{k}) + 30 = 0 \)
Given planes may be written in cartesian form as
\( 2x - 3y + 6z - 4 = 0 \dots (i) \)
\( 6x - 9y + 18z + 30 = 0 \dots (ii) \)
Let \( P(x_1, y_1, z_1) \) be a point on plane (i)
\( \therefore 2x_1 - 3y_1 + 6z_1 - 4 = 0 \)

\( \Rightarrow \) \( 2x_1 - 3y_1 + 6z_1 = 4 \dots (iii) \)
The length of the perpendicular from \( P(x_1, y_1, z_1) \) to plane (ii)
\( = \left| \frac{6x_1 - 9y_1 + 18z_1 + 30}{\sqrt{6^2 + (-9)^2 + 18^2}} \right| = \left| \frac{3(2x_1 - 3y_1 + 6z_1) + 30}{\sqrt{36 + 81 + 324}} \right| \)
\( = \left| \frac{3 \times 4 + 30}{\sqrt{441}} \right| = \left| \frac{42}{21} \right| = 2 \quad [\text{Using } (iii)] \)

Question. Write the equation of a plane which is at a distance of \( 5\sqrt{3} \) units from origin and the normal to which is equally inclined to coordinate axes. 
Answer: Obviously, a vector equally inclined to co-ordinate axes is given by \( \hat{i} + \hat{j} + \hat{k} \)
\( \therefore \) Unit vector equally inclined to co-ordinate axes = \( \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \)
Therefore, required equation of plane is
\( \vec{r} \cdot \left\{ \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \right\} = 5\sqrt{3} \)

\( \Rightarrow \) \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 15 \) or \( x + y + z = 15 \)

Question. If a line makes angles \( 90^\circ \) and \( 60^\circ \) respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Answer: Let the angle made by line with positive direction of z-axis be \( \theta \) then,
We know that
\( \cos^2 90^\circ + \cos^2 60^\circ + \cos^2 \theta = 1 \)

\( \Rightarrow \) \( 0 + \left( \frac{1}{2} \right)^2 + \cos^2 \theta = 1 \)

\( \Rightarrow \) \( \frac{1}{4} + \cos^2 \theta = 1 \)

\( \Rightarrow \) \( \cos^2 \theta = 1 - \frac{1}{4} \)

\( \Rightarrow \) \( \cos^2 \theta = \frac{3}{4} \)

\( \Rightarrow \) \( \cos \theta = \pm \frac{\sqrt{3}}{2} \)

\( \Rightarrow \) \( \theta = 60^\circ \) or \( \frac{\pi}{3} \) if \( \cos \theta = \frac{\sqrt{3}}{2} \) and \( \theta = 150^\circ \) or \( \frac{5\pi}{6} \) if \( \cos \theta = -\frac{\sqrt{3}}{2} \)

Question. Find the distance between the planes \( 2x - y + 2z = 5 \) and \( 5x - 2.5y + 5z = 20 \).
Answer: Let \( P(x_1, y_1, z_1) \) be any point on plane \( 2x - y + 2z = 5 \).

\( \Rightarrow \) \( 2x_1 - y_1 + 2z_1 = 5 \)
Now distance of point \( P(x_1, y_1, z_1) \) from plane \( 5x - 2.5y + 5z = 20 \) is given by
\( d = \left| \frac{5x_1 - 2.5y_1 + 5z_1 - 20}{\sqrt{5^2 + (2.5)^2 + (5)^2}} \right| = \left| \frac{2.5(2x_1 - y_1 + 2z_1 - 8)}{\sqrt{25 + 6.25 + 25}} \right| = \left| \frac{2.5(5 - 8)}{\sqrt{56.25}} \right| \)
\( = \frac{7.5}{7.5} = 1 \) unit

Question. Cartesian equation of a line AB is \( \frac{2x - 1}{2} = \frac{4 - y}{7} = \frac{z + 1}{2} \). Write the direction ratios of a line parallel to AB. 
Answer: We have equations of line
\( \frac{2x - 1}{2} = \frac{4 - y}{7} = \frac{z - (-1)}{2} \)

\( \Rightarrow \) \( \frac{x - \frac{1}{2}}{1} = \frac{y - 4}{-7} = \frac{z - (-1)}{2} \)
Direction ratios of given line are \( 1, -7, 2 \).
Hence, direction ratios of any parallel line are \( 1, -7, 2 \) or any multiples of ratios.

Question. Find the equation of a plane that cuts the coordinates axes at \( (a, 0, 0), (0, b, 0) \) and \( (0, 0, c) \). 
Answer: The equation of such plane is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \)

Question. Find the angle between the line \( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda (3\hat{i} - \hat{j} + 2\hat{k}) \) and the plane \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 3 \).
Answer: We have equation of line
\( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda (3\hat{i} - \hat{j} + 2\hat{k}) \)
\( \therefore \vec{b} = 3\hat{i} - \hat{j} + 2\hat{k} \)
Equation of plane \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 3 \)
\( \vec{n} = \hat{i} + \hat{j} + \hat{k} \)
Let \( \theta \) be the required angle
\( \therefore \cos \theta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} \right| = \left| \frac{3 - 1 + 2}{\sqrt{9 + 1 + 4} \sqrt{1 + 1 + 1}} \right| = \left| \frac{4}{\sqrt{14} \sqrt{3}} \right| = \frac{4}{\sqrt{42}} \)

\( \Rightarrow \) \( \theta = \cos^{-1} \left( \frac{4}{\sqrt{42}} \right) \).

Short Answer Questions

Question. Write the vector equation of the following line.
\( \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{6 - z}{2} \)
Answer: Cartesian form of the line is given as
\( \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{-2} \)
The standard form of line's equation
\( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \)
We get by comparing that the given line passes through the point \( (x_1, y_1, z_1) \) i.e., \( (5, -4, 6) \) and direction ratios are \( (a, b, c) \) i.e., \( (3, 7, -2) \).
Now, we can write vector equation of line as
\( \vec{a} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} - 2\hat{k}) \)

Question. If the x-coordinate of a point P on the join of Q(2, 2, 1) and R(5, 1, -2) is 4, then find its z-coordinate.
Answer: Let P divides QR in the ratio \( \lambda : 1 \)
Then coordinates of P are \( \left( \frac{5\lambda + 2}{\lambda + 1}, \frac{\lambda + 2}{\lambda + 1}, \frac{-2\lambda + 1}{\lambda + 1} \right) \)
It is given that x-coordinate of P is 4.
\( \therefore \frac{5\lambda + 2}{\lambda + 1} = 4 \)

\( \Rightarrow \) \( 5\lambda + 2 = 4\lambda + 4 \)

\( \Rightarrow \) \( \lambda = 2 \)
So, z-coordinate of P = \( \frac{-2\lambda + 1}{\lambda + 1} = \frac{-4 + 1}{2 + 1} = -1 \).

Question. Show that the points \( (\hat{i} - \hat{j} + 3\hat{k}) \) and \( 3(\hat{i} + \hat{j} + \hat{k}) \) are equidistant from the plane \( \vec{r} \cdot (5\hat{i} + 2\hat{j} - 7\hat{k}) + 9 = 0 \) and lies on opposite side of it. 
Answer: To show that these given points \( (\hat{i} - \hat{j} + 3\hat{k}) \) and \( 3(\hat{i} + \hat{j} + \hat{k}) \) are equidistant from the plane \( \vec{r} \cdot (5\hat{i} + 2\hat{j} - 7\hat{k}) + 9 = 0 \), we first find out the mid-point of the points which is \( 2\hat{i} + \hat{j} + 3\hat{k} \).
On substituting \( \vec{r} \) by the mid-point in plane, we get
LHS = \( (2\hat{i} + \hat{j} + 3\hat{k}) \cdot (5\hat{i} + 2\hat{j} - 7\hat{k}) + 9 = 10 + 2 - 21 + 9 = 0 \) = RHS
Hence, the two points lie on opposite sides of the plane are equidistant from the plane.

Question. If the plane \( ax + by = 0 \) is rotated about its line of intersection with the plane \( z = 0 \) through an angle \( \alpha \), then prove that the equation of the plane in its new position is \( ax + by \pm (\sqrt{a^2 + b^2} \tan \alpha)z = 0 \). 
Answer: Given, planes are \( ax + by = 0 \dots (i) \)
and \( z = 0 \dots (ii) \)
Therefore, the equation of any plane passing through the line of intersection of planes (i) and (ii) may be taken as \( ax + by + k = 0 \). ... (iii)
Then, direction cosines of a normal to the plane (iii) are
\( \frac{a}{\sqrt{a^2 + b^2 + k^2}}, \frac{b}{\sqrt{a^2 + b^2 + k^2}}, \frac{k}{\sqrt{a^2 + b^2 + k^2}} \) and direction cosines of the normal to the plane (i) are \( \frac{a}{\sqrt{a^2 + b^2}}, \frac{b}{\sqrt{a^2 + b^2}}, 0 \).
Since, the angle between the planes (i) and (ii) is \( \alpha \),
\( \therefore \cos \alpha = \frac{a \cdot a + b \cdot b + k \cdot 0}{\sqrt{a^2 + b^2 + k^2} \sqrt{a^2 + b^2}} = \frac{a^2 + b^2}{\sqrt{a^2 + b^2 + k^2} \sqrt{a^2 + b^2}} \)

\( \Rightarrow \) \( k^2 \cos^2 \alpha = a^2 (1 - \cos^2 \alpha) + b^2 (1 - \cos^2 \alpha) \)

\( \Rightarrow \) \( k^2 = \frac{(a^2 + b^2)\sin^2 \alpha}{\cos^2 \alpha} \)

\( \Rightarrow \) \( k = \pm \sqrt{a^2 + b^2} \tan \alpha \)
On putting this value in plane (iii), we get the equation of the plane as
\( ax + by + z\sqrt{a^2 + b^2} \tan \alpha = 0 \)

Long Answer Questions

Question. Find the shortest distance between the lines whose vector equations are:
\( \vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k}) \). 
Answer: Comparing the given equations with equations
\( \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \)
We get \( \vec{a}_1 = \hat{i} + \hat{j}, \vec{b}_1 = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{a}_2 = 2\hat{i} + \hat{j} - \hat{k}, \vec{b}_2 = 3\hat{i} - 5\hat{j} + 2\hat{k} \)
Therefore, \( \vec{a}_2 - \vec{a}_1 = (\hat{i} - \hat{k}) \) and
\( \vec{b}_1 \times \vec{b}_2 = (2\hat{i} - \hat{j} + \hat{k}) \times (3\hat{i} - 5\hat{j} + 2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = 3\hat{i} - \hat{j} - 7\hat{k} \)
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{9 + 1 + 49} = \sqrt{59} \)
Hence, the shortest distance between the given lines is given by
\( d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{3 - 0 + 7}{\sqrt{59}} \right| = \frac{10}{\sqrt{59}} \) units.

Question. Find the distance between the lines \( l_1 \) and \( l_2 \) given by
\( l_1: \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \); \( l_2: \vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(4\hat{i} + 6\hat{j} + 12\hat{k}) \) 
Answer: Given lines are
\( l_1: \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \)
\( l_2: \vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(4\hat{i} + 6\hat{j} + 12\hat{k}) \)
After observation, we get \( l_1 \parallel l_2 \)
Therefore, it is sufficient to find the perpendicular distance of a point of line \( l_1 \) to line \( l_2 \).
The coordinate of a point of \( l_1 \) is \( P(1, 2, -4) \)
Also the cartesian form of line \( l_2 \) is
\( \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12} \dots (i) \)
Let \( Q(\alpha, \beta, \gamma) \) be foot of perpendicular drawn from P to line \( l_2 \)
\( \therefore Q(\alpha, \beta, \gamma) \) lie on line \( l_2 \)
\( \therefore \frac{\alpha - 3}{4} = \frac{\beta - 3}{6} = \frac{\gamma + 5}{12} = \lambda \) (say)

\( \Rightarrow \) \( \alpha = 4\lambda + 3, \beta = 6\lambda + 3, \gamma = 12\lambda - 5 \)
Again, \( \because \vec{PQ} \) is perpendicular to line \( l_2 \).

\( \Rightarrow \) \( \vec{PQ} \cdot \vec{b} = 0 \), where \( \vec{b} \) is parallel vector of \( l_2 \)

\( \Rightarrow \) \( (\alpha - 1) . 4 + (\beta - 2) . 6 + (\gamma + 4) . 12 = 0 \quad [\because \vec{PQ} = (\alpha - 1)\hat{i} + (\beta - 2)\hat{j} + (\gamma + 4)\hat{k}] \)

\( \Rightarrow \) \( 4\alpha - 4 + 6\beta - 12 + 12\gamma + 48 = 0 \)

\( \Rightarrow \) \( 4\alpha + 6\beta + 12\gamma + 32 = 0 \)

\( \Rightarrow \) \( 4(4\lambda + 3) + 6(6\lambda + 3) + 12(12\lambda - 5) + 32 = 0 \)

\( \Rightarrow \) \( 16\lambda + 12 + 36\lambda + 18 + 144\lambda - 60 + 32 = 0 \)

\( \Rightarrow \) \( 196\lambda + 2 = 0 \)

\( \Rightarrow \) \( \lambda = \frac{-2}{196} = \frac{-1}{98} \)
Coordinate of \( Q = \left( 4 \times \left( -\frac{1}{98} \right) + 3, 6 \times \left( -\frac{1}{98} \right) + 3, 12 \times \left( -\frac{1}{98} \right) - 5 \right) \)
\( = \left( -\frac{2}{49} + 3, -\frac{3}{49} + 3, -\frac{6}{49} - 5 \right) = \left( \frac{145}{49}, \frac{144}{49}, -\frac{251}{49} \right) \)
Therefore required perpendicular distance is
\( \sqrt{\left(\frac{145}{49} - 1\right)^2 + \left(\frac{144}{49} - 2\right)^2 + \left(-\frac{251}{49} + 4\right)^2} = \sqrt{\left(\frac{96}{49}\right)^2 + \left(\frac{46}{49}\right)^2 + \left(-\frac{55}{49}\right)^2} \)
\( = \sqrt{\frac{96^2 + 46^2 + 55^2}{49^2}} = \sqrt{\frac{9216 + 2116 + 3025}{49^2}} \)
\( = \frac{\sqrt{14357}}{49} = \frac{7\sqrt{293}}{49} = \frac{\sqrt{293}}{7} \) units

Question. Find the coordinates of the point where the line through the points \( (3, -4, -5) \) and \( (2, -3, 1) \) crosses the plane \( 2x + y + z = 7 \). 
Answer: The equation of line passing through the points \( (3, -4, -5) \) and \( (2, -3, 1) \) is
\( \frac{x - 3}{2 - 3} = \frac{y + 4}{-3 + 4} = \frac{z + 5}{1 + 5} \)

\( \Rightarrow \) \( \frac{x - 3}{-1} = \frac{y + 4}{1} = \frac{z + 5}{6} \dots (i) \)
Let the line (i) crosses at point \( P(\alpha, \beta, \gamma) \) to the plane \( 2x + y + z = 7 \dots (ii) \)
\( \because P \) lies on line (i), therefore \( (\alpha, \beta, \gamma) \) satisfy equation (i)
\( \therefore \frac{\alpha - 3}{-1} = \frac{\beta + 4}{1} = \frac{\gamma + 5}{6} = \lambda \) (say)
\( \alpha = -\lambda + 3; \beta = \lambda - 4 \) and \( \gamma = 6\lambda - 5 \)
Also \( P(\alpha, \beta, \gamma) \) lie on plane (ii)
\( \therefore 2\alpha + \beta + \gamma = 7 \)

\( \Rightarrow \) \( 2(-\lambda + 3) + (\lambda - 4) + (6\lambda - 5) = 7 \)

\( \Rightarrow \) \( -2\lambda + 6 + \lambda - 4 + 6\lambda - 5 = 7 \)

\( \Rightarrow \) \( 5\lambda = 10 \)

\( \Rightarrow \) \( \lambda = 2 \)
Hence, the coordinate of required point P is \( (-2 + 3, 2 - 4, 6 \times 2 - 5) \) i.e., \( (1, -2, 7) \)

Question. A line passes through \( (2, -1, 3) \) and is perpendicular to the lines \( \vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \) and \( \vec{r} = (2\hat{i} - \hat{j} - 3\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \). Obtain its equation in vector and cartesian form. 
Answer: Let \( \vec{b} \) be parallel vector of required line.

\( \Rightarrow \) \( \vec{b} \) is perpendicular to both given line.

\( \Rightarrow \) \( \vec{b} = (2\hat{i} - 2\hat{j} + \hat{k}) \times (\hat{i} + 2\hat{j} + 2\hat{k}) \)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{vmatrix} = (-4 - 2)\hat{i} - (4 - 1)\hat{j} + (4 + 2)\hat{k} = -6\hat{i} - 3\hat{j} + 6\hat{k} \).
Hence, the equation of line in vector form is
\( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(-6\hat{i} - 3\hat{j} + 6\hat{k}) \quad \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) - 3\lambda(2\hat{i} + \hat{j} - 2\hat{k}) \)
\( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \mu(2\hat{i} + \hat{j} - 2\hat{k}) \quad [\mu = -3\lambda] \)
Equation in cartesian form is
\( \frac{x - 2}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} \)

Question. Find the shortest distance between the following lines :
\( \frac{x - 3}{1} = \frac{y - 5}{-2} = \frac{z - 7}{1} \) and \( \frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1} \)
Answer: Let \( \frac{x - 3}{1} = \frac{y - 5}{-2} = \frac{z - 7}{1} = \lambda \) and \( \frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1} = k \)
Now, let's take a point on first line as
\( A(\lambda + 3, -2\lambda + 5, \lambda + 7) \) and let
\( B(7k - 1, -6k - 1, k - 1) \) be point on the second line
The direction ratio of the line AB
\( 7k - \lambda - 4, -6k + 2\lambda - 6, k - \lambda - 8 \)
Now, as AB is the shortest distance between line 1 and line 2 so,
\( (7k - \lambda - 4) \times 1 + (-6k + 2\lambda - 6) \times (-2) + (k - \lambda - 8) \times 1 = 0 \dots (i) \)
and \( (7k - \lambda - 4) \times 7 + (-6k + 2\lambda - 6) \times (-6) + (k - \lambda - 8) \times 1 = 0 \dots (ii) \)
Solving equation (i) and (ii), we get
\( \lambda = 0 \) and \( k = 0 \)
\( \therefore A = (3, 5, 7) \) and \( B = (-1, -1, -1) \)
Hence, \( AB = \sqrt{(3 + 1)^2 + (5 + 1)^2 + (7 + 1)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} \text{ units} = 2\sqrt{29} \text{ units} \)

Question. Find the equation of planes passing through the intersection of the planes \( \vec{r} \cdot (2\hat{i} + 6\hat{j}) + 12 = 0 \) and \( \vec{r} \cdot (3\hat{i} - \hat{j} + 4\hat{k}) = 0 \) and are at a unit distance from the origin. 
Answer: We are given planes:
\( \vec{r} \cdot (2\hat{i} + 6\hat{j}) + 12 = 0 \dots (i) \)
\( \vec{r} \cdot (3\hat{i} - \hat{j} + 4\hat{k}) = 0 \dots (ii) \)
So equation of the required plane can be written as:
\( (\vec{r} \cdot (2\hat{i} + 6\hat{j}) + 12) + \lambda(\vec{r} \cdot (3\hat{i} - \hat{j} + 4\hat{k})) = 0 \)

\( \Rightarrow \) \( \vec{r} \cdot ((2 + 3\lambda)\hat{i} + (6 - \lambda)\hat{j} + 4\lambda\hat{k}) + 12 = 0 \dots (iii) \)
In cartesian form
\( (2 + 3\lambda)x + (6 - \lambda)y + 4\lambda z + 12 = 0 \dots (iv) \)
Since direction ratios of the normal to the plane are \( (2 + 3\lambda), (6 - \lambda), 4\lambda \); the direction cosines of it are:
\( \frac{2 + 3\lambda}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}}, \frac{6 - \lambda}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}}, \frac{4\lambda}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}} \)
So the distance of the plane from the origin is \( \frac{12}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}} \)
We are given that distance from origin is unity
\( \therefore \frac{12}{\sqrt{(2 + 3\lambda)^2 + (6 - \lambda)^2 + (4\lambda)^2}} = 1 \)

\( \Rightarrow \) \( \frac{144}{4 + 9\lambda^2 + 12\lambda + 36 - 12\lambda + \lambda^2 + 16\lambda^2} = 1 \)

\( \Rightarrow \) \( 144 = 26\lambda^2 + 40 \) (Squaring both sides)

\( \Rightarrow \) \( 26\lambda^2 = 104 \)

\( \Rightarrow \) \( \lambda^2 = \frac{104}{26} = 4 \)

\( \Rightarrow \) \( \lambda = \pm 2 \)
\( \therefore \) Required equation of the plane is \( 8x + 4y + 8z + 12 = 0 \).
In vector form
\( \vec{r} \cdot (8\hat{i} + 4\hat{j} + 8\hat{k}) + 12 = 0 \)

Question. Find the vector equation of the plane determined by the points A(3, -1, 2), B(5, 2, 4) and C(-1, -1, 6). Hence, find the distance of the plane, thus obtained from the origin. 
Answer: Required equation of plane is given by:
\( \begin{vmatrix} x - 3 & y + 1 & z - 2 \\ 2 & 1 & 2 \\ -4 & 0 & 4 \end{vmatrix} = 0 \)
\( = (x - 3) 4 - (y + 1) 16 + (z - 2) 4 = 0 \)

\( \Rightarrow \) \( 4x - 16y + 4z - 12 - 16 - 8 = 0 \)

\( \Rightarrow \) \( 4x - 16y + 4z - 36 = 0 \)

\( \Rightarrow \) \( x - 4y + z - 9 = 0 \)
Vector form:
\( \vec{r} \cdot (\hat{i} - 4\hat{j} + \hat{k}) = 9 \).

Question. Find the vector and cartesian equations of the line passing through the point \( (2, 1, 3) \) and perpendicular to the lines \( \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \) and \( \frac{x}{-3} = \frac{y}{2} = \frac{z}{5} \). 
Answer: Let the cartesian equation of the line passing through \( (2, 1, 3) \) be
\( \frac{x - 2}{a} = \frac{y - 1}{b} = \frac{z - 3}{c} \dots (i) \)
Since, line (i) is perpendicular to given line
\( \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \dots (ii) \)
and \( \frac{x}{-3} = \frac{y}{2} = \frac{z}{5} \dots (iii) \)
\( \therefore a + 2b + 3c = 0 \dots (iv) \)
\( -3a + 2b + 5c = 0 \dots (v) \)
From equation (iv) and (v), we get
\( \frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 + 6} \)

\( \Rightarrow \) \( \frac{a}{4} = \frac{b}{-14} = \frac{c}{8} = \lambda \) (say)

\( \Rightarrow \) \( a = 4\lambda, b = -14\lambda, c = 8\lambda \)
Putting the value of \( a, b \) and \( c \) in (i), we get
\( \frac{x - 2}{4\lambda} = \frac{y - 1}{-14\lambda} = \frac{z - 3}{8\lambda} \)

\( \Rightarrow \) \( \frac{x - 2}{4} = \frac{y - 1}{-14} = \frac{z - 3}{8} \)

\( \Rightarrow \) \( \frac{x - 2}{2} = \frac{y - 1}{-7} = \frac{z - 3}{4} \), which is the cartesian form.
The vector form is \( \vec{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda (2\hat{i} - 7\hat{j} + 4\hat{k}) \).

Question. Find the equation of the perpendicular drawn from the point \( (1, -2, 3) \) to the plane \( 2x - 3y + 4z + 9 = 0 \). Also, find the coordinates of the foot of the perpendicular.
Answer: Let the foot of the perpendicular on the plane be A.
PA perpendicular to the plane \( 2x - 3y + 4z + 9 = 0 \).
dr's of PA = \( 2, -3, 4 \)
Equation of PA can be written as
\( \frac{x - 1}{2} = \frac{y + 2}{-3} = \frac{z - 3}{4} = \lambda \)
General points of line PA = \( (2\lambda + 1, -3\lambda - 2, 4\lambda + 3) \)
The point is on the plane hence \( 2(2\lambda + 1) - 3(-3\lambda - 2) + 4(4\lambda + 3) + 9 = 0 \).

\( \Rightarrow \) \( 29\lambda + 29 = 0 \) or \( \lambda = -1 \)
\( \therefore \) Coordinates of foot of perpendicular are \( (-1, 1, -1) \).

Question. Find the cartesian equation of the plane passing through the points \( A (0, 0, 0) \) and \( B (3, -1, 2) \) and parallel to the line \( \frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7} \).
Answer: Equation of plane is given by
\( a (x - x_1) + b (y - y_1) + c (z - z_1) = 0 \)
Given plane passes through \( (0, 0, 0) \)
\( \therefore a (x - 0) + b (y - 0) + c (z - 0) = 0 \dots (i) \)
Plane (i), passes through \( (3, -1, 2) \)
\( \therefore 3a - b + 2c = 0 \dots (ii) \)
Also, plane (i) is parallel to the line \( \frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7} \).
\( \Rightarrow \) \( a - 4b + 7c = 0 \dots (iii) \)
Eliminating a, b, c from equations (i), (ii) and (iii), we get
\( \begin{vmatrix} x & y & z \\ 3 & -1 & 2 \\ 1 & -4 & 7 \end{vmatrix} = 0 \)

\( \Rightarrow \) \( x \begin{vmatrix} -1 & 2 \\ -4 & 7 \end{vmatrix} - y \begin{vmatrix} 3 & 2 \\ 1 & 7 \end{vmatrix} + z \begin{vmatrix} 3 & -1 \\ 1 & -4 \end{vmatrix} = 0 \)

\( \Rightarrow \) \( x (-7 + 8) - y (21 - 2) + z (-12 + 1) = 0 \)

\( \Rightarrow \) \( x - 19y - 11z = 0 \), which is the required equation.

Question. Find the distance of the point \( P (6, 5, 9) \) from the plane determined by the points \( A (3, -1, 2) \), \( B (5, 2, 4) \) and \( C (-1, -1, 6) \). 
Answer: Plane determined by the points \( A (3, -1, 2) \), \( B (5, 2, 4) \) and \( C (-1, -1, 6) \) is
\( \begin{vmatrix} x - 3 & y + 1 & z - 2 \\ 5 - 3 & 2 + 1 & 4 - 2 \\ -1 - 3 & -1 + 1 & 6 - 2 \end{vmatrix} = 0 \)

\( \Rightarrow \) \( \begin{vmatrix} x - 3 & y + 1 & z - 2 \\ 2 & 3 & 2 \\ -4 & 0 & 4 \end{vmatrix} = 0 \)

\( \Rightarrow \) \( (x - 3)\begin{vmatrix} 3 & 2 \\ 0 & 4 \end{vmatrix} - (y + 1)\begin{vmatrix} 2 & 2 \\ -4 & 4 \end{vmatrix} + (z - 2)\begin{vmatrix} 2 & 3 \\ -4 & 0 \end{vmatrix} = 0 \)

\( \Rightarrow \) \( 12x - 36 - 16y - 16 + 12z - 24 = 0 \)

\( \Rightarrow \) \( 3x - 4y + 3z - 19 = 0 \)
Distance of this plane from point \( P ( 6, 5, 9) \) is
\( \frac{|(3 \times 6) - (4 \times 5) + (3 \times 9) - 19|}{\sqrt{(3)^2 + (4)^2 + (3)^2}} = \left| \frac{18 - 20 + 27 - 19}{\sqrt{9 + 16 + 9}} \right| = \frac{6}{\sqrt{34}} \) units.

Question. 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. Also find their point of intersection. 
Answer: Given lines are
\( \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \dots (i) \)
\( \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} \dots (ii) \)
Let two lines (i) and (ii) intersect at a point \( P(\alpha, \beta, \gamma) \).

\( \Rightarrow \) \( (\alpha, \beta, \gamma) \) satisfy line (i)

\( \Rightarrow \) \( \frac{\alpha + 1}{3} = \frac{\beta + 3}{5} = \frac{\gamma + 5}{7} = \lambda \) (say)

\( \Rightarrow \) \( \alpha = 3\lambda - 1, \beta = 5\lambda - 3, \gamma = 7\lambda - 5 \dots (iii) \)
Again \( (\alpha, \beta, \gamma) \) also lie on (ii), we get
\( \frac{\alpha - 2}{1} = \frac{\beta - 4}{3} = \frac{\gamma - 6}{5} \)

\( \Rightarrow \) \( \frac{3\lambda - 1 - 2}{1} = \frac{5\lambda - 3 - 4}{3} = \frac{7\lambda - 5 - 6}{5} \)

\( \Rightarrow \) \( \frac{3\lambda - 3}{1} = \frac{5\lambda - 7}{3} = \frac{7\lambda - 11}{5} \)
I II III
From I and II
\( \frac{3\lambda - 3}{1} = \frac{5\lambda - 7}{3} \)

\( \Rightarrow \) \( 9\lambda - 9 = 5\lambda - 7 \)

\( \Rightarrow \) \( 4\lambda = 2 \)

\( \Rightarrow \) \( \lambda = \frac{1}{2} \)
From II and III
\( \frac{5\lambda - 7}{3} = \frac{7\lambda - 11}{5} \)

\( \Rightarrow \) \( 25\lambda - 35 = 21\lambda - 33 \)

\( \Rightarrow \) \( 4\lambda = 2 \)

\( \Rightarrow \) \( \lambda = \frac{1}{2} \)
Since, the value of \( \lambda \) in both the cases is same

\( \Rightarrow \) Both lines intersect each other at a point.
\( \therefore \) Intersecting point = \( (\alpha, \beta, \gamma) = \left( \frac{3}{2} - 1, \frac{5}{2} - 3, \frac{7}{2} - 5 \right) \) [From (iii)]
\( = \left( \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2} \right) \)

Question. Find the vector and cartesian equations of the line passing through the point (1, 2, -4) and perpendicular to the two lines \( \frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7} \) and \( \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5} \).
OR
Find the equation of a line passing through the point \( (1, 2, -4) \) and perpendicular to two lines \( \vec{r} = (8\hat{i} - 19\hat{j} + 10\hat{k}) + \lambda (3\hat{i} - 16\hat{j} + 7\hat{k}) \) and \( \vec{r} = (15\hat{i} + 29\hat{j} + 5\hat{k}) + \mu (3\hat{i} + 8\hat{j} - 5\hat{k}) \).
Answer: Let the cartesian equation of line passing through \( (1, 2, -4) \) be
\( \frac{x - 1}{a} = \frac{y - 2}{b} = \frac{z + 4}{c} \dots (i) \)
Given lines are
\( \frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7} \dots (ii) \)
\( \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5} \dots (iii) \)
Obviously parallel vectors \( \vec{b}_1, \vec{b}_2 \) and \( \vec{b}_3 \) of (i), (ii) and (iii) respectively are given as
\( \vec{b}_1 = a\hat{i} + b\hat{j} + c\hat{k} \); \( \vec{b}_2 = 3\hat{i} - 16\hat{j} + 7\hat{k} \); \( \vec{b}_3 = 3\hat{i} + 8\hat{j} - 5\hat{k} \)
According to question
(i) \( \perp \) (ii)
\( \Rightarrow \) \( \vec{b}_1 \perp \vec{b}_2 \)
\( \Rightarrow \) \( \vec{b}_1 \cdot \vec{b}_2 = 0 \)
(i) \( \perp \) (iii)
\( \Rightarrow \) \( \vec{b}_1 \perp \vec{b}_3 \)
\( \Rightarrow \) \( \vec{b}_1 \cdot \vec{b}_3 = 0 \)
Hence, \( 3a - 16b + 7c = 0 \dots (iv) \)
and \( 3a + 8b - 5c = 0 \dots (v) \)
From equation (iv) and (v), we get
\( \frac{a}{80 - 56} = \frac{b}{21 + 15} = \frac{c}{24 + 48} \)

\( \Rightarrow \) \( \frac{a}{24} = \frac{b}{36} = \frac{c}{72} \)

\( \Rightarrow \) \( \frac{a}{2} = \frac{b}{3} = \frac{c}{6} = \lambda \) (say)

\( \Rightarrow \) \( a = 2\lambda, b = 3\lambda, c = 6\lambda \)
Putting the value of \( a, b, c \) in (i), we get the required cartesian equation of line as
\( \frac{x - 1}{2\lambda} = \frac{y - 2}{3\lambda} = \frac{z + 4}{6\lambda} \)

\( \Rightarrow \) \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \)
Hence, vector equation is
\( \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \)

Question. A line passing through the point A with position vector \( \vec{a} = 4\hat{i} + 2\hat{j} + 2\hat{k} \) is parallel to the vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \). Find the length of the perpendicular drawn on this line from a point P with position vector \( \vec{r}_1 = \hat{i} + 2\hat{j} + 3\hat{k} \). 
Answer: The equation of line passing through the point A and parallel to \( \vec{b} \) is given in cartesian form as
\( \frac{x - 4}{2} = \frac{y - 2}{3} = \frac{z - 2}{6} \dots (i) \)
Let \( Q(\alpha, \beta, \gamma) \) be foot of perpendicular drawn from point P to the line (i).
Co-ordinate or point \( P \equiv (1, 2, 3) \quad [\because P.V. \text{ of } P \text{ is } \hat{i} + 2\hat{j} + 3\hat{k}] \)
Since, Q lie on line (i)
\( \frac{\alpha - 4}{2} = \frac{\beta - 2}{3} = \frac{\gamma - 2}{6} = \lambda \)

\( \Rightarrow \) \( \alpha = 2\lambda + 4, \beta = 3\lambda + 2, \gamma = 6\lambda + 2 \)
Now, \( \vec{PQ} = (\alpha - 1)\hat{i} + (\beta - 2)\hat{j} + (\gamma - 3)\hat{k} \)
Obviously, \( \vec{PQ} \perp \vec{b} \)
\( \therefore \vec{PQ} \cdot \vec{b} = 0 \)

\( \Rightarrow \) \( 2(\alpha - 1) + 3(\beta - 2) + 6(\gamma - 3) = 0 \)

\( \Rightarrow \) \( 2\alpha - 2 + 3\beta - 6 + 6\gamma - 18 = 0 \)

\( \Rightarrow \) \( 2\alpha + 3\beta + 6\gamma - 26 = 0 \)
Putting the value of \( \alpha, \beta, \gamma \); we get
\( 2(2\lambda + 4) + 3(3\lambda + 2) + 6(6\lambda + 2) - 26 = 0 \)

\( \Rightarrow \) \( 4\lambda + 8 + 9\lambda + 6 + 36\lambda + 12 - 26 = 0 \)

\( \Rightarrow \) \( 49\lambda = 0 \)

\( \Rightarrow \) \( \lambda = 0 \)
Hence, the co-ordinate of Q = (4, 2, 2)
\( \therefore \) Length of perpendicular PQ = \( \sqrt{(4 - 1)^2 + (2 - 2)^2 + (2 - 3)^2} \)
\( = \sqrt{9 + 0 + 1} = \sqrt{10} \) units.

Question. Find the coordinates of the point, where the line \( \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2} \) intersects the plane \( x - y + z - 5 = 0 \). Also find the angle between the line and the plane.
Answer: Let the given line
\( \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2} \dots (i) \)
intersect at point \( P(\alpha, \beta, \gamma) \) to the plane \( x - y + z - 5 = 0 \dots (ii) \)
\( \because P(\alpha, \beta, \gamma) \) lie on line (i)
\( \therefore \frac{\alpha - 2}{3} = \frac{\beta + 1}{4} = \frac{\gamma - 2}{2} = \lambda \) (say)
\( \alpha = 3\lambda + 2; \beta = 4\lambda - 1; \gamma = 2\lambda + 2 \)
Also, \( P(\alpha, \beta, \gamma) \) lies on plane (ii)
\( \therefore (3\lambda + 2) - (4\lambda - 1) + (2\lambda + 2) - 5 = 0 \)

\( \Rightarrow \) \( 3\lambda + 2 - 4\lambda + 1 + 2\lambda + 2 - 5 = 0 \)

\( \Rightarrow \) \( \lambda = 0 \)
\( \therefore \alpha = 2, \beta = -1, \gamma = 2 \)
Hence, co-ordinate of required point = (2, -1, 2)
Now, find angle between line (i) and plane (ii)
If \( \theta \) be the required angle, then
\( \sin \theta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} \right| = \left| \frac{1}{\sqrt{9 + 16 + 4} \cdot \sqrt{1^2 + (-1)^2 + 1^2}} \right| = \left| \frac{1}{\sqrt{29} \cdot \sqrt{3}} \right| \quad \left[ \because \vec{b} = 3\hat{i} + 4\hat{j} + 2\hat{k}, \vec{n} = \hat{i} - \hat{j} + \hat{k}, \therefore \vec{b} \cdot \vec{n} = 3 - 4 + 2 = 1 \right] \)
\( \sin \theta = \frac{1}{\sqrt{87}} \)

\( \Rightarrow \) \( \theta = \sin^{-1} \left( \frac{1}{\sqrt{87}} \right) \)

Question. Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane. 
Answer: Let \( P(\alpha, \beta, \gamma) \) be the point at which the given line crosses the XZ plane.
Now the equation of given line AB is
\( \frac{x - 3}{2} = \frac{y - 4}{-3} = \frac{z - 1}{5} \dots (i) \)
Since \( P(\alpha, \beta, \gamma) \) lies on line (i)
\( \therefore \frac{\alpha - 3}{2} = \frac{\beta - 4}{-3} = \frac{\gamma - 1}{5} = \lambda \) (say)

\( \Rightarrow \) \( \alpha = 2\lambda + 3; \beta = -3\lambda + 4 \) and \( \gamma = 5\lambda + 1 \)
Also \( P(\alpha, \beta, \gamma) \) lie on XZ plane, i.e., \( y = 0 \) (\( 0x + 1y + 0z = 0 \))
\( 0\alpha + 1 \cdot \beta + 0 \cdot \gamma = 0 \)

\( \Rightarrow \) \( \beta = 0 \)

\( \Rightarrow \) \( -3\lambda + 4 = 0 \)

\( \Rightarrow \) \( \lambda = \frac{4}{3} \)
Hence, the co-ordinates of required point P is
\( \alpha = 2 \times \frac{4}{3} + 3 = \frac{8}{3} + 3 = \frac{17}{3}; \beta = -3 \times \frac{4}{3} + 4 = 0; \gamma = 5 \times \frac{4}{3} + 1 = \frac{23}{3} \)
\( \therefore \) Co-ordinate of required point is \( \left( \frac{17}{3}, 0, \frac{23}{3} \right) \).
Let \( \theta \) be the angle made by line AB with XZ plane.
\( \therefore \sin \theta = \left| \frac{\vec{n} \cdot \vec{b}}{|\vec{n}| |\vec{b}|} \right| \)
Here \( \vec{n} = \hat{j} \) and \( \vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k} \)
\( |\vec{n}| = 1 \) and \( |\vec{b}| = \sqrt{4 + 9 + 25} = \sqrt{38} \)

\( \Rightarrow \) \( \sin \theta = \left| \frac{\hat{j} \cdot (2\hat{i} - 3\hat{j} + 5\hat{k})}{1 \cdot \sqrt{38}} \right| = \left| \frac{-3}{\sqrt{38}} \right| \)

\( \Rightarrow \) \( \sin \theta = \frac{3}{\sqrt{38}} \)

\( \Rightarrow \) \( \theta = \sin^{-1} \left( \frac{3}{\sqrt{38}} \right) \)

Question. Find the vector equation of the line passing through the point A(1, 2, -1) and parallel to the line \( 5x - 25 = 14 - 7y = 35z \).
Answer: Given line is
\( 5x - 25 = 14 - 7y = 35z \)

\( \Rightarrow \) \( \frac{x - 5}{\frac{1}{5}} = \frac{2 - y}{\frac{1}{7}} = \frac{z - 0}{\frac{1}{35}} \)

\( \Rightarrow \) \( \frac{x - 5}{\frac{1}{5}} = \frac{y - 2}{-\frac{1}{7}} = \frac{z - 0}{\frac{1}{35}} \)
\( = \frac{x - 5}{7} = \frac{y - 2}{-5} = \frac{z - 0}{1} \dots (i) \)
Hence, parallel vector of given line i.e., \( \vec{b} = 7\hat{i} - 5\hat{j} + \hat{k} \)
Since required line is parallel to given line (i)

\( \Rightarrow \) \( \vec{b} = 7\hat{i} - 5\hat{j} + \hat{k} \) will also be parallel vector of required line which passes through A(1, 2, -1).
Therefore, required vector equation of line is
\( \vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(7\hat{i} - 5\hat{j} + \hat{k}) \)

Question. Find the co-ordinates of the point where the line \( \vec{r} = (-\hat{i} - 2\hat{j} - 3\hat{k}) + \lambda(3\hat{i} + 4\hat{j} + 3\hat{k}) \) meets the plane which is perpendicular to the vector \( \vec{n} = \hat{i} + \hat{j} + 3\hat{k} \) and at a distance of \( \frac{4}{\sqrt{11}} \) from origin. 
Answer: We know that the equation of plane is
\( \vec{r} \cdot \hat{n} = d \); where \( \hat{n} \) is normal unit vector and d is perpendicular distance from origin.
Here, \( \hat{n} = \frac{\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{1^2 + 1^2 + 3^2}} = \frac{1}{\sqrt{11}}(\hat{i} + \hat{j} + 3\hat{k}) \) and \( d = \frac{4}{\sqrt{11}} \)
\( \therefore \) Equation of plane
\( \vec{r} \cdot \frac{1}{\sqrt{11}}(\hat{i} + \hat{j} + 3\hat{k}) = \frac{4}{\sqrt{11}} \)

\( \Rightarrow \) \( \vec{r} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 4 \)

\( \Rightarrow \) \( x + y + 3z = 4 \dots (i) \)
Equation of given line
\( \vec{r} = (-\hat{i} - 2\hat{j} - 3\hat{k}) + \lambda(3\hat{i} + 4\hat{j} + 3\hat{k}) \)
Its cartesian form is
\( \frac{x + 1}{3} = \frac{y + 2}{4} = \frac{z + 3}{3} \dots (ii) \)
Let \( Q(\alpha, \beta, \gamma) \) be the point of intersection of (i) & (ii)
\( \because Q \) lies on (ii)
\( \therefore \frac{\alpha + 1}{3} = \frac{\beta + 2}{4} = \frac{\gamma + 3}{3} = \lambda \)

\( \Rightarrow \) \( \alpha = 3\lambda - 1, \beta = 4\lambda - 2, \gamma = 3\lambda - 3 \)
Also, Q lies on (i)
\( \therefore \alpha + \beta + 3\gamma = 4 \)

\( \Rightarrow \) \( 3\lambda - 1 + 4\lambda - 2 + 9\lambda - 9 = 4 \)

\( \Rightarrow \) \( 16\lambda = 16 \)

\( \Rightarrow \) \( \lambda = 1 \)
\( \therefore \alpha = 2, \beta = 2, \gamma = 0 \)
\( \therefore \) Required point of intersection = (2, 2, 0)

Question. A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2} \). 
Answer: Let the given variable plane meets X, Y and Z axes at \( A(a, 0, 0), B(0, b, 0), C(0, 0, c) \).
Therefore the equation of given plane is given by
\( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \dots (i) \)
Let \( (\alpha, \beta, \gamma) \) be the coordinates of the centroid of triangle ABC. Then
\( \alpha = \frac{a + 0 + 0}{3} = \frac{a}{3} \)

\( \Rightarrow \) \( a = 3\alpha \); \( \beta = \frac{0 + b + 0}{3} = \frac{b}{3} \)

\( \Rightarrow \) \( b = 3\beta \)
\( \gamma = \frac{0 + 0 + c}{3} = \frac{c}{3} \)

\( \Rightarrow \) \( c = 3\gamma \)
\( \because \) 3p is the distance from origin to the plane (i)

\( \Rightarrow \) \( 3p = \frac{0 \cdot \frac{1}{a} + 0 \cdot \frac{1}{b} + 0 \cdot \frac{1}{c} - 1}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2 + \left(\frac{1}{c}\right)^2}} \)

\( \Rightarrow \) \( \sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2 + \left(\frac{1}{c}\right)^2} = \frac{1}{3p} \)
Squaring both sides, we have
\( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9p^2} \)

\( \Rightarrow \) \( \frac{1}{9\alpha^2} + \frac{1}{9\beta^2} + \frac{1}{9\gamma^2} = \frac{1}{9p^2} \) [Putting value of \( a = 3\alpha, b = 3\beta, c = 3\gamma \)]

\( \Rightarrow \) \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{1}{p^2} \)
Therefore, locus of \( (\alpha, \beta, \gamma) \) is \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2} \) Hence proved.

Question. Find the image P' of the point P having position vector \( \hat{i} + 3\hat{j} + 4\hat{k} \) in the plane \( \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) + 3 = 0 \). Hence find the length of PP'.
Answer: Let given point be P(1, 3, 4) and the equation of given plane in cartesian form be
\( 2x - y + z + 3 = 0 \dots (i) \)
Let \( R (x_1, y_1, z_1) \) be the foot of perpendicular and \( Q (\alpha, \beta, \gamma) \) be the image of P
Since, \( R(x_1, y_1, z_1) \) lie on plane (i)
\( 2x_1 - y_1 + z_1 + 3 = 0 \dots (ii) \)
Also, normal vector \( \vec{n} \) of plane (i) is \( \vec{n} = 2\hat{i} - \hat{j} + \hat{k} \)
and \( \vec{PR} = (x_1 - 1)\hat{i} + (y_1 - 3)\hat{j} + (z_1 - 4)\hat{k} \)
\( \therefore \vec{PR} \parallel \vec{n} \)

\( \Rightarrow \) \( \frac{x_1 - 1}{2} = \frac{y_1 - 3}{-1} = \frac{z_1 - 4}{1} = \lambda \)

\( \Rightarrow \) \( x_1 = 2\lambda + 1, y_1 = -\lambda + 3, z_1 = \lambda + 4 \)
Putting \( x_1, y_1, z_1 \) in (ii), we get

\( \Rightarrow \) \( 2(2\lambda + 1) - (-\lambda + 3) + (\lambda + 4) + 3 = 0 \)

\( \Rightarrow \) \( 4\lambda + 2 + \lambda - 3 + \lambda + 4 + 3 = 0 \)

\( \Rightarrow \) \( 6\lambda + 6 = 0 \)

\( \Rightarrow \) \( \lambda = -1 \)
\( \therefore R = (x_1, y_1, z_1) = (-1, 4, 3) \)
Since R is the mid point of PQ
\( \therefore -1 = \frac{\alpha + 1}{2} \)

\( \Rightarrow \) \( \alpha = -3 \)
\( 4 = \frac{\beta + 3}{2} \)

\( \Rightarrow \) \( \beta = 5 \)
\( 3 = \frac{\gamma + 4}{2} \)

\( \Rightarrow \) \( \gamma = 2 \)
Hence, image Q = (-3, 5, 2)
\( PP' = \sqrt{(-3 - 1)^2 + (5 - 3)^2 + (2 - 4)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \) units

Question. Find the equation 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 \( l_1, l_2 \) be given lines as
\( l_1: \frac{x + 2}{1} = \frac{y - 3}{2} = \frac{z + 1}{4} \); \( l_2: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \)
Let \( l \) be the required line, which passes through \( P(1, 1, 1) \) and intersect \( l_1 \) and \( l_2 \) at \( Q (\alpha_1, \beta_1, \gamma_1) \) and \( R (\alpha_2, \beta_2, \gamma_2) \) respectively.
Now, \( Q (\alpha_1, \beta_1, \gamma_1) \) lie on line \( l_1 \)
\( \therefore \frac{\alpha_1 + 2}{1} = \frac{\beta_1 - 3}{2} = \frac{\gamma_1 + 1}{4} = \lambda \) (say)
\( \alpha_1 = \lambda - 2, \beta_1 = 2\lambda + 3, \gamma_1 = 4\lambda - 1 \)
Similarly, \( R (\alpha_2, \beta_2, \gamma_2) \) lie on line \( l_2 \)
\( \frac{\alpha_2 - 1}{2} = \frac{\beta_2 - 2}{3} = \frac{\gamma_2 - 3}{4} = \mu \) (say)

\( \Rightarrow \) \( \alpha_2 = 2\mu + 1, \beta_2 = 3\mu + 2, \gamma_2 = 4\mu + 3 \)
\( \therefore \vec{PQ} = (\alpha_1 - 1)\hat{i} + (\beta_1 - 1)\hat{j} + (\gamma_1 - 1)\hat{k} \)
\( = (\lambda - 3)\hat{i} + (2\lambda + 2)\hat{j} + (4\lambda - 2)\hat{k} \)
Similarly, \( \vec{PR} = 2\mu\hat{i} + (3\mu + 1)\hat{j} + (4\mu + 2)\hat{k} \)
\( \because \vec{PQ} \parallel \vec{PR} \)

\( \Rightarrow \) \( \frac{\lambda - 3}{2\mu} = \frac{2\lambda + 2}{3\mu + 1} = \frac{4\lambda - 2}{4\mu + 2} = M \) (say)
Now, \( \frac{\lambda - 3}{2\mu} = M \)

\( \Rightarrow \) \( \lambda - 3 = 2M\mu \)

\( \Rightarrow \) \( M\mu = \frac{\lambda - 3}{2} \)
Also, \( \frac{2\lambda + 2}{3\mu + 1} = M \)

\( \Rightarrow \) \( 2\lambda + 2 = 3M\mu + M \)

\( \Rightarrow \) \( 2\lambda + 2 = \frac{3\lambda - 9}{2} + M \)

\( \Rightarrow \) \( 2\lambda + 2 - \frac{3\lambda - 9}{2} = M \)

\( \Rightarrow \) \( \frac{4\lambda + 4 - 3\lambda + 9}{2} = M \)

\( \Rightarrow \) \( \frac{\lambda + 13}{2} = M \)
Also, \( \frac{4\lambda - 2}{4\mu + 2} = M \)

\( \Rightarrow \) \( 4\lambda - 2 = 4M\mu + 2M \)

\( \Rightarrow \) \( 4\lambda - 2 = 4 \left( \frac{\lambda - 3}{2} \right) + 2M \)

\( \Rightarrow \) \( 4\lambda - 2 = 2\lambda - 6 + 2M \)

\( \Rightarrow \) \( 2\lambda + 4 = 2M \)

\( \Rightarrow \) \( \lambda + 2 = M \)

\( \Rightarrow \) \( \frac{\lambda + 13}{2} = \lambda + 2 \)

\( \Rightarrow \) \( \lambda + 13 = 2\lambda + 4 \)

\( \Rightarrow \) \( \lambda = 9 \)

\( \Rightarrow \) \( M = 11 \)

\( \Rightarrow \) \( \mu = \frac{3}{11} \)
\( \therefore \vec{PQ} = 6\hat{i} + 20\hat{j} + 34\hat{k} \)
Hence, equation of required line is \( \frac{x - 1}{6} = \frac{y - 1}{20} = \frac{z - 1}{34} \)

\( \Rightarrow \) \( \frac{x - 1}{3} = \frac{y - 1}{10} = \frac{z - 1}{17} \)