CBSE Class 12 Mathematics Three Dimensional Geometry VBQs Set A

BASIC CONCEPTS

1. Distance between two given points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) is
\( |PQ| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \).

2. Direction ratios of a line joining the points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are \( x_2 - x_1, y_2 - y_1, z_2 - z_1 \).

3. Angle between two lines, whose direction ratios are \( a_1, b_1, c_1 \) and \( a_2, b_2, c_2 \) is given by
\( \cos \theta = \left| \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \right| \).
(i) If lines are perpendicular, then \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \).
(ii) If lines are parallel, then \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).

4. Vector equation of a straight line passing through a fixed point with the position vector \( \vec{a} \) and parallel to a given vector \( \vec{b} \) is \( \vec{r} = \vec{a} + \lambda\vec{b} \), where \( \lambda \) is a parameter and \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).

5. Cartesian equation (symmetrical form) of the straight line passing through a fixed point \( (x_1, y_1, z_1) \) having the direction ratios \( a, b, c \) is given by \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \).

6. The parametric equations of the line \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \) are
\( x = x_1 + a\lambda, y = y_1 + b\lambda, z = z_1 + c\lambda \), where \( \lambda \) is a parameter.

7. The coordinates of any point on the line \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \) are \( (x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda) \), where \( \lambda \in R \).

8. Equation of straight line passing through the point \( (x_1, y_1, z_1) \) having direction cosines \( l, m, n \) is \( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \).

9. Vector equation of two straight lines passing through two given points with position vector \( \vec{a} \) and \( \vec{b} \) is \( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \), where \( \lambda \) is a parameter.

10. Cartesian equation of a straight line passing through two given points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) is given by \( \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1} \).

11. Angle between two straight lines: Angle between two straight lines whose vector equations are \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \lambda\vec{b}_2 \), is equal to the angle between \( \vec{b}_1 \) and \( \vec{b}_2 \), because \( \vec{b}_1 \) and \( \vec{b}_2 \) are parallel vectors to the lines \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \lambda\vec{b}_2 \) respectively.
If \( \theta \) is the angle between both lines, then \( \cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1||\vec{b}_2|} \).
(i) If \( \vec{b}_1 \cdot \vec{b}_2 = 0 \), then \( \cos \theta = 0 \Rightarrow \theta = 90^\circ \Rightarrow \vec{b}_1 \perp \vec{b}_2 \). Both lines are perpendicular to each other.
(ii) If \( \vec{b}_1 = \lambda\vec{b}_2 \), then \( \cos \theta = 1 \Rightarrow \theta = 0^\circ \Rightarrow \vec{b}_1 \parallel \vec{b}_2 \). Both lines are parallel to each other.

12. Shortest distance between two lines: Let \( l_1 \) and \( l_2 \) be two skew lines given by \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu\vec{b}_2 \) respectively, where \( \vec{a}_1 \) and \( \vec{a}_2 \) are position vectors of points on \( l_1 \) and \( l_2 \) then shortest distance between two given points is given by \( d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \).
Note: If two lines are intersecting, then shortest distance between them is zero, i.e., \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0 \).

13. Shortest distance between two parallel lines: Let \( l_1 \) and \( l_2 \) be two parallel lines given by \( \vec{r} = \vec{a}_1 + \lambda\vec{b} \) and \( \vec{r} = \vec{a}_2 + \mu\vec{b} \) respectively. Then shortest distance between them is \( d = \left| \frac{\vec{b} \times (\vec{a}_2 - \vec{a}_1)}{|\vec{b}|} \right| \).

14. Shortest distance between two skew lines in cartesian form: Let \( \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \) and \( \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2} \) are two skew lines, then shortest distance between them is given by \( d = \frac{\left| \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} \right|}{\sqrt{(b_1c_2 - b_2c_1)^2 + (c_1a_2 - c_2a_1)^2 + (a_1b_2 - a_2b_1)^2}} \).
Note: If \( \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \), then lines are intersecting.

15. Equation of a plane passing through a given point, whose position vector is \( \vec{a} \) and perpendicular to a given vector \( \vec{n} \), is \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \) or \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \) or \( (\vec{r} - \vec{a}) \cdot \hat{n} = 0 \) when \( \hat{n} = \frac{\vec{n}}{|\vec{n}|} \).

Determination of plane under given conditions:

• (i) If \( a_1 a_2 + b_1b_2 + c_1c_2 = 0 \), then planes are perpendicular to each other.
• (ii) If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) then given planes are parallel to each other.

An equation of first degree in \( x, y, z \) of the form \( ax + by + cz + d = 0 \) where at least one of \( a, b, c \) is non-zero real number, i.e., \( a^2 + b^2 + c^2 \neq 0 \) represents a plane.

Equation of plane in intercept form: \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) where \( a, b, c \) are intercepts made by the plane on the x-axis, y-axis and z-axis respectively.

Equation of plane passing through three given points:
Case I. Vector equation of the plane passing through three given points having position vector \( \vec{a}, \vec{b} \) and \( \vec{c} \) is given by
\[ [(\vec{r} - \vec{a}) \quad (\vec{b} - \vec{a}) \quad (\vec{c} - \vec{a})] = 0 \]
or \( (\vec{r} - \vec{a}) \cdot ((\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})) = 0 \)
Case II. Cartesian equation of the plane passing through points \( A(x_1, y_1, z_1), B(x_2, y_2, z_2) \) and \( C(x_3, y_3, z_3) \) is given by
\[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \]

Condition for coplanarity of two lines:
Case I. When lines are in vector form:
(i) Let \( \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \) be two lines then these lines are coplanar if
\( [\vec{a}_2 - \vec{a}_1 \quad \vec{b}_1 \quad \vec{b}_2] = 0 \) i.e \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0 \)
(ii) Equation of plane containing these two lines is
\( (\vec{r} - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0 \) or \( (\vec{r} - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2) = 0 \).
Case II. When lines are in cartesian form:
(i) Let \( \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \) and \( \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2} \) be the two lines then these lines are coplanar iff
\[ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \]
(ii) Equation of plane containing these lines is
\( \frac{x - x_1}{b_1c_2 - b_2c_1} = \frac{y - y_1}{a_2c_1 - a_1c_2} = \frac{z - z_1}{a_1b_2 - a_2b_1} \) or \( \frac{x - x_2}{b_1c_2 - b_2c_1} = \frac{y - y_2}{a_2c_1 - a_1c_2} = \frac{z - z_2}{a_1b_2 - a_2b_1} \)
(iii) The length of perpendicular from a point having position vector \( \vec{a} \) to the plane \( \vec{r} \cdot \vec{n} = d \) is given by \( |d - \vec{a} \cdot \hat{n}| \).
(iv) The length of perpendicular from a point \( P(x_1, y_1, z_1) \) to the plane \( ax + by + cz + d = 0 \) is given by
\[ \left| \frac{ax_1 + by_1 + cz_1 + d}{\sqrt{a^2 + b^2 + c^2}} \right| \]

Questions-

Question. If a line makes angles \( 90^\circ, 135^\circ, 45^\circ \) with the x, y and z axis respectively, find its direction cosines.
Answer: Since the line makes angle \( 90^\circ, 135^\circ, 45^\circ \) with the \( x, y \) and \( z \) axis respectively
then \( \alpha = 90^\circ, \beta = 135^\circ \) and \( \gamma = 45^\circ \)
\( l = \cos 90^\circ = 0, m = \cos 135^\circ = \cos (180 - 45)^\circ = -\cos 45^\circ = -\frac{1}{\sqrt{2}} \) and \( n = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
Thus, direction cosines of the line are \( 0, -\frac{1}{\sqrt{2}} \) and \( \frac{1}{\sqrt{2}} \).

Question. Show that the line through the points (1, –1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Answer: Let \( A(1, -1, 2) \) and \( B(3, 4, -2) \) be given points.
Direction ratios of AB are
\( (3 - 1), \{4 - (-1)\}, (-2 - 2) \) i.e., \( 2, 5, -4 \).
Let \( C(0, 3, 2) \) and \( D(3, 5, 6) \) be given points.
Direction ratios of CD are
\( (3 - 0), (5 - 3), (6 - 2) \) i.e., \( 3, 2, 4 \).
We know that two lines with direction ratios \( a_1, b_1, c_1 \) and \( a_2, b_2, c_2 \) are perpendicular if \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \).
\( \therefore 2 \times 3 + 5 \times 2 + (-4) \times 4 = 6 + 10 - 16 = 0 \), which is true.
It will shows that lines AB and CD are perpendicular.

Question. Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and parallel to the line given by \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \).
Answer: The equation of given line is \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \).
The direction ratios of the given line are \( 3, 5, 6 \). Since the required line is parallel to given line, so, the direction ratios of required line are proportional i.e., \( 3, 5, 6 \).
Now the equation of the line passing through point \( (-2, 4, -5) \) and having direction ratios \( 3, 5, 6 \) is
\( \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} \)
which is equation of required line.

Question. Find the angle between the following pair of lines:
\( \frac{x}{2} = \frac{y}{2} = \frac{z}{1} \) and \( \frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8} \).
Answer: Here the equation of given lines are
\( \frac{x}{2} = \frac{y}{2} = \frac{z}{1} \) and \( \frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8} \).
\( \therefore \) Direction ratios of two lines are \( 2, 2, 1 \) and \( 4, 1, 8 \).
Let \( \theta \) be the angle between two given lines then
\( \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \)
\( = \frac{2 \times 4 + 2 \times 1 + 1 \times 8}{\sqrt{(2)^2 + (2)^2 + (1)^2} \cdot \sqrt{(4)^2 + (1)^2 + (8)^2}} = \frac{8 + 2 + 8}{\sqrt{4 + 4 + 1} \sqrt{16 + 1 + 64}} \)
\( \therefore \cos \theta = \frac{2}{3} \Rightarrow \theta = \cos^{-1} \frac{2}{3} \).

Question. Find the value of \( p \) so that the lines \( \frac{1-x}{3} = \frac{7y-14}{2p} = \frac{5z-10}{11} \) and \( \frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5} \) are perpendicular to each other.
Answer: The given lines \( \frac{1-x}{3} = \frac{7y-14}{2p} = \frac{5z-10}{11} \) and \( \frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5} \) are rearranged to get
\( \frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-2}{11/5} \) ...(i)
\( \frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5} \) ...(ii)
Direction ratios of lines are \( -3, \frac{2p}{7}, \frac{11}{5} \) and \( \frac{-3p}{7}, 1, -5 \)
As the lines are perpendicular, we get
\( -3 \left( \frac{-3p}{7} \right) + \frac{2p}{7} \times 1 + \frac{11}{5} (-5) = 0 \)
\( \Rightarrow \frac{9p}{7} + \frac{2p}{7} - 11 = 0 \Rightarrow \frac{11}{7}p = 11 \Rightarrow p = 7 \).

Question. Find the shortest distance between the lines :
\( \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu (2\hat{i} + \hat{j} + 2\hat{k}) \).
Answer: Given lines are
\( \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}) \) ... (i)
\( \vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu (2\hat{i} + \hat{j} + 2\hat{k}) \) ... (ii)
Comparing the equation (i) and (ii) with \( \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \), we get
\( \vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}, \vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k} \)
\( \vec{b}_1 = \hat{i} - \hat{j} + \hat{k}, \vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k} \)
Now, \( \vec{a}_2 - \vec{a}_1 = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} - 3\hat{j} - 2\hat{k} \)
\( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = (-2-1)\hat{i} - (2-2)\hat{j} + (1+2)\hat{k} = -3\hat{i} + 3\hat{k} \)
\( \therefore |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + (3)^2} = 3\sqrt{2} \)
Shortest distance = \( \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{(\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 0\hat{j} + 3\hat{k})}{3\sqrt{2}} \right| \)
\( = \left| \frac{-3 - 0 - 6}{3\sqrt{2}} \right| = \frac{9}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{9\sqrt{2}}{3 \times 2} = \frac{3\sqrt{2}}{2} \).

Question. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the plane is \( 3\hat{i} + 5\hat{j} - 6\hat{k} \).
Answer: Normal vector of the plane is \( \vec{n} = 3\hat{i} + 5\hat{j} - 6\hat{k} \)
\( \therefore |\vec{n}| = \sqrt{(3)^2 + (5)^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70} \)
\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{1}{\sqrt{70}}(3\hat{i} + 5\hat{j} - 6\hat{k}) = \frac{3}{\sqrt{70}}\hat{i} + \frac{5}{\sqrt{70}}\hat{j} - \frac{6}{\sqrt{70}}\hat{k} \)
The required equation of plane is \( \vec{r} \cdot \hat{n} = 7 \)
\( \therefore \vec{r} \cdot \left( \frac{3}{\sqrt{70}}\hat{i} + \frac{5}{\sqrt{70}}\hat{j} - \frac{6}{\sqrt{70}}\hat{k} \right) = 7 \).

Question. Find the vector equation of the plane passing through the intersection of the planes \( \vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) = 7, \vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9 \) and through the point (2, 1, 3).
Answer: Let the equation of plane passing through the intersection of two planes be.
\( \vec{r} \cdot [(2\hat{i} + 2\hat{j} - 3\hat{k}) + \lambda (2\hat{i} + 5\hat{j} + 3\hat{k})] = 7 + 9\lambda \)
\( \vec{r} \cdot [(2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k}] = 7 + 9\lambda \) ...(i)
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot [(2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k}] = 7 + 9\lambda \)
\( (2 + 2\lambda)x + (2 + 5\lambda)y + (-3 + 3\lambda)z = 7 + 9\lambda \)
It contains point (2, 1, 3).
\( \therefore (2 + 2\lambda) \times 2 + (2 + 5\lambda) \times 1 + (-3 + 3\lambda) \times 3 = 7 + 9\lambda \)
\( 4 + 4\lambda + 2 + 5\lambda - 9 + 9\lambda = 7 + 9\lambda \Rightarrow 18\lambda - 3 = 7 + 9\lambda \Rightarrow 18\lambda - 9\lambda = 7 + 3 \)
\( 9\lambda = 10 \Rightarrow \lambda = \frac{10}{9} \)
Put in equation (i), we get
\( \vec{r} \cdot \left( \frac{38}{9}\hat{i} + \frac{68}{9}\hat{j} + \frac{3}{9}\hat{k} \right) = 17 \)
\( \vec{r} \cdot (38\hat{i} + 68\hat{j} + 3\hat{k}) = 153 \) is the required vector equation in plane.

Question. Find the equation of the plane through the line of intersection of the planes \( x + y + z = 1 \) and \( 2x + 3y + 4z = 5 \) which is perpendicular to the plane \( x - y + z = 0 \).
Answer: Let the equation of the plane passing through the intersection of two planes be
(equation of (i) plane) + \(\lambda\) (equation of (ii) plane) = \(d_1 + \lambda d_2\)
\( (x + y + z) + \lambda (2x + 3y + 4z) = 1 + 5\lambda \) ... (i)
\( x(1 + 2\lambda) + y(1 + 3\lambda) + z(1 + 4\lambda) = 1 + 5\lambda \)
\( a_1= (1 + 2\lambda), b_1 = (1 + 3\lambda), c_1= (1 + 4\lambda) \)
This plane is perpendicular to the plane \( x - y + z = 0 \).
\( a_2 = 1, b_2 = - 1, c_2 = 1 \)
As plane is perpendicular to another plane then, \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
\( (1 + 2\lambda) \times 1 + (1 + 3\lambda) \times (-1) + (1 + 4\lambda) \times 1 = 0 \)
\( 1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0 \Rightarrow 3\lambda = - 1 \Rightarrow \lambda = -\frac{1}{3} \)
Put value of \(\lambda\) in equation (i), we get
\( (x + y + z) - \frac{1}{3} (2x + 3y + 4z) = 1 + 5 \left( -\frac{1}{3} \right) \)
\( \frac{3x + 3y + 3z - (2x + 3y + 4z)}{3} = \frac{3 - 5}{3} \)
\( x - z = - 2 \Rightarrow x - z + 2 = 0 \) is the required equation of the plane.

Question. Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane \( 2x + y + z = 7 \).
Answer: The equation of given plane is \( 2x + y + z = 7 \) ...(i)
Equation of the line passing through points (3, – 4, – 5) and (2, – 3, 1) is
\( \frac{x-3}{2-3} = \frac{y+4}{-3+4} = \frac{z+5}{1+5} \Rightarrow \frac{x-3}{-1} = \frac{y+4}{1} = \frac{z+5}{6} = \lambda \text{ (say)} \)
\( \therefore \frac{x-3}{-1} = \lambda \Rightarrow x-3 = -\lambda \Rightarrow x = 3-\lambda \)
\( \frac{y+4}{1} = \lambda \Rightarrow y+4 = \lambda \Rightarrow y = -4+\lambda \)
\( \frac{z+5}{6} = \lambda \Rightarrow z+5 = 6\lambda \Rightarrow z = -5+6\lambda \)
Putting value of \( x, y \) and \( z \) in (i), we have
\( 2(3-\lambda) + (-4+\lambda) + (-5+6\lambda) = 7 \)
\( \Rightarrow 6-2\lambda-4+\lambda-5+6\lambda = 7 \)
\( \Rightarrow 5\lambda = 7+3 \Rightarrow \lambda = 2 \)
\( \therefore x = 3-2 = 1, y = -4+2 = -2 \) and \( z = -5+6 \times 2 = -5+12 = 7 \)
Thus, coordinates of required point are (1, – 2, 7).

Question. Find the equation of the plane passing through the line of intersection of the planes \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1 \) and \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) + 4 = 0 \) and parallel to x-axis.
Answer: Here the equations of given planes are
\( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 = 0 \) and \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) + 4 = 0 \)
The equation of a plane passing through the intersection of the given planes is
\( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 + \lambda [\vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) + 4] = 0 \)
\( \Rightarrow \vec{r} \cdot [(2\lambda + 1)\hat{i} + (3\lambda + 1)\hat{j} + (1 - \lambda)\hat{k}] + 4\lambda - 1 = 0 \) ...(i)
Since the above plane is parallel to x-axis i.e., \( \hat{i} + 0\hat{j} + 0\hat{k} \).
\( \therefore [(2\lambda + 1)\hat{i} + (3\lambda + 1)\hat{j} + (1 - \lambda)\hat{k}] \cdot (\hat{i} + 0\hat{j} + 0\hat{k}) = 0 \Rightarrow 2\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{2} \)
Putting value of \(\lambda\) in (i), we have
\( \vec{r} \cdot \left[ \left( 2 \times (-\frac{1}{2}) + 1 \right)\hat{i} + \left( 3 \times (-\frac{1}{2}) + 1 \right)\hat{j} + \left( 1 - (-\frac{1}{2}) \right)\hat{k} \right] + 4 \times (-\frac{1}{2}) - 1 = 0 \)
\( \Rightarrow \vec{r} \cdot \left( -\frac{1}{2}\hat{j} + \frac{3}{2}\hat{k} \right) - 3 = 0 \Rightarrow \vec{r} \cdot (-\hat{j} + 3\hat{k}) - 6 = 0 \)
Putting \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), we have
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (-\hat{j} + 3\hat{k}) - 6 = 0 \Rightarrow -y + 3z - 6 = 0 \Rightarrow y - 3z + 6 = 0 \)
which is required equation of the plane.

Question. Find the equation of the plane which contains the line of intersection of the planes \( \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) - 4 = 0 \) and \( \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) + 5 = 0 \) and which is perpendicular to the plane \( \vec{r} \cdot (5\hat{i} + 3\hat{j} - 6\hat{k}) + 8 = 0 \).
Answer: The given planes are
\( \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) - 4 = 0 \) ...(i)
and \( \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) + 5 = 0 \) ...(ii)
Therefore, a plane which contains the line of intersection of the planes (i) and (ii) is
\( \Rightarrow \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) - 4 + \lambda \{ \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) + 5 \} = 0 \)
\( \Rightarrow \vec{r} \cdot [(1+2\lambda)\hat{i} + (2+\lambda)\hat{j} + (3-\lambda)\hat{k}] - 4 + 5\lambda = 0 \) ...(iii)
Now, the plane (iii) is perpendicular to the plane
\( \vec{r} \cdot (5\hat{i} + 3\hat{j} - 6\hat{k}) + 8 = 0 \) ...(iv)
Therefore from (iii) and (iv), we get
\( (1 + 2\lambda) \cdot 5 + (2 + \lambda) \cdot 3 + (3 - \lambda) \cdot (-6) = 0 \) [ \( \because \vec{n}_1 \cdot \vec{n}_2 = 0 \) ]
\( \Rightarrow 5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0 \)
\( \Rightarrow 19\lambda - 7 = 0 \Rightarrow \lambda = \frac{7}{19} \)
Now, putting the value of \(\lambda\) in (iii), we get
\( \vec{r} \cdot \left[ \left(1 + \frac{14}{19}\right)\hat{i} + \left(2 + \frac{7}{19}\right)\hat{j} + \left(3 - \frac{7}{19}\right)\hat{k} \right] - 4 + 5 \times \frac{7}{19} = 0 \)
\( \vec{r} \cdot \left[ \frac{33}{19}\hat{i} + \frac{45}{19}\hat{j} + \frac{50}{19}\hat{k} \right] + \frac{35-76}{19} = 0 \)
\( \Rightarrow \vec{r} \cdot (33\hat{i} + 45\hat{j} + 50\hat{k}) - 41 = 0 \), which is the required equation.

Question. Find the distance of the point (–1, –5, –10), from the point of intersection of the line \( \vec{r} = (2\hat{i} - \hat{j} + 2\hat{k}) + \lambda (3\hat{i} + 4\hat{j} + 2\hat{k}) \) and the plane \( \vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 \).
Answer: Given line and plane are
\( \vec{r} = (2\hat{i} - \hat{j} + 2\hat{k}) + \lambda (3\hat{i} + 4\hat{j} + 2\hat{k}) \) ...(i)
\( \vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 \) ...(ii)
For intersection point, we solve equations (i) and (ii) by putting the value of \(\vec{r}\) from (i) in (ii).
\( [(2\hat{i} - \hat{j} + 2\hat{k}) + \lambda (3\hat{i} + 4\hat{j} + 2\hat{k})] \cdot (\hat{i} - \hat{j} + \hat{k}) = 5 \)
\( \Rightarrow (2 + 1 + 2) + \lambda (3 - 4 + 2) = 5 \Rightarrow 5 + \lambda = 5 \Rightarrow \lambda = 0 \)
Hence, position vector of intersecting point is \( 2\hat{i} - \hat{j} + 2\hat{k} \)
i.e., coordinates of intersection of line and plane is (2, –1, 2).
Hence, required distance
\( = \sqrt{(2-(-1))^2 + (-1-(-5))^2 + (2-(-10))^2} = \sqrt{9 + 16 + 144} \)
\( = \sqrt{169} = 13 \) units

Question. Find the vector equation of the line passing through (1, 2, – 4) and perpendicular to the two lines:
\( \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \) and \( \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} \)
Answer: Equation of any line through the point (1, 2, – 4) is
\( \frac{x-1}{a} = \frac{y-2}{b} = \frac{z+4}{c} \) ...(i)
where \( a, b \) and \( c \) are direction ratios of line (i).
Now the line (i) is perpendicular to the lines
\( \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \) and \( \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} \)
having direction ratios \( 3, -16, 7 \) and \( 3, 8, -5 \) respectively.
\( \therefore 3a - 16b + 7c = 0 \) ...(ii)
\( 3a + 8b - 5c = 0 \) ...(iii)
Solving (ii) and (iii) by cross-multiplication method, we have
\( \frac{a}{80 - 56} = \frac{b}{21 + 15} = \frac{c}{24 + 48} \Rightarrow \frac{a}{24} = \frac{b}{36} = \frac{c}{72} \)
\( \Rightarrow \frac{a}{2} = \frac{b}{3} = \frac{c}{6} \)
Let \( \frac{a}{2} = \frac{b}{3} = \frac{c}{6} = \lambda \Rightarrow a = 2\lambda, b = 3\lambda \) and \( c = 6\lambda \)
The equation of required line which passes through point (1, 2, –4) and parallel to vector \( 2\hat{i} + 3\hat{j} + 6\hat{k} \) is \( \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k}) \).

Question. Prove that if a plane has the intercepts \( a, b, c \) and is at a distance of \( p \) units from the origin then \( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2} \).
Answer: Let the equation of plane be \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) ...(i)
\( \therefore \) Length of perpendicular from origin to plane (i) is
\[ \frac{\left| \frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1 \right|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2 + (\frac{1}{c})^2}} = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} \]
It is given that
\( p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} \Rightarrow \sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} = \frac{1}{p} \Rightarrow \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2} \) (on squaring both sides)

Multiple Choice Questions

Question. The co-ordinates of the foot of the perpendicular drawn from the point (2, –3, 4) on the y-axis is
(a) (2, 3, 4)
(b) (– 2, – 3, – 4)
(c) (0, –3, 0)
(d) (2, 0, 4)
Answer: (c)

Question. The two planes \( x - 2y + 4z = 10 \) and \( 18x + 17y + kz = 50 \) are perpendicular, if k is equal to
(a) –4
(b) 4
(c) 2
(d) – 2
Answer: (b)

Question. Distance of the point \( (\alpha, \beta, \gamma) \) from y-axis is
(a) \( \beta \)
(b) \( |\beta| \)
(c) \( |\beta| + |\gamma| \)
(d) \( \sqrt{\alpha^2 + \gamma^2} \)
Answer: (d)

Question. If the direction cosines of a line are k, k, k then 
(a) \( k > 0 \)
(b) \( 0 < k < 1 \)
(c) \( k = 1 \)
(d) \( k = \frac{1}{\sqrt{3}} \) or \( -\frac{1}{\sqrt{3}} \)
Answer: (d)

Question. The distance of the plane \( \vec{r} \cdot \left(\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}\right) = 1 \) from the origin is
(a) 1
(b) 7
(c) \( \frac{1}{7} \)
(d) none of these
Answer: (a)

Question. The sine of the angle between the straight line \( \frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5} \) and the plane \( 2x - 2y + z = 5 \) is 
(a) \( \frac{10}{6\sqrt{5}} \)
(b) \( \frac{4}{5\sqrt{2}} \)
(c) \( \frac{2\sqrt{3}}{5} \)
(d) \( \frac{\sqrt{2}}{10} \)
Answer: (d)

Question. The reflection of the point \( (\alpha, \beta, \gamma) \) in the xy-plane is
(a) \( (\alpha, \beta, 0) \)
(b) \( (0, 0, \gamma) \)
(c) \( (-\alpha, -\beta, \gamma) \)
(d) \( (\alpha, \beta, -\gamma) \)
Answer: (d)

Question. P is a point on the line segment joining the points (3, 2, –1) and (6, 2, –2). If x co-ordinate of P is 5, then its y co-ordinate is 
(a) 2
(b) 1
(c) –1
(d) –2
Answer: (a)

Question. If \( \alpha, \beta, \gamma \) are the angles that a line makes with the positive direction of x, y, z axis, respectively, then the direction cosines of the line are 
(a) \( \sin \alpha, \sin \beta, \sin \gamma \)
(b) \( \cos \alpha, \cos \beta, \cos \gamma \)
(c) \( \tan \alpha, \tan \beta, \tan \gamma \)
(d) \( \cos^2 \alpha, \cos^2 \beta, \cos^2 \gamma \)
Answer: (b)

Question. The distance of a point P (a, b, c) from x-axis is
(a) \( \sqrt{a^2 + c^2} \)
(b) \( \sqrt{a^2 + b^2} \)
(c) \( \sqrt{b^2 + c^2} \)
(d) \( b^2 + c^2 \)
Answer: (c)

Question. The equations of x-axis in space are 
(a) \( x = 0, y = 0 \)
(b) \( x = 0, z = 0 \)
(c) \( x = 0 \)
(d) \( y = 0, z = 0 \)
Answer: (d)

Question. A line makes equal angles with co-ordinate axis. Direction cosines of this line are 
(a) \( \pm (1, 1, 1) \)
(b) \( \pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \)
(c) \( \pm \left( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right) \)
(d) \( \pm \left( \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \right) \)
Answer: (b)

Question. P is the point on the line segment joining the points (3, 2, –1) and (6, 2, –2). If x co-ordinate of P is 5, then its y co-ordinate is
(a) 2
(b) 1
(c) –1
(d) –2
Answer: (a)

Question. The sine of the angle between the straight line \( \frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5} \) and the plane \( 2x - 2y + z = 5 \) is
(a) \( \frac{10}{6\sqrt{5}} \)
(b) \( \frac{4}{5\sqrt{2}} \)
(c) \( \frac{2\sqrt{3}}{5} \)
(d) \( \frac{\sqrt{2}}{10} \)
Answer: (d)

Question. The area of the quadrilateral ABCD where A (0, 4, 1), B (2, 3, –1), C (4, 5, 0) and D (2, 6, 2) is equal to
(a) 9 sq units
(b) 18 sq units
(c) 27 sq units
(d) 81 sq units
Answer: (a)

Question. The intercepts made by the plane \( 2x - 3y + 5z + 4 = 0 \) on the coordinate axes are
(a) \( -2, \frac{4}{3} \) and \( -\frac{4}{5} \)
(b) \( -2, -\frac{4}{3} \) and \( \frac{4}{5} \)
(c) \( \frac{4}{3}, -\frac{4}{3} \) and \( \frac{7}{3} \)
(d) \( -2, -\frac{4}{3} \) and \( -\frac{4}{5} \)
Answer: (a)

Question. The shortest distance between the lines given by \( \vec{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k} \) and \( \vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k}) \) is
(a) 7 units
(b) 2 units
(c) 14 units
(d) 3 units
Answer: (c)

Question. The image of the point (1, 6, 3) in the line \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{2} \) is
(a) (2, 0, 5)
(b) (1, 3, 4)
(c) (1, 0, 7)
(d) (– 3, – 2, 0 )
Answer: (c)

Question. The coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane passing through three points (2, 2, 1), (3, 0, 1) and (4, – 1, 0) are
(a) (0, – 2, 7)
(b) (3, – 2, 5)
(c) (1, – 2, – 7)
(d) (1, – 2, 7)
Answer: (d)

Question. The co-ordinates of the foot of perpendicular drawn from point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1) are
(a) \( \left(-\frac{7}{3}, \frac{2}{3}, \frac{11}{3}\right) \)
(b) \( \left(-\frac{5}{3}, \frac{2}{3}, \frac{19}{3}\right) \)
(c) \( \left(\frac{4}{3}, \frac{2}{3}, \frac{11}{3}\right) \)
(d) None of these
Answer: (b)