CBSE Class 12 Mathematics Three Dimensional Geometry Worksheet Set 01

Selected NCERT Questions

Question. If a line makes angles 90°, 135°, 45° with the x, y and z axis respectively, find its direction cosines.
Answer: Since the line makes angle 90°, 135°, 45° 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 \( 0, \frac{-1}{\sqrt{2}} \text{ and } \frac{1}{\sqrt{2}} \).

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
\( (3 - 1), \{(4 - (-1)\}, (-2 -2) \text{ i.e., } 2, 5, - 4 \).
Let C (0, 3, 2) and D (3, 5, 6) be given points.
Direction ratios of CD are
\( (3 - 0), (5 - 3), (6 - 2) \text{ i.e., } 3, 2, 4 \).
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 which 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
\[ \cos \theta = \frac{a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} \]
\[ = \frac{2 \times 4 + 2 \times 1 + 1 \times 8}{\sqrt{(2)^{2} + (2)^{2} + (1)^{2}} \cdot \sqrt{(4)^{2} + (1)^{2} + (8)^{2}}} = \frac{8 + 2 + 8}{\sqrt{4 + 4 + 1} \sqrt{16 + 1 + 64}} \]
\( \therefore \quad \cos \theta = \frac{2}{3} \)
\( \implies \theta = \cos^{-1}\frac{2}{3} \).

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
\( \therefore \quad -3\left(\frac{-3p}{7}\right) + \frac{2p}{7} \times 1 + \frac{11}{5} (-5) = 0 \)
\( \implies \frac{9p}{7} + \frac{2p}{7} - 11 = 0 \)
\( \implies \frac{11}{7}p = 11 \)
\( \implies p = 7 \)

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 \( \vec{r} = \vec{a}_{2} + \lambda \vec{b}_{2} \), we get
\( \vec{a}_{1} = \hat{i} + 2\hat{j} + \hat{k} \quad \quad \vec{a}_{2} = 2\hat{i} - \hat{j} - \hat{k} \)
\( \vec{b}_{1} = \hat{i} - \hat{j} + \hat{k} \quad \quad \vec{b}_{2} = 2\hat{i} + \hat{j} + 2\hat{k} \)
Now, \( \vec{a}_{2} - \vec{a}_{1} = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} - 3\hat{j} - 2\hat{k} \)
\( \vec{b}_{1} \times \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = (-2 - 1)\hat{i} - (2 - 2)\hat{j} + (1 + 2)\hat{k} = -3\hat{i} + 3\hat{k} \)
\( \therefore \quad |\vec{b}_{1} \times \vec{b}_{2}| = \sqrt{(-3)^{2} + (3)^{2}} = 3\sqrt{2} \)
\( \therefore \quad \text{Shortest distance} = \left| \frac{(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} \times \vec{b}_{2})}{|\vec{b}_{1} \times \vec{b}_{2}|} \right| = \left| \frac{(\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 0\hat{j} + 3\hat{k})}{|\vec{b}_{1} \times \vec{b}_{2}|} \right| \)
\( = \left| \frac{-3 - 0 - 6}{3\sqrt{2}} \right| = \frac{9}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{9\sqrt{2}}{3 \times 2} = \frac{3\sqrt{2}}{2} \)

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
\( \vec{n} = 3\hat{i} + 5\hat{j} - 6\hat{k} \)
\( \therefore \quad |\vec{n}| = \sqrt{(3)^{2} + (5)^{2} + (-6)^{2}} = \sqrt{9 + 25 + 36} = \sqrt{70} \)
\( \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{1}{\sqrt{70}}(3\hat{i} + 5\hat{j} - 6\hat{k}) = \frac{3}{\sqrt{70}}\hat{i} + \frac{5}{\sqrt{70}}\hat{j} - \frac{6}{\sqrt{70}}\hat{k} \).
The required equation of plane is \( \vec{r} \cdot \hat{n} = 7 \).
\( \therefore \quad \vec{r} \cdot \left( \frac{3}{\sqrt{70}}\hat{i} + \frac{5}{\sqrt{70}}\hat{j} - \frac{6}{\sqrt{70}}\hat{k} \right) = 7 \).

Question. Find the intercepts cut off by the planes \( 2x + y - z = 5 \).
Answer: The equation of the given plane is
\( 2x + y - z = 5 \)
\( \therefore \quad \frac{2}{5}x + \frac{y}{5} - \frac{z}{5} = 1 \)
\( \implies \frac{x}{5/2} + \frac{y}{5} + \frac{z}{-5} = 1 \)
which is of the form \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) where \( a, b \) and \( c \) are intercepts on \( x, y \) and \( z \) axis. Thus,
intercepts on \( x, y \) and \( z \) axis are \( \frac{5}{2}, 5 \) and \( -5 \) respectively.

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.
\( \vec{r} \cdot [ (2\hat{i} + 2\hat{j} - 3\hat{k}) + \lambda(2\hat{i} + 5\hat{j} + 3\hat{k}) ] = 7 + 9\lambda \)
\( \vec{r} \cdot [ (2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k} ] = 7 + 9\lambda \quad \dots (i) \)
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot [ (2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k} ] = 7 + 9\lambda \)
\( (2 + 2\lambda)x + (2 + 5\lambda)y + (-3 + 3\lambda)z = 7 + 9\lambda \)
\( \because \quad \) It contains point (2, 1, 3).
\( \therefore \quad (2 + 2\lambda) \times 2 + (2 + 5\lambda) \times 1 + (-3 + 3\lambda) \times 3 = 7 + 9\lambda \)
\( 4 + 4\lambda + 2 + 5\lambda - 9 + 9\lambda = 7 + 9\lambda \)
\( \implies 18\lambda - 3 = 7 + 9\lambda \)
\( \implies 18\lambda - 9\lambda = 7 + 3 \)
\( 9\lambda = 10 \)
\( \implies \lambda = \frac{10}{9} \)
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, \( a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0 \)
\( (1 + 2\lambda) \times 1 + (1 + 3\lambda) \times (-1) + (1 + 4\lambda) \times 1 = 0 \)
\( 1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0 \)
\( \implies -\lambda + 4\lambda = -1 \)
\( \implies 3\lambda = -1 \)
\( \lambda = -\frac{1}{3} \)
Put value of \( \lambda \) in equation (i), we get
\( (x + y + z) - \frac{1}{3} (2x + 3y + 4z) = 1 + 5\left(-\frac{1}{3}\right) \)
\( \frac{3x + 3y + 3z - (2x + 3y + 4z)}{3} = \frac{3 - 5}{3} \)
\( x - z = -2 \)
\( \implies x - z + 2 = 0 \) is the required equation of the plane.

Question. If the line \( \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} \) and \( \frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5} \) are perpendicular, find the value of k.
Answer: Here the given lines are \( \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} \) and \( \frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5} \).
The direction ratios of two lines are \( -3, 2k, 2 \) and \( 3k, 1, -5 \).
we know that two lines are perpendicular to each other if \( a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0 \)
\( \therefore \quad -3 \times 3k + 2k \times 1 + 2 \times (-5) = 0 \)
\( \implies -9k + 2k - 10 = 0 \)
\( \implies -7k - 10 = 0 \)
\( \implies k = -\frac{10}{7} \).

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
\( \frac{x - 3}{2 - 3} = \frac{y + 4}{-3 + 4} = \frac{z + 5}{1 + 5} \)
\( \implies \frac{x - 3}{-1} = \frac{y + 4}{1} = \frac{z + 5}{6} = \lambda \) (say)
\( \therefore \quad \frac{x - 3}{-1} = \lambda \)
\( \implies x - 3 = -\lambda \)
\( \implies x = 3 - \lambda \)
\( \frac{y + 4}{1} = \lambda \)
\( \implies y + 4 = \lambda \)
\( \implies y = -4 + \lambda \)
\( \frac{z + 5}{6} = \lambda \)
\( \implies z + 5 = 6\lambda \)
\( \implies z = -5 + 6\lambda \)
Putting value of \(x, y\) and \(z\) in (i), we have
\( 2(3 - \lambda) + (-4 + \lambda) + (-5 + 6\lambda) = 7 \)
\( \implies 6 - 2\lambda - 4 + \lambda - 5 + 6\lambda = 7 \)
\( \implies 5\lambda = 7 + 3 \)
\( \implies \lambda = 2 \)
\( \therefore \quad x = 3 - 2 = 1, y = -4 + 2 = -2 \) 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
\( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1 \) 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
\( [ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 ] + \lambda [ \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) + 4 ] = 0 \)
\( \implies \vec{r} \cdot [ (2\lambda + 1)\hat{i} + (3\lambda + 1)\hat{j} + (1 - \lambda)\hat{k} ] - 1 + 4\lambda = 0 \) ... (i)
Since the above plane is parallel to x-axis i.e., \( \hat{i} + 0\hat{j} + 0\hat{k} \).
\( \therefore \quad [ (2\lambda + 1)\hat{i} + (3\lambda + 1)\hat{j} + (1 - \lambda)\hat{k} ] \cdot (\hat{i} + 0\hat{j} + 0\hat{k}) = 0 \)
\( \implies 2\lambda + 1 = 0 \)
\( \implies \lambda = -\frac{1}{2} \)
Putting value of \( \lambda \) in (i), we have
\( \vec{r} \cdot \left[ \left(2 \times \left(-\frac{1}{2}\right) + 1\right)\hat{i} + \left(3 \times \left(-\frac{1}{2}\right) + 1\right)\hat{j} + \left(1 + \frac{1}{2}\right)\hat{k} \right] + 4 \times \left(-\frac{1}{2}\right) - 1 = 0 \)
\( \implies \vec{r} \cdot \left( -\frac{1}{2}\hat{j} + \frac{3}{2}\hat{k} \right) - 3 = 0 \)
\( \implies \vec{r} \cdot ( -\hat{j} + 3\hat{k} ) - 6 = 0 \)
Putting \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), we have
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (-\hat{j} + 3\hat{k}) - 6 = 0 \)
\( \implies -y + 3z - 6 = 0 \)
\( \implies y - 3z + 6 = 0 \)
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
\( \implies \vec{r} \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) - 4 + \lambda \{ \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) + 5 \} = 0 \)
\( \implies \vec{r} \cdot [ (1 + 2\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 - \lambda)\hat{k} ] - 4 + 5\lambda = 0 \) ... (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
\( (1 + 2\lambda) \cdot 5 + (2 + \lambda) \cdot 3 + (3 - \lambda) \cdot (-6) = 0 \quad \quad [ \because \vec{n_{1}} \cdot \vec{n_{2}} = 0 ] \)
\( \implies 5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0 \)
\( \implies 19\lambda - 7 = 0 \)
\( \implies \lambda = \frac{7}{19} \)
Now, putting the value of \( \lambda \) in (iii), we get
\( \vec{r} \cdot \left[ \left(1 + \frac{14}{19}\right)\hat{i} + \left(2 + \frac{7}{19}\right)\hat{j} + \left(3 - \frac{7}{19}\right)\hat{k} \right] - 4 + 5 \times \frac{7}{19} = 0 \)
\( \implies \vec{r} \cdot \left[ \frac{33}{19}\hat{i} + \frac{45}{19}\hat{j} + \frac{50}{19}\hat{k} \right] + \frac{35 - 76}{19} = 0 \)
\( \implies \vec{r} \cdot (33\hat{i} + 45\hat{j} + 50\hat{k}) - 41 = 0 \), which is the required equation.

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.
\( \therefore \quad 3a - 16b + 7c = 0 \) ... (ii)
\( 3a + 8b - 5c = 0 \) ... (iii)
Solving (ii) and (iii) by cross-multiplication method, we have
\( \frac{a}{80 - 56} = \frac{b}{21 + 15} = \frac{c}{24 + 48} \)
\( \implies \frac{a}{24} = \frac{b}{36} = \frac{c}{72} \)
\( \implies \frac{a}{2} = \frac{b}{3} = \frac{c}{6} \)
Let \( \frac{a}{2} = \frac{b}{3} = \frac{c}{6} = \lambda \)
\( \implies a = 2\lambda, b = 3\lambda \text{ and } c = 6\lambda \)
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
\( \frac{\left| \frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1 \right|}{\sqrt{\left(\frac{1}{a}\right)^{2} + \left(\frac{1}{b}\right)^{2} + \left(\frac{1}{c}\right)^{2}}} = \frac{|-1|}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} \)
It is given that
\( p = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} \)
\( \implies \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} = \frac{1}{p} \)
\( \implies \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}} \). (on squaring both sides)

Long Answer Questions

Question. Find the equation of the plane that contains the point (1, -1, 2) and is perpendicular to both the planes \( 2x + 3y - 2z = 5 \) and \( x + 2y - 3z = 8 \). Hence find the distance of point P(-2, 5, 5) from the plane obtained above. 
Answer: Equation of plane containing the point (1, -1, 2) is given by
\( a(x - 1) + b(y + 1) + c(z - 2) = 0 \dots (i) \)
\( \because \) (i) is perpendicular to plane \( 2x + 3y - 2z = 5 \)
\( \therefore 2a + 3b - 2c = 0 \dots (ii) \)
Also, (i) is perpendicular to plane \( x + 2y - 3z = 8 \)
\( a + 2b - 3c = 0 \dots (iii) \)
From (ii) and (iii), we get
\( \frac{a}{-9 + 4} = \frac{b}{-2 + 6} = \frac{c}{4 - 3} \)

\( \Rightarrow \) \( \frac{a}{-5} = \frac{b}{4} = \frac{c}{1} = \lambda \) (say)

\( \Rightarrow \) \( a = -5\lambda, b = 4\lambda, c = \lambda \)
Putting these values in (i), we get
\( -5\lambda (x - 1) + 4\lambda (y + 1) + \lambda (z - 2) = 0 \)

\( \Rightarrow \) \( -5 (x - 1) + 4 (y + 1) + (z - 2) = 0 \)

\( \Rightarrow \) \( -5x + 5 + 4y + 4 + z - 2 = 0 \)

\( \Rightarrow \) \( -5x + 4y + z + 7 = 0 \)

\( \Rightarrow \) \( 5x - 4y - z - 7 = 0 \dots (iv) \) is the required equation of plane.
Again, if d be the distance of point P(-2, 5, 5) to plane (iv), then
\( d = \left| \frac{5 \times (-2) - 4 \times 5 - 5 - 7}{\sqrt{5^2 + (-4)^2 + (-1)^2}} \right| = \left| \frac{-10 - 20 - 5 - 7}{\sqrt{25 + 16 + 1}} \right| = \frac{42}{\sqrt{42}} = \sqrt{42} \) units

Question. Find the vector and Cartesian equations of a plane containing the two lines.
\( \vec{r} = (2\hat{i} + \hat{j} - 3\hat{k}) + \lambda (\hat{i} + 2\hat{j} + 5\hat{k}) \) and \( \vec{r} = (3\hat{i} + \hat{j} + 2\hat{k}) + \mu (3\hat{i} - 2\hat{j} + 5\hat{k}) \) 
Answer: Given lines are
\( \vec{r} = (2\hat{i} + \hat{j} - 3\hat{k}) + \lambda (\hat{i} + 2\hat{j} + 5\hat{k}) \dots (i) \)
\( \vec{r} = (3\hat{i} + \hat{j} + 2\hat{k}) + \mu (3\hat{i} - 2\hat{j} + 5\hat{k}) \dots (ii) \)
Here \( \vec{a}_1 = 2\hat{i} + \hat{j} - 3\hat{k}; \vec{a}_2 = 3\hat{i} + \hat{j} + 2\hat{k} \)
\( \vec{b}_1 = \hat{i} + 2\hat{j} + 5\hat{k}; \vec{b}_2 = 3\hat{i} - 2\hat{j} + 5\hat{k} \)
Now, \( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 5 \\ 3 & -2 & 5 \end{vmatrix} \)
\( = (10 + 10)\hat{i} - (5 - 15)\hat{j} + (-2 - 6)\hat{k} = 20\hat{i} + 10\hat{j} - 8\hat{k} \)
Hence, vector equation of required plane is
\( (\vec{r} - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0 \)

\( \Rightarrow \) \( \vec{r} \cdot (\vec{b}_1 \times \vec{b}_2) = \vec{a}_1 \cdot (\vec{b}_1 \times \vec{b}_2) \)

\( \Rightarrow \) \( \vec{r} \cdot (20\hat{i} + 10\hat{j} - 8\hat{k}) = (2\hat{i} + \hat{j} - 3\hat{k}) \cdot (20\hat{i} + 10\hat{j} - 8\hat{k}) \)

\( \Rightarrow \) \( \vec{r} \cdot (20\hat{i} + 10\hat{j} - 8\hat{k}) = 40 + 10 + 24 \)

\( \Rightarrow \) \( \vec{r} \cdot (20\hat{i} + 10\hat{j} - 8\hat{k}) = 74 \)

\( \Rightarrow \) \( \vec{r} \cdot (20\hat{i} + 10\hat{j} - 8\hat{k}) = 74 \)

\( \Rightarrow \) \( \vec{r} \cdot (10\hat{i} + 5\hat{j} - 4\hat{k}) = 37 \)
Therefore, Cartesian equation is \( 10x + 5y - 4z = 37 \)

Question. Find the equation of the plane passing through the point \( (-1, 3, 2) \) and perpendicular to each of the planes \( x + 2y + 3z = 5 \) and \( 3x + 3y + z = 0 \). 
Answer: The equation of plane through (-1, 3, 2) can be expressed as
\( A (x + 1) + B (y - 3) + C (z - 2) = 0 \)
As the required plane is perpendicular to \( x + 2y + 3z = 5 \) and \( 3x + 3y + z = 0 \), we get
\( A + 2B + 3C = 0 \) and \( 3A + 3B + C = 0 \)

\( \Rightarrow \) \( \frac{A}{2 - 9} = \frac{B}{9 - 1} = \frac{C}{3 - 6} \)

\( \Rightarrow \) \( \frac{A}{-7} = \frac{B}{8} = \frac{C}{-3} \)
\( \therefore \) Direction ratios of normal to the required plane are -7, 8, -3.
Hence, equation of the plane will be
\( -7(x + 1) + 8 (y - 3) - 3 (z - 2) = 0 \)

\( \Rightarrow \) \( -7x - 7 + 8y - 24 - 3z + 6 = 0 \)
or \( 7x - 8y + 3z + 25 = 0 \)

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 \( x - y + z = 0 \). Also find the distance of the plane obtained above, from the origin. 
Answer: The equation of a plane passing through the intersection of the given planes is
\( (x + y + z - 1) + \lambda (2x + 3y + 4z - 5) = 0 \)

\( \Rightarrow \) \( (1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z - (1 + 5\lambda) = 0 \dots (i) \)
Since, (i) is perpendicular to \( x - y + z = 0 \)
\( \Rightarrow \) \( (1 + 2\lambda) 1 + (1 + 3\lambda) (-1) + (1 + 4\lambda) 1 = 0 \)

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

\( \Rightarrow \) \( 3\lambda + 1 = 0 \)

\( \Rightarrow \) \( \lambda = -\frac{1}{3} \)
Putting the value of \( \lambda \) in (i), we get
\( \left(1 - \frac{2}{3}\right)x + \left(1 - \frac{3}{3}\right)y + \left(1 - \frac{4}{3}\right)z - \left(1 - \frac{5}{3}\right) = 0 \)

\( \Rightarrow \) \( \frac{x}{3} - \frac{z}{3} + \frac{2}{3} = 0 \)

\( \Rightarrow \) \( x - z + 2 = 0 \), it is required plane.
Let d be the distance of this plane from origin.
\( \therefore d = \left| \frac{0 \cdot x + 0 \cdot y + 0 \cdot (-z) + 2}{\sqrt{1^2 + 0^2 + (-1)^2}} \right| = \left| \frac{2}{\sqrt{2}} \right| = \sqrt{2} \) units.
[Note: The distance of the point \( (\alpha, \beta, \gamma) \) to the plane \( ax + by + cz + d = 0 \) is given by \( \left| \frac{a\alpha + b\beta + c\gamma + d}{\sqrt{a^2 + b^2 + c^2}} \right| \) ]

Question. Find the direction ratios of the normal to the plane, which passes through the points (1, 0, 0) and (0, 1, 0) and makes angle \( \frac{\pi}{4} \) with the plane \( x + y = 3 \). Also find the equation of the plane. 
Answer: Let the equation of plane passing through the point (1, 0, 0) be
\( a (x - 1) + b (y - 0) + c (z - 0) = 0 \)

\( \Rightarrow \) \( ax - a + by + cz = 0 \)

\( \Rightarrow \) \( ax + by + cz = a \dots (i) \)
Since, (i) also passes through (0, 1, 0)

\( \Rightarrow \) \( 0 + b + 0 = a \)

\( \Rightarrow \) \( b = a \dots (ii) \)
Given, the angle between plane (i) and plane \( x + y = 3 \) is \( \frac{\pi}{4} \).
\( \therefore \cos \frac{\pi}{4} = \frac{a \cdot 1 + b \cdot 1 + c \cdot 0}{\sqrt{a^2 + b^2 + c^2} \sqrt{1^2 + 1^2}} \)

\( \Rightarrow \) \( \frac{1}{\sqrt{2}} = \frac{a + b}{\sqrt{a^2 + b^2 + c^2} \sqrt{1 + 1}} \)

\( \Rightarrow \) \( \frac{1}{\sqrt{2}} = \frac{a + b}{\sqrt{a^2 + b^2 + c^2} \sqrt{2}} \)

\( \Rightarrow \) \( 1 = \frac{a + b}{\sqrt{a^2 + b^2 + c^2}} \)

\( \Rightarrow \) \( \sqrt{a^2 + b^2 + c^2} = \pm (a + b) \)

\( \Rightarrow \) \( a^2 + b^2 + c^2 = (a + b)^2 \)

\( \Rightarrow \) \( a^2 + b^2 + c^2 = a^2 + b^2 + 2ab \)

\( \Rightarrow \) \( c^2 = 2ab \)

\( \Rightarrow \) \( c^2 = 2a^2 \) [From (ii)]

\( \Rightarrow \) \( c = \pm \sqrt{2} a \)
Now, equation (i) becomes \( ax + ay \pm \sqrt{2} az = a \).

\( \Rightarrow \) \( x + y \pm \sqrt{2} z = 1 \), is the required equation of plane.
Therefore, required direction ratios are 1, 1, \( \pm \sqrt{2} \).

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 whose intercept on x-axis is equal to that of on y-axis. 
Answer: Given planes are \( \vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) - 4 = 0 \) and \( \vec{r} \cdot (-2\hat{i} + \hat{j} + \hat{k}) + 5 = 0 \)
These can be written in cartesian form as
\( x - 2y + 3z - 4 = 0 \dots (i) \)
and \( -2x + y + z + 5 = 0 \dots (ii) \)
Now the equation of plane containing the line of intersection of the planes (i) and (ii) is given by
\( (x - 2y + 3z - 4) + \lambda(-2x + y + z + 5) = 0 \dots (iii) \)

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

\( \Rightarrow \) \( (1 - 2\lambda) x - (2 - \lambda) y + (3 + \lambda) z = 4 - 5\lambda \)

\( \Rightarrow \) \( \frac{x}{\frac{4 - 5\lambda}{1 - 2\lambda}} + \frac{y}{\frac{4 - 5\lambda}{-2 + \lambda}} + \frac{z}{\frac{4 - 5\lambda}{3 + \lambda}} = 1 \)
According to question \( \frac{4 - 5\lambda}{1 - 2\lambda} = \frac{4 - 5\lambda}{-2 + \lambda} \)

\( \Rightarrow \) \( 1 - 2\lambda = -2 + \lambda \)

\( \Rightarrow \) \( 3\lambda = 3 \)

\( \Rightarrow \) \( \lambda = 1 \)
Putting the value of \( \lambda = 1 \) in (iii), we get
\( (x - 2y + 3z - 4) + 1(-2x + y + z + 5) = 0 \)

\( \Rightarrow \) \( -x - y + 4z + 1 = 0 \)

\( \Rightarrow \) \( x + y - 4z - 1 = 0 \)
Its vector form is \( \vec{r} \cdot (\hat{i} + \hat{j} - 4\hat{k}) - 1 = 0 \)

Question. If \( l_1, m_1, n_1 \), \( l_2, m_2, n_2 \) and \( l_3, m_3, n_3 \) are the direction cosines of three mutually perpendicular lines, then prove that the line whose direction cosines are proportional to \( l_1 + l_2 + l_3 \), \( m_1 + m_2 + m_3 \) and \( n_1 + n_2 + n_3 \) makes equal angles with them. 
Answer: Let \( \vec{a} = l_1\hat{i} + m_1\hat{j} + n_1\hat{k} \); \( \vec{b} = l_2\hat{i} + m_2\hat{j} + n_2\hat{k} \); \( \vec{c} = l_3\hat{i} + m_3\hat{j} + n_3\hat{k} \)
\( \vec{d} = (l_1 + l_2 + l_3)\hat{i} + (m_1 + m_2 + m_3)\hat{j} + (n_1 + n_2 + n_3)\hat{k} \)
Also, let \( \alpha, \beta \) and \( \gamma \) are the angles between \( \vec{a} \) and \( \vec{d} \), \( \vec{b} \) and \( \vec{d} \), \( \vec{c} \) and \( \vec{d} \).
\( \therefore \cos \alpha = l_1 (l_1 + l_2 + l_3) + m_1 (m_1 + m_2 + m_3) + n_1 (n_1 + n_2 + n_3) \)
\( = l_1^2 + l_1 l_2 + l_1 l_3 + m_1^2 + m_1 m_2 + m_1 m_3 + n_1^2 + n_1 n_2 + n_1 n_3 \)
\( = (l_1^2 + m_1^2 + n_1^2) + (l_1 l_2 + m_1 m_2 + n_1 n_2) + (l_1 l_3 + m_1 m_3 + n_1 n_3) \)
\( = 1 + 0 + 0 = 1 \)
\( [\because l_1^2 + m_1^2 + n_1^2 = 1 \text{ and } l_1 l_2 + m_1 m_2 + n_1 n_2 = 0 \text{ and } l_1 l_3 + m_1 m_3 + n_1 n_3 = 0] \)
Similarly, \( \cos \beta = l_2 (l_1 + l_2 + l_3) + m_2 (m_1 + m_2 + m_3) + n_2 (n_1 + n_2 + n_3) \)
\( = 1 + 0 \text{ and } \cos \gamma = 1 + 0 \)

\( \Rightarrow \) \( \cos \alpha = \cos \beta = \cos \gamma \)

\( \Rightarrow \) \( \alpha = \beta = \gamma \)
So, the line whose direction cosines are proportional to \( l_1 + l_2 + l_3 \), \( m_1 + m_2 + m_3 \), \( n_1 + n_2 + n_3 \) makes equal angles with the three mutually perpendicular lines whose direction cosines are \( l_1, m_1, n_1 \), \( l_2, m_2, n_2 \) and \( l_3, m_3, n_3 \) respectively.

Question. Find the value of p, so that the lines \( l_1 = \frac{1 - x}{3} = \frac{7y - 14}{p} = \frac{z - 3}{2} \) and \( l_2 = \frac{7 - 7x}{3p} = \frac{y - 5}{1} = \frac{6 - z}{5} \) are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, -4) and parallel to line \( l_1 \).
Answer: Given line \( l_1 \) and \( l_2 \) are
\( l_1 \equiv \frac{1 - x}{3} = \frac{7y - 14}{p} = \frac{z - 3}{2} \)

\( \Rightarrow \) \( \frac{x - 1}{-3} = \frac{y - 2}{\frac{p}{7}} = \frac{z - 3}{2} \)
\( l_2 \equiv \frac{7 - 7x}{3p} = \frac{y - 5}{1} = \frac{6 - z}{5} \)

\( \Rightarrow \) \( \frac{x - 1}{-\frac{3p}{7}} = \frac{y - 5}{1} = \frac{z - 6}{-5} \)
Since \( l_1 \perp l_2 \)

\( \Rightarrow \) \( (-3) \left(-\frac{3p}{7}\right) + \frac{p}{7} \times 1 + 2 \times (-5) = 0 \)

\( \Rightarrow \) \( \frac{9p}{7} + \frac{p}{7} - 10 = 0 \)

\( \Rightarrow \) \( \frac{10p}{7} = 10 \)

\( \Rightarrow \) \( p = \frac{7 \times 10}{10} \)

\( \Rightarrow \) \( p = 7 \)
The equation of line passing through (3, 2, -4) and parallel to \( l_1 \) is given by
\( \frac{x - 3}{-3} = \frac{y - 2}{\frac{p}{7}} = \frac{z + 4}{2} \) i.e., \( \frac{x - 3}{-3} = \frac{y - 2}{1} = \frac{z + 4}{2} \) (\( \because p = 7 \))

Question. A plane meets the coordinate axes in A, B, C, such that the centroid of the triangle ABC is the point \( (\alpha, \beta, \gamma) \). Show that the equation of the plane is \( \frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3 \). 
Answer: Let the equation of required plane be
\( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \dots (i) \)
Then the coordinates of A, B, C are (a, 0, 0), (0, b, 0) and (0, 0, c) respectively. So, the centroid of triangle ABC is \( \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right) \). But the coordinates of the centroid are \( (\alpha, \beta, \gamma) \) as given in problem.
\( \alpha = \frac{a}{3}, \beta = \frac{b}{3} \) and \( \gamma = \frac{c}{3} \)

\( \Rightarrow \) \( a = 3\alpha, b = 3\beta, c = 3\gamma \)
Substituting the values of a, b and c in equation (i), we get the required equation of the plane as follows
\( \frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1 \)

\( \Rightarrow \) \( \frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3 \).