NCERT Solutions Class 12 Maths (Three Dimensional Geometry) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-11 (Three Dimensional Geometry) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Q1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

Answer. Let OA be the line joining the origin, O (0, 0, 0) and the point A(2, 1, 1) Also, let BC be the line joining the points, B (3, 5, -1) and C (4, 3, -1). The direction ratios of OA are 2, 1, and 1 and of BC are (4 - 3) = 1, (3 - 5) = -2 and (-1 + 1) = 0 OA is perpendicular to BC, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ $\therefore a_1{a}_2+b_1{b}_2+c_1{c}_2=2\times 1+1(-2)+1\times 0=2-2=0$ Thus, OA is perpendicular to BC.

Q2. If $l_1,m_1,n_1\text{and}{l}_2,m_2,n_2$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_1{n}_2-m_2{n}_1,n_1{l}_2-n_2{l}_1,l_1{m}_2-l_2{m}_1$.

Answer. It is given that $l_1,m_1,n_1\text{and}{l}_2,m_2,n_2$ are the direction cosines of two mutually perpendicular lines. Therefore, $\begin{array}{c}{l}_1{l}_2+m_1{m}_2+n_1{n}_2=0\\ l_1^2+m_1^2+n_1^2=1\\ l_2^2+m_2^2+n_2^2=1\end{array}$ Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction cosines $l_1,m_{1,},n_1\text{and}{l}_2,m_2,n_2$. $\begin{array}{c}\therefore \parallel {}_1+m{m}_1+m{n}_1=0\\ \parallel {}_2+m{m}_2+m_2=0\end{array}$ $\therefore \frac{l}{m_1{n}_2-m_2{n}_1}=\frac{m}{n_1{l}_2-n_2{I}_1}=\frac{n}{l_1{m}_2-l_2{m}_l}$ $\Rightarrow \frac{l^2}{{(m_1{n}_2-m_2{n}_1)}^2}=\frac{m^2}{{(n_1{l}_2-n_2{I}_1)}^2}=\frac{n^2}{{(l_1{m}_2-l_2{m}_i)}^2}$ $\Rightarrow \frac{l^2}{{(m_1{n}_2-m_2{n}_1)}^2}=\frac{m^2}{{(n_1{l}_2-n_2{I}_1)}^2}=\frac{n^2}{{(l_1{m}_2-l_2{m}_2)}^2}$ $=\frac{t^2+m^2+n^2}{{(m_1{n}_2-m_2{n}_1)}^2+{(n_1{l}_2-n_2{l}_1)}^2+{(l_1{m}_2-l_2{m}_i)}^2}$ l, m, n are the direction cosines of the line. $\therefore l^2+m^2+n^2=1\dots \left(5\right)$ It is known that, $(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)-{(l_1{l}_2+m_1{m}_2+n_1{n}_2)}^2$ $={(m_1{n}_2-m_2{n}_1)}^2+{(n_1{l}_2-n_2{l}_1)}^2+{(l_1{m}_2-l_2{m}_1)}^2$ From (1), (2), and (3), we obtain $\Rightarrow 1.1-0={(m_1{n}_2+m_2{n}_1)}^2+{(n_1{l}_2-n_2{l}_1)}^2+{(l_1{m}_2-l_2{m}_1)}^2$ ${(m_1{n}_2-m_2{n}_1)}^2+{(n_1{l}_2-n_2{I}_1)}^2+{(l_1{m}_2-l_2{m}_1)}^2=1$ Substituting the values from equations (5) and (6) in equation (4), we obtain $\begin{array}{c}\frac{l^2}{{(m_1{n}_2-m_2{n}_1)}^2}=\frac{m^2}{{(n_2{I}_2-n_2{l}_1)}^2}=\frac{n^2}{{(l_1{m}_2-l_2{m}_1)}^2}=1\\ \Rightarrow l=m_1{n}_2-m_2{n}_1,m=n_1{l}_2-n_2{I}_{11},n=l_1{m}_2-l_2{m}_1\end{array}$ Thus, the direction cosines of the required line are $m_1{n}_2-m_2{n}_1,n_1{l}_2-n_2{l}_1,\text{and}{l}_1{m}_2-l_2{m}_1$.

Q3. Find the angle between the lines whose direction ratios a, b, c and $b-c,c-a,a-b$.

Answer. The angle Q between the lines with direction cosines a, b, c and b-c, c-a, a-b, is given by, $\begin{array}{c}\cos Q=\left|\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}+\sqrt{(b-c{)}^2+(c-a{)}^2+(a-b{)}^2}}\right|\\ \Rightarrow Q={\cos }^{-1}0\\ \Rightarrow Q={90}^{\circ }\end{array}$ Thus, the angle between the lines is ${90}^{\circ }$

Q4. Find the equation of a line parallel to x-axis and passing through the origin.

Answer. The line parallel to x-axis and passing through the origin is x-axis itself. Let A be a point on x-axis . Therefore, the coordinates of A are given by (a, 0, 0) where $a\in R$ Direction ratios of OA are (a - 0) = a, 0, 0 The equation of OA is given by, $\begin{array}{c}\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}\\ \Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a\end{array}$ Thus, the equation of line parallel to x-axis and passing through origin is $\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

Q5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer. The coordinates of A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively. The direction ratios of AB are (4, -1) = 3, (5 - 2)=3, and (7 - 3) = 4 The direction ratios of CD are $(2-(-4\left)\right)=6,(9-3)=6,\text{and}(2-(-6\left)\right)=8$ It can be seen that, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}$ Therefore, AB is parallel to CD. Thus, the angle between AB and CD is either $0^{\circ }\text{or}{180}^{\circ }$.

Q6. If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\text{and}\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular, find the value of k.

Answer. The direction of ratios of the lines, $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\text{and}\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are -3, 2k, 2 and 3k, 1, -5 respectively. It is known that two lines with direction ratios, $a_1,b_1,c_1\text{and}{a}_2,b_2,c_2$ are perpendicular, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=$ = 0. $\begin{array}{c}\therefore -3\left(3k\right)+2k\times 1+2(-5)=0\\ \Rightarrow -9k+2k-10=0\\ \Rightarrow 7k=-10\\ \Rightarrow k=\frac{-10}{7}\end{array}$ Therefore, for $k=-\frac{10}{7}$, the given lines are perpendicular to each other.

Q7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}-5\hat{k})+9=0$.

Answer. The position vector of the point (1, 2, 3) is ${\overrightarrow{r}}_1=\hat{i}+2\hat{j}+3\hat{k}$ The direction ratios of the normal to the plane, $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}-5\hat{k})+9=0$ are 1, 2, and -5 and the normal vector is $\overline{N}=\hat{i}+2\hat{j}-5\hat{k}$. The equation of a line passing through a point and perpendicular to the given plane is given by, $\overrightarrow{l}=\overrightarrow{r}+\lambda \overrightarrow{N},\lambda \in R$ $\Rightarrow \overrightarrow{l}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}+2\hat{j}-5\hat{k})$

Q8. Find the equation of the plane passing through (a, b, c) and parallel to the plane $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=2$.

Answer. Any plane parallel to the plane, ${\overrightarrow{r}}_1\cdot (\hat{i}+\hat{j}+\hat{k})=2$ is of form $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=\lambda$ The plane passes through the point (a, b, c). Therefore, the position vector of this point is $\overrightarrow{r}=a\hat{i}+b\hat{j}+c\hat{k}$ Therefore, equation (1) becomes $\begin{array}{c}(a\hat{i}+b\hat{j}+c\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=\lambda \\ \Rightarrow a+b+c=\lambda \end{array}$ Substituting $\lambda =a+b+c$ in equation (1), we obtain $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=a+b+c$ This is the vector equation of the required plane. Substituting $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in equation (2), we obtain $\begin{array}{c}(x\hat{i}+y\hat{j}+z\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})=a+b+c\\ \Rightarrow x+y+z=a+b+c\end{array}$

Q9. Find the shortest distance between lines $\begin{array}{c}\overrightarrow{r}=6\hat{i}+2\hat{j}+2\hat{k}+\lambda (\hat{i}-2\hat{j}+2\hat{k})\text{and}\overrightarrow{r}=-4\hat{i}-\hat{k}+\mu (3\hat{i}-2\hat{j}-2\hat{k})\end{array}$

Answer. The given lines are $\begin{array}{cc}\overrightarrow{r}=6\hat{i}+2\hat{j}+2\hat{k}+\lambda (\hat{i}-2\hat{j}+2\hat{k})& \dots \left(1\right)\\ \overrightarrow{r}=-4\hat{i}-\hat{k}+\mu (3\hat{i}-2\hat{j}-2\hat{k})& \dots \left(2\right)\end{array}$ It is known that the shortest distance between two lines, $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_{1\text{and}}\overrightarrow{r}={\overrightarrow{a}}_2+\lambda {\overrightarrow{b}}_2$, is given by $d=\left|\frac{({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\cdot ({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}\right|$ Comparing to equations (1) and (2), we obtain $\begin{array}{c}{\overrightarrow{a}}_1=6\hat{i}+2\hat{j}+2\hat{k}\\ {\overrightarrow{b}}_1=\hat{i}-2\hat{j}+2\hat{k}\\ {\overrightarrow{a}}_2=-4\hat{i}-\hat{k}\\ {\overrightarrow{b}}_2=3\hat{i}-2\hat{j}-2\hat{k}\end{array}$ $\Rightarrow {\overrightarrow{a}}_2-{\overrightarrow{a}}_1=(-4\hat{i}-\hat{k})-(6\hat{i}+2\hat{j}+2\hat{k})=-10\hat{i}-2\hat{j}-3\hat{k}$ $\Rightarrow {\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 2\\ 3& -2& -2\end{array}\right|$ $=(4+4)\hat{i}-(-2-6)\hat{j}+(-2+6)\hat{k}=8\hat{i}+8\hat{j}+4\hat{k}$ $\therefore |{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|=\sqrt{(8{)}^2+(8{)}^2+(4{)}^2}=12$ $({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)\cdot ({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)=(8\hat{i}+8\hat{j}+4\hat{k})\cdot (-10\hat{i}-2\hat{j}-3\hat{k})=-80-16-12=-108$ Substituting all the values in equation (1) we obtain $d=\left|\frac{-108}{12}\right|=9$ Therefore, the shortest distance between the two given lines is 9 units.

Q10. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane.

Answer. It is known that the equation of the line passing through the points, $(x_1,y_1,z_1)\text{and}(x_2,y_2,z_2)$ is $\begin{array}{cc}\frac{x-x_1}{x_2-x_1}& =\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\end{array}$ The line passing through the points (5, 1, 6) and (3, 4, 1) is given by, $\begin{array}{c}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ \Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k\left(\text{say}\right)\\ \Rightarrow x=5-2k,y=3k+1,z=6-5k\end{array}$ Any point on the line is of the form $(5-2k,3k+1,6-5k)$. The equation of YZ-plane is x = 0 Since the line passes through YZ-plane, $\begin{array}{c}5-2k=0\\ \Rightarrow k=\frac{5}{9}\\ \Rightarrow 3k+1=3\times \frac{5}{2}+1=\frac{17}{2}\\ 6-5k=6-5\times \frac{5}{2}=\frac{-13}{2}\end{array}$ Therefore, the required point is $(0,\frac{17}{2},\frac{-13}{2})$.

Q11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

Answer. It is known that the equation of the line passing through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$ The line passing through the points (5, 1, 6) and (3, 4, 1) is given by, $\begin{array}{c}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ \Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k\left(\text{say}\right)\\ \Rightarrow x=5-2k,y=3k+1,z=6-5k\end{array}$ Any point on the line is of the form $(5-2k,3k+1,6-5k)$. Since the line passes through ZX-plane, $\begin{array}{c}3k+1=0\\ \Rightarrow k=-\frac{1}{3}\\ \Rightarrow 5-2k=5-2(-\frac{1}{3})=\frac{17}{3}\\ 6-5k=6-5(-\frac{1}{3})=\frac{23}{3}\end{array}$ Therefore, the required point is $(\frac{17}{3},0,\frac{23}{3})$.

Q12. 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. It is known that the equation of the line passing through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$ The line passes through the points (3, -4, -5) and (2, -3, 1), its equation is given by, $\begin{array}{c}\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\\ \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k\left(\text{say}\right)\\ \Rightarrow x=3-k,y=k-4,z=6k-5\end{array}$ Therefore, any point on the line is of the form $(3-k,k-4,6k-5)$ This point lies on the plane, 2x + y + z = 7 $\begin{array}{c}\therefore 2(3-k)+(k-4)+(6k-5)=7\\ \Rightarrow 5k-3=7\\ \Rightarrow k=2\end{array}$ Hence, the coordinates of the required point are $(3-2,2-4,6\times 2-5)$i.e., (1, -2, 7).

Q13. 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 the plane passing through the point (-1, 3, 2) is $a(x+1)+b(y-3)+c(z-2)=0\dots \left(1\right)$ where a, b, c are the direction ratios of normal to the plane. It is known that two planes, $a_{1}x+b_{1}y+c_{1}z+d_1=0\text{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0,$ are perpendicular, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$ Plane (1) is perpendicular to the plane, x + 2y + 3z = 5 $\begin{array}{c}\therefore a\cdot 1+b\cdot 2+c\cdot 3=0\\ \Rightarrow a+2b+3c=0\end{array}$ Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0 $\begin{array}{c}\therefore a\cdot 3+b\cdot 3+c\cdot 1=0\\ \Rightarrow 3a+3b+c=0\end{array}$ From equations (2) and (3) we obtain $\begin{array}{c}\frac{a}{2\times 1-3\times 3}=\frac{b}{3\times 3-1\times 1}=\frac{c}{1\times 3-2\times 3}\\ \Rightarrow \frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=k\left(\text{say}\right)\\ \Rightarrow a=-7k,b=8k,c=-3k\end{array}$ Substituting the values of a, b, c in equation (1), we obtain $\begin{array}{c}-7k(x+1)+8k(y-3)-3k(z-2)=0\\ \Rightarrow (-7x-7)+(8y-24)-3z+6=0\\ \Rightarrow -7x+8y-3z-25=0\\ \Rightarrow 7x-8y+3z+25=0\end{array}$ This is the required equation of the plane.

Q14. If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane $\overrightarrow{r}\cdot (3\hat{i}+4\hat{j}-12\hat{k})+13=0$ then find the value of p.

Answer. The position vector through the point (1, 1, p ) is ${\overrightarrow{a}}_1=\hat{i}+\hat{j}+p\hat{k}$ Similarly, the position vector through the point (-3, 0, 1) is ${\overrightarrow{a}}_2=-4\hat{i}+\hat{k}$ The equation of the given plane is $\overrightarrow{r}\cdot (3\hat{i}+4\hat{j}-12\hat{k})+13=0$ It is known that the perpendicular distance between a point whose position vector is ${\overline{a}}_{\text{and the plane,}}\overrightarrow{r}\cdot \overrightarrow{N}=d_{\text{is given by}}D=\frac{|\overrightarrow{a}\cdot \overrightarrow{N}-d|}{|\overrightarrow{N}|}$ Here, $\overline{N}=3\hat{i}+4\hat{j}-12\hat{k}\text{and}d=-13$ Therefore, the distance between the point (1, 1, p) and the given plane is $\begin{array}{c}{D}_1=\frac{\left(\right(\hat{i}+\hat{j}+p\hat{k})\cdot (3\hat{i}+4\hat{j}-12\hat{k})+13}{3\hat{i}+4\hat{j}-12\hat{k}}|\\ \Rightarrow D_1=\frac{|3+4-12p+13|}{\sqrt{3^2+4^2+(-12{)}^2}}\\ \Rightarrow D_1=\frac{|20-12p|}{13}\end{array}$ Similarly, the distance between the point (-3, 0, 1) and the given plane is $\begin{array}{c}{D}_2=\frac{|(-3\hat{i}+\hat{k})\cdot (3\hat{i}+4\hat{j}-12\hat{k})+13}{|3\hat{i}+4\hat{j}-12\hat{k}|}\\ \Rightarrow D_2=\frac{|-9-12+13|}{\sqrt{3^2+4^2+(-12{)}^2}}\\ \Rightarrow D_2=\frac{8}{13}\end{array}$ It is given that the distance between the required plane and the points (1, 1, p) and (-3, 0, 1) is equal. $\begin{array}{c}\therefore D_1=D_2\\ \Rightarrow \frac{|20-12p|}{13}=\frac{8}{13}\\ \Rightarrow 12-12p=8\text{or}-(20-12p)=8\\ \Rightarrow 12p=12\text{or}12p=28\\ \Rightarrow p=1\text{or}p=\frac{7}{3}\end{array}$.

Q15. Find the equation of the plane passing through the line of intersection of the planes $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=1\text{and}\overrightarrow{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4=0$ and parallel to x-axis.

Answer. The given planes are $\begin{array}{c}\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=1\\ \Rightarrow \overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})-1=0\\ \overrightarrow{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4=0\end{array}$ The equation of any plane passing through the line of intersection of these planes is $\begin{array}{c}[\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})-1]+\lambda [\overrightarrow{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4]=0\\ \overrightarrow{r}\cdot \left[\right(2\lambda +1)\hat{i}+(3\lambda +1)\hat{j}+(1-\lambda \left)\hat{k}\right]+(4\lambda +1)=0\end{array}$ Its direction ratios are $(2\lambda +1),(3\lambda +1),\text{and}(1-\lambda )$ The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis The direction ratios of x-axis are 1, 0, and 0. $\begin{array}{c}\therefore 1.(2\lambda +1)+0(3\lambda +1)+0(1-\lambda )=0\\ \Rightarrow 2\lambda +1=0\\ \Rightarrow \lambda =-\frac{1}{2}\end{array}$ Substituting $\lambda =-\frac{1}{2}$ in equation (1), we obtain $\begin{array}{c}\Rightarrow \overrightarrow{r}\cdot [-\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k}]+(-3)=0\\ \Rightarrow \overrightarrow{r}(\hat{j}-3\hat{k})+6=0\end{array}$ Therefore, its cartesian equation is y - 3z + 6 = 0 This is the equation of the required plane.

Q16. If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

Answer. The coordinates of the points, O and P are (0, 0, 0) and (1, 2, -3) respectively. Therefore, the direction ratios of OP are $(1-0)=1,(2-0)=2,\text{and}(-3-0)=-3$ It is known that the equation of the plane passing through the point $(x_1,y_1,z_1)$ is $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ where a, b, c are the direction ratios of normal. Here, the direction ratios of normal are 1, 2, and -3 and the point P is (1, 2, -3). Thus, the equation of the required plane is $\begin{array}{c}1(x-1)+2(y-2)-3(z+3)=0\\ \Rightarrow x+2y-3z-14=0\end{array}$

Q17. Find the equation of the plane which contains the line of intersection of the planes $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+3\hat{k})-4=0,\overrightarrow{r}\cdot (2\hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $\overrightarrow{r}\cdot (5\hat{i}+3\hat{j}-6\hat{k})+8=0$

Answer. The equations of the given planes are $\begin{array}{cc}\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+3\hat{k})-4=0& \dots \left(1\right)\\ \overrightarrow{r}\cdot (2\hat{i}+\hat{j}-\hat{k})+5=0& \dots \left(2\right)\end{array}$ The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is $\begin{array}{c}[\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+3\hat{k})-4]+\lambda [\overrightarrow{r}\cdot (2\hat{i}+\hat{j}-\hat{k})+5]=0\\ \overrightarrow{r}\cdot \left[\right(2\lambda +1)\hat{i}+(\lambda +2)\hat{j}+(3-\lambda \left)\hat{k}\right]+(5\lambda -4)=0\end{array}$ The plane in equation (3) is perpendicular to the plane, $\overrightarrow{r}\cdot (5\hat{i}+3\hat{j}-6\hat{k})+8=0$ $\begin{array}{c}\therefore 5(2\lambda +1)+3(\lambda +2)-6(3-\lambda )=0\\ \Rightarrow 19\lambda -7=0\\ \Rightarrow \lambda =\frac{7}{19}\end{array}$ Substituting $\lambda =\frac{7}{19}$ in equation (3), we obtain $\begin{array}{c}\Rightarrow \overrightarrow{r}\cdot [\frac{33}{19}\hat{i}+\frac{45}{19}\hat{j}+\frac{50}{19}\dot{k}]\frac{-41}{19}=0\\ \Rightarrow \overrightarrow{r}\cdot (33\hat{i}+45\hat{j}+50\hat{k})-41=0\end{array}$ This is the vector equation of the required plane. The Cartesian equation of this plane can be obtained by substituting $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in equation (3). $\begin{array}{c}(x\hat{i}+y\hat{j}+z\hat{k})\cdot (33\hat{i}+45\hat{j}+50\hat{k})-41=0\\ \Rightarrow 33x+45y+50z-41=0\end{array}$

Q18. Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line $\overrightarrow{r}=2\hat{i}-\hat{j}+2\hat{k}+\lambda (3\hat{i}+4\hat{j}+2\hat{k})\text{and the plane}\overrightarrow{r}\cdot (\hat{i}-\hat{j}+\hat{k})=5$.

Answer. The equation of the given line is $\overrightarrow{r}:=2\hat{i}-\hat{j}+2\hat{k}+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$ The equation of the given plane is $\overrightarrow{r}\cdot (\hat{i}-\hat{j}+\hat{k})=5$ Substituting the value of $\overrightarrow{r}$ from equation (1) in equation (2) , we obtain $\begin{array}{c}[2\hat{i}-\hat{j}+2\hat{k}+\lambda (3\hat{i}+4\hat{j}+2\hat{k}\left)\right]\cdot (i-\hat{j}+\hat{k})=5\\ \Rightarrow \left[\right(3\lambda +2)\hat{i}+(4\lambda -1)\hat{j}+(2\lambda +2\left)\hat{k}\right]\cdot (\hat{i}-\hat{j}+\hat{k})=5\\ \Rightarrow (3\lambda +2)-(4\lambda -1)+(2\lambda +2)=5\\ \Rightarrow \lambda =0\end{array}$ Substituting this value in equation (1), we obtain the equation of the line as $\overrightarrow{r}=2\hat{i}-\hat{j}+2\hat{k}$ This means that the position vector of the point of intersection of the line and the plane is $\overrightarrow{r}=2\hat{i}-\hat{j}+2\hat{k}$ This shows that the point of intersection of the given line and plane is given by the coordinates, $(2,-1,2).\text{The point is}(-1,-5,-10)$ The distance d between the points $(2,-1,2)\text{and}(-1,-5,-10)$ is $d=\sqrt{(-1-2{)}^2+(-5+1{)}^2+(-10-2{)}^2}=\sqrt{9+16+144}=\sqrt{169}=13$

Q19. Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes $\overrightarrow{r}\cdot (\hat{i}-\hat{j}+2\hat{k})=5\text{and}\overrightarrow{r}\cdot (3\hat{i}+\hat{j}+\hat{k})=6$.

Answer. Let the required line be parallel to vector $\overrightarrow{b}$ given by, $\overrightarrow{b}=b_1\hat{i}+b_2\hat{ȷ}+b_3\hat{k}$ The position vector of the point $(1,2,3)\text{is}\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$ The equation of line passing through (1, 2, 3) is $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$ The equation of line passing through (1, 2, 3) and parallel to $\overrightarrow{b}$ is given by, $\begin{array}{c}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \Rightarrow \overrightarrow{r}(\hat{i}+2\hat{j}+3\hat{k})+\lambda (b\hat{i}+b_2\hat{j}+b,\hat{k})\\ \text{The equations of the given planes are}\\ \overrightarrow{r}\cdot (\hat{i}-\hat{j}+2\hat{k})=5\\ \overrightarrow{r}\cdot (3\hat{i}+\hat{j}+\hat{k})=6\end{array}$ The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular. $\begin{array}{c}\Rightarrow (\hat{i}-\hat{j}+2\hat{k})\cdot \lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})=0\\ \Rightarrow \lambda (b_1-b_2+2b_3)=0\\ \Rightarrow b_1-b_2+2b_3=0\end{array}$ Similarly, $(3\hat{i}+\hat{j}+\hat{k})\cdot \lambda (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})=0$ $\begin{array}{c}\Rightarrow \lambda (3b_1+b_2+b_3)=0\\ \Rightarrow 3b_1+b_2+b_3=0\end{array}$ From equations (4) and (5), we obtain $\begin{array}{c}\frac{b_1}{(-1)\times 1-1\times 2}=\frac{b_2}{2\times 3-1\times 1}=\frac{b_3}{1\times 1-3(-1)}\\ \Rightarrow \frac{b_1}{-3}=\frac{b_2}{5}=\frac{b_3}{4}\end{array}$ Therefore, the direction ratios of $\overrightarrow{b}$ are -3, 5, and 4. $\therefore \overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}=-3\hat{i}+5\hat{j}+4\hat{k}$ Substituting the value of $\overrightarrow{b}$ in equation (1) we obtain $\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (-3\hat{i}+5\hat{j}+4\hat{k})$ This is the equation of the required line.

Q20. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: $\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\text{and}\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$.

Answer. Let the required line be parallel to the vector $\overrightarrow{b}$ given by, $\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ The position vector of the point (1, 2, -4) is $\overrightarrow{a}=\hat{i}+2\hat{j}-4\hat{k}$ The equation of the line passing through (1, 2, -4) and parallel to vector $\overrightarrow{b}$ is $\begin{array}{c}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \Rightarrow \overrightarrow{r}(\hat{i}+2\hat{j}-4\hat{k})+\lambda (b_i\hat{i}+b_2\hat{j}+b_j\hat{k})\end{array}$ The equations of the lines are $\begin{array}{c}\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\\ \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\end{array}$ Line (1) and line (2) are perpendicular to each other. $\therefore 3b_1-16b_2+7b_3=0$ Also, line (1) and line (3) are perpendicular to each other: $\therefore 3b_1+8b_2-5b_3=0$ From equations (4) and (5), we obtain $\begin{array}{c}\frac{b_1}{(-16)(-5)-8\times 7}=\frac{b_2}{7\times 3-3(-5)}=\frac{b_3}{3\times 8-3(-16)}\\ \Rightarrow \frac{b_1}{24}=\frac{b_2}{36}=\frac{b_1}{72}\\ \Rightarrow \frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}\end{array}$ $\begin{array}{c}\therefore \text{Direction ratios of}\overrightarrow{b}\text{are}2,3,\text{and}6\\ \therefore \overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}\end{array}$ Substituting $\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}$ in equation (1) we obtain $\overrightarrow{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$ This is the equation of the required line.

Q21. 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. The equation of a plane having intercepts a, b, c with x, y, and z axis respectively is given by, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ The distance (p) of the plane from the origin is given by, $p=\left|\frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2+{\left(\frac{1}{c}\right)}^2}}\right|$ $\begin{array}{cc}\Rightarrow p& \Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\\ \Rightarrow p^2& =\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\\ \Rightarrow \frac{1}{p^2}& =\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\end{array}$

Q22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is $\left(A\right)2\text{units}\text{(B) 4 units}\text{(C) 8 units}\text{(D)}\frac{2}{\sqrt{29}}$

Answer. The equations of the planes are $\begin{array}{c}2x+3y+4z=4\\ 4x+6y+8z=12\\ \Rightarrow 2x+3y+4z=6\end{array}$ It can be seen that the given planes are parallel. It is known that the distance between two parallel planes $ax+by+cz=d_1\text{and}ax+by+cz=$ $d_2$ is given by, $D=\left|\frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}\right|$ $\begin{array}{c}D=\left|\frac{6-4}{\sqrt{(2{)}^2+(3{)}^2+(4{)}^2}}\right|\\ D=\frac{2}{\sqrt{29}}\end{array}$ Thus, the distance between the lines is $\frac{2}{\sqrt{29}}$ units. Hence, the correct answer is D.

Q23. The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are $\begin{array}{cc}\text{(A) Perpendicular}& \text{(B) Parallel}\\ \text{(C) intersect y-axis}& \text{(D) passes through}(0,0,\frac{5}{4})\end{array}$

Answer. The equations of the planes are $\begin{array}{c}2x-y+4z=5\dots \left(1\right)\\ 5x-2.5y+10z=6\dots \left(2\right)\end{array}$ It can be seen that, $\begin{array}{cc}\frac{a_1}{a_2}& =\frac{2}{5}\\ \frac{b_1}{b_2}& =\frac{-1}{-2.5}=\frac{2}{5}\\ \frac{c_1}{c_2}& =\frac{4}{10}=\frac{2}{5}\\ \frac{a_1}{a_2}& =\frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$ Therefore, the given planes are parallel. Hence, the correct answer is B.