# NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 12 - Three Dimensional Geometry

Three Dimensional Geometry is the 11th chapter of NCERT Class 12 Mathematics. The chapter covers a few concepts from lines and planes in three dimensions. Concepts like equations of a line, plane and distance between them etc. A the end of the chapter the miscellaneous exercise comes. Here the NCERT solutions for Class 12 Maths chapter 11 miscellaneous exercise is given. The NCERT book class 12 maths chapter 11 miscellaneous exercise solutions covers questions related to all the concepts of the chapter. These Class 12 Maths chapter 11 miscellaneous solutions are designed by mathematics expert faculties. Along with the miscellaneous exercise, the following exercise is also present.

• Three Dimensional Geometry 11.1

• Three Dimensional Geometry 11.2

• Three Dimensional Geometry 11.3

## Three Dimensional Geometry Class 12th Chapter 11-Miscellaneous Exercise

Question:1 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).

We can assume the line joining the origin, be OA where $O(0,0,0)$ and the point $A(2,1,1)$ and PQ be the line joining the points $P(3,5,-1)$ and $Q(4,3,-1)$ .

Then the direction ratios of the line OA will be $(2-0),(1-0),\ and\ (1-0) = 2,1,1$ and that of line PQ will be

$(4-3),(3-5),\ and\ (-1+1) = 1,-2,0$

So to check whether line OA is perpendicular to line PQ then,

Applying the relation we know,

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

$\Rightarrow 2(1)+1(-2)+1(0) = 2-2+0 = 0$

Therefore OA is perpendicular to line PQ.

Question:2 If l 1 , m 1 , n 1 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 $m_{1}n_{2}-m_{2}n_{1}, n_{1}l_{2}-n_{2}l_{1},l_{1}m_{2}-l_{2}m_{1}$ .

Given that $l_1,m_1,n_1\ and\ l_2,m_2,n_2$ are the direction cosines of two mutually perpendicular lines.

Therefore, we have the relation:

$l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2} = 0$ .........................(1)

$l_{1}^2+m_{1}^2+n_{1}^2 =1\ and\ l_{2}^2+m_{2}^2+n_{2}^2 =1$ .............(2)

Now, let us assume $l,m,n$ be the new direction cosines of the lines which are perpendicular to the line with direction cosines. $l_1,m_1,n_1\ and\ l_2,m_2,n_2$

Therefore we have, $ll_{1}+mm_{1}+nn_{1} = 0 \and\ ll_{2}+mm_{2}+nn_{2} = 0$

Or, $\frac{l}{m_{1}n_{2}-m_{2}n_{1}} = \frac{m}{n_{1}l_{2}-n_{2}l_{1}} = \frac{n}{l_{1}m_{2}-l_{2}m_{1}}$

$\Rightarrow \frac{l^2}{(m_{1}n_{2}-m_{2}n_{1})^2} = \frac{m^2}{(n_{1}l_{2}-n_{2}l_{1})^2} = \frac{n^2}{(l_{1}m_{2}-l_{2}m_{1})^2}$

$\Rightarrow \frac{l^2+m^2+n^2}{(m_{1}n_{2}-m_{2}n_{1})^2 +(n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2}$ ......(3)

So, l,m,n are the direction cosines of the line.

where, $l^2+m^2+n^2 =1$ ........................(4)

Then we know that,

$\Rightarrow (l_{1}^2+m_{1}^2+n_{1}^2)(l_{2}^2+m_{2}^2+n_{2}^2) - (l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2})^2$

$= (m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2$

So, from the equation (1) and (2) we have,

$(1)(1) -(0) =(m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2$

Therefore, $(m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2 =1$ ..(5)

Now, we will substitute the values from the equation (4) and (5) in equation (3), to get

$\frac{l^2}{(m_{1}n_{2}-m_{2}n_{1})^2} = \frac{m^2}{(n_{1}l_{2}-n_{2}l_{1})^2} = \frac{n^2}{(l_{1}m_{2}-l_{2}m_{1})^2} =1$

Therefore we have the direction cosines of the required line as;

$l =m_{1}n_{2} - m_{2}n_{1}$

$m =n_{1}l_{2} - n_{2}l_{1}$

$n =l_{1}m_{2} - l_{2}m_{1}$

Question:3 Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Given direction ratios $a,b,c$ and $b-c,\ c-a,\ a-b$ .

Thus the angle between the lines A is given by;

$A = \left | \frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}.\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} \right |$

$\Rightarrow \cos A = 0$

$\Rightarrow A = \cos^{-1}(0) = 90^{\circ}$ a

Thus, the angle between the lines is $90^{\circ}$

Question:4 Find the equation of a line parallel to x-axis and passing through the origin.

Equation of a line parallel to the x-axis and passing through the origin $(0,0,0)$ is itself x-axis .

So, let A be a point on the x-axis.

Therefore, the coordinates of A are given by $(a,0,0)$ , where $a\epsilon R$ .

Now, the direction ratios of OA are $(a-0) =a,0 , 0$

So, the equation of OA is given by,

$\frac{x-0}{a} = \frac{y-0}{0} = \frac{z-0}{0}$

or $\Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a$

Thus, the equation of the line parallel to the x-axis and passing through origin is

$\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$

Question:5 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.

Direction ratios of AB are $(4-1),(5-2),(7-3) = 3,3,4$

and Direction ratios of CD are $(2-(-4)), (9-3), (2-(-6)) = 6,6,8$

So, it can be noticed 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, we can easily say the angle between AB and CD which is either $0^{\circ}\ or\ 180^{\circ}$ .

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

Given both lines are perpendicular so we have the relation; $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

For the two lines whose direction ratios are known,

$a_{1},b_{1},c_{1}\ and\ a_{2},b_{2},c_{2}$

We have the direction ratios of the lines, $\frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are $-3,2k,2$ and $3k,1,-5$ respectively.

Therefore applying the formula,

$-3(3k)+2k(1)+2(-5) = 0$

$\Rightarrow -9k +2k -10 = 0$

$\Rightarrow7k=-10$ or $k= \frac{-10}{7}$

$\therefore$ For, $k= \frac{-10}{7}$ the lines are perpendicular.

Question:7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0$

Given that the plane is passing through the point $A (1,2,3)$ so, the position vector of the point A is $\vec{r_{A}} = \widehat{i}+2\widehat{j}+3\widehat{k}$ and perpendicular to the plane $\overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0$ whose direction ratios are $1,2,\ and\ -5$ and the normal vector is $\vec{n} = \widehat{i}+2\widehat{j}-5\widehat{k}$

So, the equation of a line passing through a point and perpendicular to the given plane is given by,

$\vec{l} = \vec{r} + \lambda\vec{n}$ , where $\lambda \epsilon R$

$\Rightarrow \vec{l} = (\widehat{i}+2\widehat{j}+3\widehat{k}) + \lambda(\widehat{i}+2\widehat{j}-5\widehat{k})$

Question:8 Find the equation of the plane passing through (a, b, c) and parallel to the plane $\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2$ .

Given that the plane is passing through $(a,b,c)$ and is parallel to the plane $\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2$

So, we have

The position vector of the point $A(a,b,c)$ is, $\vec{r_{A}} = a\widehat{i}+b\widehat{j}+c\widehat{k}$

and any plane which is parallel to the plane, $\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2$ is of the form,

$\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=\lambda$ . .......................(1)

Therefore the equation we get,

$( a\widehat{i}+b\widehat{j}+c\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=\lambda$

Or, $a+b+c = \lambda$

So, now substituting the value of $\lambda = a+b+c$ in equation (1), we get

$\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c$ .................(2)

So, this is the required equation of the plane .

Now, substituting $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$ in equation (2), we get

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c$

Or, $x+y+z = a+b+c$

Question:9 Find the shortest distance between lines $\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k})$ and $\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k})$ .

Given lines are;

$\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k})$ and

$\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k})$

So, we can find the shortest distance between two lines $\vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}}$ and $\vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}}$ by the formula,

$d = \left | \frac{(\vec{b_{1}}\times\vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})}{\left | \vec{b_{1}}\times\vec{b_{2}} \right |} \right |$ ...........................(1)

Now, we have from the comparisons of the given equations of lines.

$\vec{a_{1}} = 6\widehat{i}+2\widehat{j}+2\widehat{k}$ $\vec{b_{1}} = \widehat{i}-2\widehat{j}+2\widehat{k}$

$\vec{a_{2}} = -4\widehat{i}-\widehat{k}$ $\vec{b_{2}} = 3\widehat{i}-2\widehat{j}-2\widehat{k}$

So, $\vec{a_{2}} -\vec{a_{1}} = (-4\widehat{i}-\widehat{k}) -(6\widehat{i}+2\widehat{j}+2\widehat{k}) = -10\widehat{i}-2\widehat{j}-3\widehat{k}$

and $\Rightarrow \vec{b_{1}}\times \vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 &-2 &2 \\ 3& -2 &-2 \end{vmatrix} = (4+4)\widehat{i}-(-2-6)\widehat{j}+(-2+6)\widehat{k}$

$=8\widehat{i}+8\widehat{j}+4\widehat{k}$

$\therefore \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{8^2+8^2+4^2} =12$

$(\vec{b_{1}}\times \vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}}) = (8\widehat{i}+8\widehat{j}+4\widehat{k}).(-10\widehat{i}-2\widehat{j}-3\widehat{k}) = -80-16-12 =-108$ Now, substituting all values in equation (3) we get,

$d = | \frac{-108}{12}| = 9$

Hence the shortest distance between the two given lines is 9 units.

Question:10 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane.

We know that the equation of the line that passes through the points $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ is given by the relation;

$\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}$

and the line passing through the points, $\frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}$

$\implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k\ (say)$

$\implies x=5-2k,\ y=3k+1,\ z=6-5k$

And any point on the line is of the form $(5-2k,3k+1,6-5k)$ .

So, the equation of the YZ plane is $x=0$

Since the line passes through YZ- plane,

we have then,

$5-2k = 0$

$\Rightarrow k =\frac{5}{2}$

or $3k+1 = 3(\frac{5}{2})+1 = \frac{17}{2}$ and $6-5k= 6-5(\frac{5}{2}) = \frac{-13}{2}$

So, therefore the required point is $\left ( 0,\frac{17}{2},\frac{-13}{2} \right )$

Question : 11 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

We know that the equation of the line that passes through the points $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ is given by the relation;

$\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}$

and the line passing through the points, $\frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}$

$\implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k\ (say)$

$\implies x=5-2k,\ y=3k+1,\ z=6-5k$

And any point on the line is of the form $(5-2k,3k+1,6-5k)$ .

So, the equation of ZX plane is $y=0$

Since the line passes through YZ- plane,

we have then,

$3k+1 = 0$

$\Rightarrow k =-\frac{1}{3}$

or $5-2k = 5-2\left ( -\frac{1}{3} \right ) = \frac{17}{3}$ and $6-5k= 6-5(\frac{-1}{3}) = \frac{23}{3}$

So, therefore the required point is $\left ( \frac{17}{3},0,\frac{23}{3} \right )$

Question:12 Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

We know that the equation of the line that passes through the points $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ is given by the relation;

$\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}$

and the line passing through the points, $(3,-4,-5)\ and\ (2,-3,1)$ .

$\implies \frac{x-3}{2-3} = \frac{y+4}{-3+4} = \frac{z+5}{1+5} = k\ (say)$

$\implies \frac{x-3}{-1} = \frac{y+4}{-1} = \frac{z+5}{6} = k\ (say)$

$\implies x=3-k,\ y=k-4,\ z=6k-5$

And any point on the line is of the form. $(3-k,k-4,6k-5)$

This point lies on the plane, $2x+y+z = 7$

$\therefore 2(3-k)+(k-4)+(6k-5) = 7$

$\implies 5k-3=7$

or $k =2$ .

Hence, the coordinates of the required point are $(3-2,2-4,6(2)-5)$ or $(1,-2,7)$ .

Question:13 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.

Given

two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

the normal vectors of these plane are

$n_1=\hat i+2\hat j+ 3\hat k$

$n_2=3\hat i+3\hat j+ \hat k$

Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :

$\vec n = \vec n_1\times\vec n_2$

$\vec n = (\hat i+2\hat j +3\hat k )\times (3\hat i + 3\hat j +\hat k)$

$\vec n =\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ 3& 3& 1 \end{vmatrix}=\hat i(2-9)-\hat j(1-9)+\hat k (3-6)$

$\vec n =-7\hat i+8\hat j-3\hat k$

Now, as we know

the equation of a plane in vector form is :

$\vec r\cdot\vec n=d$

$\vec r\cdot(-7\hat i+8\hat j-3\hat k)=d$

Now Since this plane passes through the point (-1,3,2)

$(-\hat i+3\hat j+2\hat k)\cdot(-7\hat i+8\hat j-3\hat k)=d$

$7+24-6=d$

$d=25$

Hence the equation of the plane is

$\vec r\cdot(-7\hat i+8\hat j-3\hat k)=25$

Question:14 If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane $\overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0$ then find the value of p.

Given that the points $A(1,1,p)$ and $B(-3,0,1)$ are equidistant from the plane

$\overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0$

So we can write the position vector through the point $(1,1,p)$ is $\vec{a_{1}} = \widehat{i}+\widehat{j}+p\widehat{k}$

Similarly, the position vector through the point $B(-3,0,1)$ is

$\vec{a_{2}} = -4\widehat{i}+\widehat{k}$

The equation of the given plane is $\overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0$

and We know that the perpendicular distance between a point whose position vector is $\vec{a}$ and the plane, $\vec{n} = 3\widehat{i}+4\widehat{j}-12\widehat{k}$ and $d =-13$

Therefore, the distance between the point $A(1,1,p)$ and the given plane is

$D_{1} = \frac{\left | (\widehat{i}+\widehat{j}+p\widehat{k}).(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 \right |}{3\widehat{i}+4\widehat{j}-12\widehat{k}}$

$D_{1} = \frac{\left | 3+4-12p+13 \right |}{\sqrt{3^2+4^2+(-12)^2}}$

$D_{1} = \frac{\left | 20-12p \right |}{13}$ nbsp; .........................(1)

Similarly, the distance between the point $B(-1,0,1)$ , and the given plane is

$D_{2} = \frac{\left | (-3\widehat{i}+\widehat{k}).(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 \right |}{3\widehat{i}+4\widehat{j}-12\widehat{k}}$

$D_{2} = \frac{\left |-9-12+13 \right |}{\sqrt{3^2+4^2+(-12)^2}}$

$D_{2} = \frac{8}{13}$ .........................(2)

And it is given that the distance between the required plane and the points, $A(1,1,p)$ and $B(-3,0,1)$ is equal.

$\therefore D_{1} =D_{2}$

$\implies \frac{\left | 20-12p \right |}{13} =\frac{8}{13}$

therefore we have,

$\implies 12p =12$

or $p =1$ or $p = \frac{7}{3}$

Question:15 Find the equation of the plane passing through the line of intersection of the planes $\overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )=1$ and $\overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4=0$ and parallel to x-axis.

So, the given planes are:

$\overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )=1$ and $\overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4=0$

The equation of any plane passing through the line of intersection of these planes is

$[\overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )-1] + \lambda \left [ \overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4\right ] = 0$

$\vec{r}.[(2\lambda+1)\widehat{i}+(3\lambda+1)\widehat{j}+(1-\lambda)\widehat{k}]+(4\lambda+1) = 0$ ..............(1)

Its direction ratios are $(2\lambda+1) , (3\lambda+1),$ and $(1-\lambda)$ = 0

The required plane is parallel to the x-axis.

Therefore, its normal is perpendicular to the x-axis.

The direction ratios of the x-axis are 1,0, and 0.

$\therefore \1.(2\lambda+1) + 0(\3\lambda+1)+0(1-\lambda) = 0$

$\implies 2\lambda+1 = 0$

$\implies \lambda = -\frac{1}{2}$

Substituting $\lambda = -\frac{1}{2}$ in equation (1), we obtain

$\implies \vec{r}.\left [ -\frac{1}{2}\widehat{j}+\frac{3}{2}\widehat{k} \right ]+(-3)= 0$

$\implies \vec{r}(\widehat{j}-3\widehat{k})+6= 0$

So, the Cartesian equation is $y -3z+6 = 0$

Question:16 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.

We have the coordinates of the points $O(0,0,0)$ and $P(1,2,-3)$ respectively.

Therefore, the direction ratios of OP are $(1-0) = 1, (2-0)=2,\ and\ (-3-0)=-3$

And we know 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

$1(x-1)+2(y-2)-3(z+3) = 0$

$\implies x+2y -3z-14 = 0$

Question:17 Find the equation of the plane which contains the line of intersection of the planes $\dpi{100} \overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0$ and which is perpendicular to the plane $\overrightarrow{r}.(5\widehat{i}+3\widehat{j}-6\widehat{k})+8=0$

The equation of the plane passing through the line of intersection of the given plane in $\overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0$

$\left [ \vec{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4 \right ]+\lambda\left [ \vec{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5 \right ] = 0$

$\vec{r}.[(2\lambda+1)\widehat{i}+(\lambda+2)\widehat{j}+(3-\lambda)\widehat{k}]+(5\lambda-4)= 0$ ,,,,,,,,,,,,,(1)

The plane in equation (1) is perpendicular to the plane, Therefore $5(2\lambda+1)+3(\lambda+2) -6(3-\lambda) = 0$

$\implies 19\lambda -7 = 0$

$\implies \lambda = \frac{7}{19}$

Substituting $\lambda = \frac{7}{19}$ in equation (1), we obtain

$\implies \vec{r}.\left [ \frac{33}{19}\widehat{i}+\frac{45}{19}\widehat{j}+\frac{50}{19}\widehat{k} \right ] -\frac{41}{19} = 0$

$\implies \vec{r}.(33\widehat{i}+45\widehat{j}+50\widehat{k}) -41 = 0$ .......................(4)

So, this is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting $\implies \vec{r}= (x\widehat{i}+y\widehat{j}+z\widehat{k})$ in equation (1).

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).(33\widehat{i}+45\widehat{j}+50\widehat{k}) -41 = 0$

Therefore we get the answer $33x+45y+50z -41 = 0$

Question:18 Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line $\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )$ and the plane $\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5$ .

Given,

Equation of a line :

$\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )$

Equation of the plane

$\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5$

Let's first find out the point of intersection of line and plane.

putting the value of $\vec r$ into the equation of a plane from the equation from line

$\left (2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) \right )\cdot (\hat i-\hat j+\hat k)=5$

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

$\lambda+5=5$

$\lambda=0$

Now, from the equation, any point p in line is

$P=(2+3\lambda,4\lambda-1,2+2\lambda)$

So the point of intersection is

$P=(2+3*0,4*0-1,2+2*0)=(2,-1,2)$

SO, Now,

The distance between the points (-1,-5,-10) and (2,-1,2) is

$d=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{9+16+144}$

$d=\sqrt{169}=13$

Hence the required distance is 13.

Question:19 Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes $\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5$ and $\overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6$ .

Given

A point through which line passes

$\vec a=\hat i+2\hat j+3\hat k$

two plane

$\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5$ And

$\overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6$

it can be seen that normals of the planes are

$\vec n_1=\hat i-\hat j+2\hat k$

$\vec n_2=3\hat i+\hat j+\hat k$
since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes.

So, a vector perpendicular to both these normal vector is

$\vec d=\vec n_1\times\vec n_2$

$\vec d=\begin{vmatrix} \hat i &\hat j & \hat k\\ 1 &-1 &2 \\ 3& 1 & 1 \end{vmatrix}=\hat i(-1-2)-\hat j(1-6)+\hat k(1+3)$

$\vec d=-3\hat i+5\hat j+4\hat k$

Now a line which passes through $\vec a$ and parallels to $\vec d$ is

$L=\vec a+\lambda\vec d$

So the required line is

$L=\vec a+\lambda\vec d$

$L=\hat i+2\hat j+3\hat k+\lambda(-3\hat i+5\hat j+4\hat k)$

$L=(1-3\lambda)\hat i+(2+5\lambda)\hat j+(3+4\lambda)\hat k$

Questio n: 20 Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$

Given

Two straight lines in 3D whose direction cosines (3,-16,7) and (3,8,-5)

Now the two vectors which are parallel to the two lines are

$\vec a= 3\hat i-16\hat j+7\hat k$ and

$\vec b= 3\hat i+8\hat j-5\hat k$

As we know, a vector perpendicular to both vectors $\vec a$ and $\vec b$ is $\vec a\times\vec b$ , so

$\vec a\times\vec b=\begin{vmatrix} \hat i &\hat j &\hat k \\ 3& -16 &7 \\ 3&8 &-5 \end{vmatrix}=\hat i(80-56)-\hat j(-15-21)+\hat k(24+48)$

$\vec a\times\vec b=24\hat i+36\hat j+72\hat k$

A vector parallel to this vector is

$\vec d=2\hat i+3\hat j+6\hat k$

Now as we know the vector equation of the line which passes through point p and parallel to vector d is

$L=\vec p+\lambda \vec d$

Here in our question, give point p = (1,2,-4) which means position vector of this point is

$\vec p = \hat i +2\hat j-4\hat k$

So, the required line is

$L=\vec p+\lambda \vec d$

$L=\hat i+2\hat j-4\hat k +\lambda (2\hat i+3\hat j+6\hat k)$

$L=(2\lambda +1)\hat i+(2+3\lambda)\hat j+(6\lambda-4)\hat k$

Question:21 Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}$ .

The equation of plane having a, b and c intercepts 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
$\\p = \left | \frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2(\frac{1}{c})^2}} \right |\\ \\ p = \left | \frac{-1}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2(\frac{1}{c})^2}} \right |\\ \\ \frac{1}{p^2}= \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$
Hence proved

Question:22 Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

(A) 2 units (B) 4 units (C) 8 units (D) $\dpi{80} \frac{2}{\sqrt{29}}unit$

Given equations are
$2x+3y+4z= 4 \ \ \ \ \ \ \ \ \ -(i)$
and
$4x+6y+8z= 12\\ 2(2x+3y+4z)= 12\\ 2x+3y+4z = 6 \ \ \ \ \ \ \ \ \ \ -(ii)$
Now, it is clear from equation (i) and (ii) that given planes are parallel
We know that the distance between two parallel planes $ax+by +cz = d_1 \ and \ ax+by +cz = d_2$ is given by
$D= \left | \frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}} \right |$
Put the values in this equation
we will get,
$D= \left | \frac{6-4}{\sqrt{2^2+3^2+4^2}} \right |$
$D= \left | \frac{2}{\sqrt{4+9+16}} \right |= \left | \frac{2}{\sqrt{29}} \right |$
Therefore, the correct answer is (D)

Question:23 The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

(A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through $\left ( 0,0,\frac{5}{4} \right )$

Given equations of planes are
$2x-y+4z=5 \ \ \ \ \ \ \ \ \ -(i)$
and
$5x-2.5y+10z=6\\ 2.5(2x-y+4z)=6\\ 2x-y+4z= 2.4 \ \ \ \ \ \ \ \ \ -(ii)$
Now, from equation (i) and (ii) it is clear that given planes are parallel to each other
$\because \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow \frac{2}{2}= \frac{-1}{-1}=\frac{4}{4}\Rightarrow 1=1=1$
Therefore, the correct answer is (B)

## More About NCERT Solutions for Class 12 Maths Chapter 11 Miscellaneous Exercise

Twenty-three questions are there in the Class 12 Maths chapter 11 miscellaneous solutions. Miscellaneous exercise chapter 11 Class 12 are important to understand the concepts explained in the chapter. The NCERT syllabus for Class 12 Maths chapter 11 miscellaneous exercise covers all the concepts of the chapter.

Also Read| Three Dimensional Geometry Class 12th Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 11 Miscellaneous Exercise

• To practise the complete chapter the miscellaneous exercise chapter 11 Class 12 is useful

• To practice for the board exam the NCERT solutions for Class 12 Maths chapter 11 miscellaneous exercise is helpful.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 11

• NCERT Solutions for Class 12 Maths Chapter 11

## NCERT Solutions Subject Wise

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