# NCERT Solutions for Exercise 11.3 Class 12 Maths Chapter 11- Three Dimensional Geometry

NCERT solutions for exercise 11.3 Class 12 Maths chapter 11 move around the topic plane. The questions in NCERT solutions for Class 12 Maths chapter 11 exercise 11.3 are related to exercise 11.3 Class 12 Maths equation of a plane in different conditions, the concept of coplanarity of two lines, the angle between two planes and the exercise 11.3 Class 12 Maths also covers the distance between a point and a plane. One should grasp the concepts well before solving Class 12 Maths chapter 11 exercise 11.3. And to get more idea about steps involved in solving the problems under the topic plane, one can go through the solved example given in the NCERT and then crack the Class 12th Maths chapter 11 exercise 11.3. The below-mentioned exercises are also present in the NCERT Book.

• Dimensional geometry 11.1

• Three Dimensional geometry 11.2

• Three Dimensional geometry miscellaneous exercise

## Three Dimensional Geometry Class 12th Chapter 11-Exercise: 11.3

Question:1(a) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2

Equation of plane Z=2, i.e. $0x+0y+z=2$

The direction ratio of normal is 0,0,1

$\therefore \, \, \, \sqrt{0^2+0^2+1^2}=1$

Divide equation $0x+0y+z=2$ by 1 from both side

We get, $0x+0y+z=2$

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

Question:1(b) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

x + y + z = 1

Given the equation of the plane is $x+y+z=1$ or we can write $1x+1y+1z=1$

So, the direction ratios of normal from the above equation are, $1,\1,\ and\ 1$ .

Therefore $\sqrt{1^2+1^2+1^2} =\sqrt{3}$

Then dividing both sides of the plane equation by $\sqrt{3}$ , we get

$\frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3}=\frac{1}{\sqrt3}$

So, this is the form of $lx+my+nz = d$ the plane, where $l,\ m,\ n$ are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

$\therefore$ The direction cosines of the given line are $\frac{1}{\sqrt3},\ \frac{1}{\sqrt3},\ \frac{1}{\sqrt3}$ and the distance of the plane from the origin is $\frac{1}{\sqrt3}$ units.

Question:1(c) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

2x + 3y - z = 5

Given the equation of plane is $2x+3y-z=5$

So, the direction ratios of normal from the above equation are, $2,\3,\ and\ -1$ .

Therefore $\sqrt{2^2+3^2+(-1)^2} =\sqrt{14}$

Then dividing both sides of the plane equation by $\sqrt{14}$ , we get

$\frac{2x}{\sqrt{14}}+\frac{3y}{\sqrt{14}}-\frac{z}{\sqrt{14}}=\frac{5}{\sqrt{14}}$

So, this is the form of $lx+my+nz = d$ the plane, where $l,\ m,\ n$ are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

$\therefore$ The direction cosines of the given line are $\frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}},\ \frac{-1}{\sqrt{14}}$ and the distance of the plane from the origin is $\frac{5}{\sqrt{14}}$ units.

Question:1(d) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

5y + 8 = 0

Given the equation of plane is $5y+8=0$ or we can write $0x-5y+0z=8$

So, the direction ratios of normal from the above equation are, $0,\ -5,\ and\ 0$ .

Therefore $\sqrt{0^2+(-5)^2+0^2} =5$

Then dividing both sides of the plane equation by $5$ , we get

$-y = \frac{8}{5}$

So, this is the form of $lx+my+nz = d$ the plane, where $l,\ m,\ n$ are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

$\therefore$ The direction cosines of the given line are $0,\ -1,\ and\ 0$ and the distance of the plane from the origin is $\frac{8}{5}$ units.

Question:2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\widehat{i}+5\widehat{j}-6\widehat{k}$ .

We have given the distance between the plane and origin equal to 7 units and normal to the vector $3\widehat{i}+5\widehat{j}-6\widehat{k}$ .

So, it is known that the equation of the plane with position vector $\vec{r}$ is given by, the relation,

$\vec{r}.\widehat{n} =d$ , where d is the distance of the plane from the origin.

Calculating $\widehat{n}$ ;

$\widehat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{(3)^2+(5)^2+(6)^2}} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}}$

$\vec{r}.\left ( \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}} \right ) = 7$ is the vector equation of the required plane.

Question:3(a) Find the Cartesian equation of the following planes:

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

Given the equation of the plane $\overrightarrow{r}.(\widehat{i}+\widehat{j}-\widehat{k})=2$

So we have to find the Cartesian equation,

Any point $A (x,y,z)$ on this plane will satisfy the equation and its position vector given by,

$\vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}$

Hence we have,

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).(\widehat{i}+\widehat{j}-\widehat{k}) =2$

Or, $x+y-z=2$

Therefore this is the required Cartesian equation of the plane.

Question:3(b) Find the Cartesian equation of the following planes:

$\overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1$

Given the equation of plane $\overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1$

So we have to find the Cartesian equation,

Any point $A (x,y,z)$ on this plane will satisfy the equation and its position vector given by,

$\vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}$

Hence we have,

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).(2\widehat{i}+3\widehat{j}-4\widehat{k}) =1$

Or, $2x+3y-4z=1$

Therefore this is the required Cartesian equation of the plane.

Question:3(c) Find the Cartesian equation of the following planes:

$\overrightarrow{r}.\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right ]=15$

Given the equation of plane $\overrightarrow{r}.\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right ]=15$

So we have to find the Cartesian equation,

Any point $A (x,y,z)$ on this plane will satisfy the equation and its position vector given by, $\vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}$

Hence we have,

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right] =15$

Or, $(s-2t)x+(3-t)y+(2s+t)z=15$

Therefore this is the required Cartesian equation of the plane.

Question:4(a) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2 x + 3y + 4 z - 12 = 0

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given a plane equation $2x+3y+4z-12=0$ ,

Or, $2x+3y+4z=12$

The direction ratios of the normal of the plane are 2, 3 and 4 .

Therefore $\sqrt{(2)^2+(3)^2+(4)^2} = \sqrt{29}$

So, now dividing both sides of the equation by $\sqrt{29}$ we will obtain,

$\frac{2}{\sqrt{29}}x+\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z = \frac{12}{\sqrt{29}}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$ .

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left [ \frac{2}{\sqrt{29}}.\frac{12}{\sqrt{29}},\frac{3}{\sqrt{29}}.\frac{12}{\sqrt{29}},\frac{4}{\sqrt{29}}.\frac{12}{\sqrt{29}} \right ]$ or $\left [ \frac{24}{29}, \frac{36}{49}, \frac{48}{29} \right ]$

Question:4(b) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

3y + 4z - 6 = 0

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given a plane equation $3y+4z-6=0$ ,

Or, $0x+3y+4z=6$

The direction ratios of the normal of the plane are 0,3 and 4 .

Therefore $\sqrt{(0)^2+(3)^2+(4)^2} = 5$

So, now dividing both sides of the equation by $5$ we will obtain,

$0x+\frac{3}{5}y+\frac{4}{5}z = \frac{6}{5}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$ .

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left (0,\frac{3}{5}.\frac{6}{5},\frac{4}{5}.\frac{6}{5} \right )$ or $\left ( 0, \frac{18}{25}, \frac{24}{25} \right )$

Question:4(c) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

x + y + z = 1

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given plane equation $x+y+z=1$ .

The direction ratios of the normal of the plane are 1,1 and 1 .

Therefore $\sqrt{(1)^2+(1)^2+(1)^2} = \sqrt3$

So, now dividing both sides of the equation by $\sqrt3$ we will obtain,

$\frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3} = \frac{1}{\sqrt3}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$ .

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left ( \frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3} \right )$ or $\left ( \frac{1}{3},\frac{1}{3},\frac{1}{3} \right )$ ..

Question: 4(d) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

5y + 8 = 0

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given plane equation $5y+8=0$ .

or written as $0x-5y+0z=8$

The direction ratios of the normal of the plane are 0, -5 and 0 .

Therefore $\sqrt{(0)^2+(-5)^2+(0)^2} = 5$

So, now dividing both sides of the equation by $5$ we will obtain,

$-y=\frac{8}{5}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$ .

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left ( 0,-1(\frac{8}{5}),0 \right )$ or $\left ( 0,\frac{-8}{5},0 \right )$ .

Question:5(a) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane is $\widehat{i}+\widehat{j}-\widehat{k}.$

Given the point $A (1,0,-2)$ and the normal vector $\widehat{n}$ which is perpendicular to the plane is $\widehat{n} = \widehat{i}+\widehat{j}-\widehat{k}$

The position vector of point A is $\vec {a} = \widehat{i}-2\widehat{k}$

So, the vector equation of the plane would be given by,

$(\vec{r}-\vec{a}).\widehat{n} = 0$

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

where $\vec{r}$ is the position vector of any arbitrary point $A(x,y,z)$ in the plane.

$\therefore$ $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$

Therefore, the equation we get,

$\left [(x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}-2\widehat{k}) \right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0$

$\Rightarrow \left [(x-1)\widehat{i}+y\widehat{j}+(z+2)\widehat{k}\right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0$

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

$\Rightarrow x+y-z-3=0$ or $x+y-z=3$

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

Question:5(b) Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is $\widehat{i}-2\widehat{j}+\widehat{k}$ .

Given the point $A (1,4,6)$ and the normal vector $\widehat{n}$ which is perpendicular to the plane is $\widehat{n} = \widehat{i}-2\widehat{j}+\widehat{k}$

The position vector of point A is $\vec {a} = \widehat{i}+4\widehat{j}+6\widehat{k}$

So, the vector equation of the plane would be given by,

$(\vec{r}-\vec{a}).\widehat{n} = 0$

Or $\left [ \vec{r}-(\widehat{i}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0$

where $\vec{r}$ is the position vector of any arbitrary point $A(x,y,z)$ in the plane.

$\therefore$ $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$

Therefore, the equation we get,

$\left [ (x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0$

$\Rightarrow \left [(x-1)\widehat{i}+(y-4)\widehat{j}+(z-6)\widehat{k}\right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0$

$(x-1)-2(y-4)+(z-6)=0$

$\Rightarrow x-2y+z+1=0$

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

Question:6(a) Find the equations of the planes that passes through three points.

(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)

The equation of the plane which passes through the three points $A(1,1,-1),\ B(6,4,-5),\ and\ C(-4,-2,3)$ is given by;

Determinant method,

$\begin{vmatrix} 1 &1 &-1 \\ 6& 4 & -5\\ -4& -2 &3 \end{vmatrix} = (12-10)-(18-20)-(-12+16)$

Or, $= 2+2-4 = 0$

Here, these three points A, B, C are collinear points.

Hence there will be an infinite number of planes possible which passing through the given points.

Question:6(b) Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

The equation of the plane which passes through the three points $A(1,1,0),\ B(1,2,1),\ and\ C(-2,2,-1)$ is given by;

Determinant method,

$\begin{vmatrix} 1 &1 &0 \\ 1& 2 & 1\\ -2& 2 &-1 \end{vmatrix} = (-2-2)-(2+2)= -8 \neq 0$

As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.

Finding the equation of the plane through the points, $(x_{1},y_{1},z_{1}), (x_{2},y_{2},z_{2})\ and\ (x_{3},y_{3},z_{3})$

$\begin{vmatrix} x-x_{1} &y-y_{1} &z-z_{1} \\ x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ x_{3}-x_{1}&y_{3}-y_{1} &z_{3}-z_{1} \end{vmatrix} = 0$

After substituting the values in the determinant we get,

$\begin{vmatrix} x-1 &y-1 &z \\ 0& 1 &1 \\ -3& 1&-1 \end{vmatrix} = 0$

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

$\Rightarrow-2x+2-3y+3+3z = 0$

$2x+3y-3z = 5$

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

Question:7 Find the intercepts cut off by the plane 2x + y – z = 5.

Given plane $2x + y-z = 5$

We have to find the intercepts that this plane would make so,

Making it look like intercept form first:

By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,

$\frac{2}{5}x+\frac{y}{5}-\frac{z}{5} =1$

$\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5} =1$

So, as we know that from the equation of a plane in intercept form, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1$ where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.

Therefore after comparison, we get the values of a,b, and c.

$a = \frac{5}{2},\ b=5,\ and\ c=-5$ .

Hence the intercepts are $\frac{5}{2},\ 5,\ and\ -5$ .

Question:8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Given that the plane is parallel to the ZOX plane.

So, we have the equation of plane ZOX as $y = 0$ .

And an intercept of 3 on the y-axis $\Rightarrow b =3$

Intercept form of a plane given by;

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1$

So, here the plane would be parallel to the x and z-axes both.

we have any plane parallel to it is of the form, $y=a$ .

Equation of the plane required is $y=3$ .

Question:9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

The equation of any plane through the intersection of the planes,

$3x-y+2z-4=0\ and\ x+y+z-2=0$

Can be written in the form of; $(3x-y+2z-4)\ +\alpha( x+y+z-2)= 0$ , where $\alpha \epsilon R$

So, the plane passes through the point $(2,2,1)$ , will satisfy the above equation.

$(3\times2-2+2\times1-4)+\alpha(2+2+1-2) = 0$

That implies $2+3\alpha= 0$

$\alpha = \frac{-2}{3}$

Now, substituting the value of $\alpha$ in the equation above we get the final equation of the plane;

$(3x-y+2z-4)\ +\alpha( x+y+z-2)= 0$

$(3x-y+2z-4)\ +\frac{-2}{3}( x+y+z-2)= 0$

$\Rightarrow 9x-3y+6z-12\ -2 x-2y-2z+4= 0$

$\Rightarrow 7x-5y+4z-8= 0$ is the required equation of the plane.

Question:10 Find the vector equation of the plane passing through the intersection of the planes $\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})=7$ , $\overrightarrow{r}(2\widehat{i}+5\widehat{j}+3\widehat{k})=9$ and through the point (2, 1, 3).

Here $\vec{n_{1}} =2 \widehat{i}+2\widehat{j}-3\widehat{k}$ and $\vec{n_{2}} = 2\widehat{i}+5\widehat{j}+3\widehat{k}$

and $d_{1} = 7$ and $d_{2} = 9$

Hence, using the relation $\vec{r}.(\vec{n_{1}}+\lambda\vec{n_{2}}) = d_{1}+\lambda d_{2}$ , we get

$\vec{r}.[2\widehat{i}+2\widehat{j}-3\widehat{k}+\lambda(2\widehat{i}+5\widehat{j}+3\widehat{k})] = 7+9\lambda$

or $\vec{r}.[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda$ ..............(1)

where, $\lambda$ is some real number.

Taking $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$ , we get

$(\vec{x\widehat{i}+y\widehat{j}+z\widehat{k}}).[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda$

or $x(2+2\lambda) + y(2+5\lambda) +z(3\lambda-3) = 7+9\lambda$

or $2x+2y-3z-7 + \lambda(2x+5y+3z-9) = 0$ .............(2)

Given that the plane passes through the point $(2,1,3)$ , it must satisfy (2), i.e.,

$(4+2-9-7) + \lambda(4+5+9-9) = 0$

or $\lambda = \frac{10}{9}$

Putting the values of $\lambda$ in (1), we get

$\vec{r}\left [\left ( 2+\frac{20}{9} \right )\widehat{i}+\left ( 2+\frac{50}{9} \right )\widehat{j}+\left ( \frac{10}{3}-3 \right )\widehat{k} \right ] = 7+10$

or $\vec{r}\left ( \frac{38}{9}\widehat{i}+\frac{68}{9}\widehat{j}+\frac{1}{3}\widehat{k} \right ) = 17$

or $\vec{r}.\left ( 38\widehat{i}+68\widehat{j}+3\widehat{k} \right ) = 153$

which is the required vector equation of the plane.

Question:11 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.

The equation of the plane through the intersection of the given two planes, $x+y+z =1$ and $2x+3y+4z =5$ is given in Cartesian form as;

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

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

So, the direction ratios of (1) plane are $a_{1},b_{1},c_{1}$ which are $(1+2\lambda),(1+3\lambda),\ and\ (1+4\lambda)$ .

Then, the plane in equation (1) is perpendicular to $x-y+z= 0$ whose direction ratios $a_{2},b_{2},c_{2}$ are $1,-1,\ and\ 1$ .

As planes are perpendicular then,

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

we get,

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

or $1+3\lambda = 0$

or $\lambda = -\frac{1}{3}$

Then we will substitute the values of $\lambda$ in the equation (1), we get

$\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3} = 0$

or $x-z+2=0$

This is the required equation of the plane.

Question:12 Find the angle between the planes whose vector equations are $\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5$ and $\overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3$ .

Given two vector equations of plane

$\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5$ and $\overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3$ .

Here, $\vec{n_{1}} = 2\widehat{i}+2\widehat{j}-3\widehat{k}$ and $\vec{n_{2}} = 3\widehat{i}-3\widehat{j}+5\widehat{k}$

The formula for finding the angle between two planes,

$\cos A = \left | \frac{\vec{n_{1}}.\vec{n_{2}}}{|\vec{n_{1}}||\vec{n_{2}}|} \right |$ .............................(1)

$\vec{n_{1}}.\vec{n_{2}} = (2\widehat{i}+2\widehat{j}-3\widehat{k})(3\widehat{i}-3\widehat{j}+5\widehat{k}) = 2(3)+2(-3)-3(5) = -15$

$|\vec{n_{1}}| =\sqrt{(2)^2+(2)^2+(-3)^2} =\sqrt{17}$

and $|\vec{n_{2}}| =\sqrt{(3)^2+(-3)^2+(5)^2} =\sqrt{43}$

Now, we can substitute the values in the angle formula (1) to get,

$\cos A = \left | \frac{-15}{\sqrt{17}\sqrt{43}} \right |$

or $\cos A =\frac{15}{\sqrt{731}}$

or $A = \cos^{-1}\left ( \frac{15}{\sqrt{731}} \right )$

Question:13(a) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $7x + 5y + 6z + 30 = 0\ and\ 3x -y - 10z + 4 = 0$

Here,

$a_{1} = 7,b_{1} = 5, c_{1} = 6$ and $a_{2} = 3,b_{2} = -1, c_{2} = -10$

So, applying each condition to check:

Parallel check: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{7}{3}, \frac{b_{1}}{b_{2}}=\frac{5}{-1},\frac{c_{1}}{c_{2}} = \frac{6}{-10}$

Clearly, the given planes are NOT parallel. $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Perpendicular check: $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

$\Rightarrow 7(3)+5(-1)+6(-10) = 21-5-60 = -44 \neq 0$ .

Clearly, the given planes are NOT perpendicular.

Then find the angle between them,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$

$= \cos^{-1}\left | \frac{-44}{\sqrt{7^2+5^2+6^2}.\sqrt{3^2+(-1)^2+(-10)^2}} \right |$

$= \cos^{-1}\left | \frac{-44}{\sqrt{110}.\sqrt{110}} \right |$

$= \cos^{-1}\left ( \frac{44}{110} \right )$

$= \cos^{-1}\left ( \frac{2}{5} \right )$

Question:13(b) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $2x + y + 3z -2 = 0\ and\ x -2y + 5 = 0$

Here,

$a_{1} = 2,b_{1} = 1, c_{1} = 3$ and $a_{2} = 1,b_{2} = -2, c_{2} = 0$

So, applying each condition to check:

Perpendicular check: $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

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

Thus, the given planes are perpendicular to each other.

Question:13(c) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $2x - 2y + 4z + 5 = 0\ and\ 3x -3y +6z -1 = 0$

Here,

$a_{1} = 2,b_{1} = -2, c_{1} = 4$ and $a_{2} = 3,b_{2} = -3, c_{2} = 6$

So, applying each condition to check:

Parallel check: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{-2}{-3}=\frac{2}{3},\ and\ \frac{c_{1}}{c_{2}} = \frac{4}{6}=\frac{2}{3}$

Thus, the given planes are parallel as $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Question:13(d) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $2x - y + 3z -1 = 0\ and\ 2x -y +3z + 3 = 0$

Here,

$a_{1} = 2,b_{1} = -1, c_{1} = 3$ and $a_{2} = 2,b_{2} = -1, c_{2} = 3$

So, applying each condition to check:

Parallel check: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{2}{2}=1, \frac{b_{1}}{b_{2}}=\frac{-1}{-1} =1,\frac{c_{1}}{c_{2}} = \frac{3}{3} = 1$

Therefore $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Thus, the given planes are parallel to each other.

Question:13(e) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

4x + 8y + z – 8 = 0 and y + z – 4 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $4x + 8y + z -8 = 0\ and\ y + z - 4 = 0$

Here,

$a_{1} = 4,b_{1} = 8, c_{1} = 1$ and $a_{2} = 0,b_{2} = 1, c_{2} = 1$

So, applying each condition to check:

Parallel check: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{4}{0}, \frac{b_{1}}{b_{2}}=\frac{8}{1},\frac{c_{1}}{c_{2}} = \frac{1}{1}$

Clearly, the given planes are NOT parallel as $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ .

Perpendicular check: $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

## NCERT solutions for class 12 maths chapter 11 three dimensional geometry-Exercise: 11.3

Question:1(a) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2

Equation of plane Z=2, i.e. $0x+0y+z=2$

The direction ratio of normal is 0,0,1

$\therefore \, \, \, \sqrt{0^2+0^2+1^2}=1$

Divide equation $0x+0y+z=2$ by 1 from both side

We get, $0x+0y+z=2$

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

Question:1(b) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

x + y + z = 1

Given the equation of the plane is $x+y+z=1$ or we can write $1x+1y+1z=1$

So, the direction ratios of normal from the above equation are, $1,\1,\ and\ 1$ .

Therefore $\sqrt{1^2+1^2+1^2} =\sqrt{3}$

Then dividing both sides of the plane equation by $\sqrt{3}$ , we get

$\frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3}=\frac{1}{\sqrt3}$

So, this is the form of $lx+my+nz = d$ the plane, where $l,\ m,\ n$ are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

$\therefore$ The direction cosines of the given line are $\frac{1}{\sqrt3},\ \frac{1}{\sqrt3},\ \frac{1}{\sqrt3}$ and the distance of the plane from the origin is $\frac{1}{\sqrt3}$ units.

Question:1(c) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

2x + 3y - z = 5

Given the equation of plane is $2x+3y-z=5$

So, the direction ratios of normal from the above equation are, $2,\3,\ and\ -1$ .

Therefore $\sqrt{2^2+3^2+(-1)^2} =\sqrt{14}$

Then dividing both sides of the plane equation by $\sqrt{14}$ , we get

$\frac{2x}{\sqrt{14}}+\frac{3y}{\sqrt{14}}-\frac{z}{\sqrt{14}}=\frac{5}{\sqrt{14}}$

So, this is the form of $lx+my+nz = d$ the plane, where $l,\ m,\ n$ are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

$\therefore$ The direction cosines of the given line are $\frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}},\ \frac{-1}{\sqrt{14}}$ and the distance of the plane from the origin is $\frac{5}{\sqrt{14}}$ units.

Question:1(d) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

5y + 8 = 0

Given the equation of plane is $5y+8=0$ or we can write $0x-5y+0z=8$

So, the direction ratios of normal from the above equation are, $0,\ -5,\ and\ 0$ .

Therefore $\sqrt{0^2+(-5)^2+0^2} =5$

Then dividing both sides of the plane equation by $5$ , we get

$-y = \frac{8}{5}$

So, this is the form of $lx+my+nz = d$ the plane, where $l,\ m,\ n$ are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

$\therefore$ The direction cosines of the given line are $0,\ -1,\ and\ 0$ and the distance of the plane from the origin is $\frac{8}{5}$ units.

Question:2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\widehat{i}+5\widehat{j}-6\widehat{k}$ .

We have given the distance between the plane and origin equal to 7 units and normal to the vector $3\widehat{i}+5\widehat{j}-6\widehat{k}$ .

So, it is known that the equation of the plane with position vector $\vec{r}$ is given by, the relation,

$\vec{r}.\widehat{n} =d$ , where d is the distance of the plane from the origin.

Calculating $\widehat{n}$ ;

$\widehat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{(3)^2+(5)^2+(6)^2}} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}}$

$\vec{r}.\left ( \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}} \right ) = 7$ is the vector equation of the required plane.

Question:3(a) Find the Cartesian equation of the following planes:

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

Given the equation of the plane $\overrightarrow{r}.(\widehat{i}+\widehat{j}-\widehat{k})=2$

So we have to find the Cartesian equation,

Any point $A (x,y,z)$ on this plane will satisfy the equation and its position vector given by,

$\vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}$

Hence we have,

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).(\widehat{i}+\widehat{j}-\widehat{k}) =2$

Or, $x+y-z=2$

Therefore this is the required Cartesian equation of the plane.

Question:3(b) Find the Cartesian equation of the following planes:

$\overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1$

Given the equation of plane $\overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1$

So we have to find the Cartesian equation,

Any point $A (x,y,z)$ on this plane will satisfy the equation and its position vector given by,

$\vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}$

Hence we have,

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).(2\widehat{i}+3\widehat{j}-4\widehat{k}) =1$

Or, $2x+3y-4z=1$

Therefore this is the required Cartesian equation of the plane.

Question:3(c) Find the Cartesian equation of the following planes:

$\overrightarrow{r}.\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right ]=15$

Given the equation of plane $\overrightarrow{r}.\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right ]=15$

So we have to find the Cartesian equation,

Any point $A (x,y,z)$ on this plane will satisfy the equation and its position vector given by, $\vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}$

Hence we have,

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right] =15$

Or, $(s-2t)x+(3-t)y+(2s+t)z=15$

Therefore this is the required Cartesian equation of the plane.

Question:4(a) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2 x + 3y + 4 z - 12 = 0

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given a plane equation $2x+3y+4z-12=0$ ,

Or, $2x+3y+4z=12$

The direction ratios of the normal of the plane are 2, 3 and 4 .

Therefore $\sqrt{(2)^2+(3)^2+(4)^2} = \sqrt{29}$

So, now dividing both sides of the equation by $\sqrt{29}$ we will obtain,

$\frac{2}{\sqrt{29}}x+\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z = \frac{12}{\sqrt{29}}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$ .

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left [ \frac{2}{\sqrt{29}}.\frac{12}{\sqrt{29}},\frac{3}{\sqrt{29}}.\frac{12}{\sqrt{29}},\frac{4}{\sqrt{29}}.\frac{12}{\sqrt{29}} \right ]$ or $\left [ \frac{24}{29}, \frac{36}{49}, \frac{48}{29} \right ]$

Question:4(b) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

3y + 4z - 6 = 0

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given a plane equation $3y+4z-6=0$ ,

Or, $0x+3y+4z=6$

The direction ratios of the normal of the plane are 0,3 and 4 .

Therefore $\sqrt{(0)^2+(3)^2+(4)^2} = 5$

So, now dividing both sides of the equation by $5$ we will obtain,

$0x+\frac{3}{5}y+\frac{4}{5}z = \frac{6}{5}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$ .

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left (0,\frac{3}{5}.\frac{6}{5},\frac{4}{5}.\frac{6}{5} \right )$ or $\left ( 0, \frac{18}{25}, \frac{24}{25} \right )$

Question:4(c) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

x + y + z = 1

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given plane equation $x+y+z=1$ .

The direction ratios of the normal of the plane are 1,1 and 1 .

Therefore $\sqrt{(1)^2+(1)^2+(1)^2} = \sqrt3$

So, now dividing both sides of the equation by $\sqrt3$ we will obtain,

$\frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3} = \frac{1}{\sqrt3}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$ .

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left ( \frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3} \right )$ or $\left ( \frac{1}{3},\frac{1}{3},\frac{1}{3} \right )$ ..

Question: 4(d) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

5y + 8 = 0

Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1},y_{1},z_{1})$

Given plane equation $5y+8=0$ .

or written as $0x-5y+0z=8$

The direction ratios of the normal of the plane are 0, -5 and 0 .

Therefore $\sqrt{(0)^2+(-5)^2+(0)^2} = 5$

So, now dividing both sides of the equation by $5$ we will obtain,

$-y=\frac{8}{5}$

This equation is similar to $lx+my+nz = d$ where, $l,\ m,\ n$ are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by $(ld,md,nd)$ .

$\therefore$ The coordinates of the foot of the perpendicular are;

$\left ( 0,-1(\frac{8}{5}),0 \right )$ or $\left ( 0,\frac{-8}{5},0 \right )$ .

Question:5(a) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane is $\widehat{i}+\widehat{j}-\widehat{k}.$

Given the point $A (1,0,-2)$ and the normal vector $\widehat{n}$ which is perpendicular to the plane is $\widehat{n} = \widehat{i}+\widehat{j}-\widehat{k}$

The position vector of point A is $\vec {a} = \widehat{i}-2\widehat{k}$

So, the vector equation of the plane would be given by,

$(\vec{r}-\vec{a}).\widehat{n} = 0$

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

where $\vec{r}$ is the position vector of any arbitrary point $A(x,y,z)$ in the plane.

$\therefore$ $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$

Therefore, the equation we get,

$\left [(x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}-2\widehat{k}) \right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0$

$\Rightarrow \left [(x-1)\widehat{i}+y\widehat{j}+(z+2)\widehat{k}\right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0$

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

$\Rightarrow x+y-z-3=0$ or $x+y-z=3$

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

Question:5(b) Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is $\widehat{i}-2\widehat{j}+\widehat{k}$ .

Given the point $A (1,4,6)$ and the normal vector $\widehat{n}$ which is perpendicular to the plane is $\widehat{n} = \widehat{i}-2\widehat{j}+\widehat{k}$

The position vector of point A is $\vec {a} = \widehat{i}+4\widehat{j}+6\widehat{k}$

So, the vector equation of the plane would be given by,

$(\vec{r}-\vec{a}).\widehat{n} = 0$

Or $\left [ \vec{r}-(\widehat{i}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0$

where $\vec{r}$ is the position vector of any arbitrary point $A(x,y,z)$ in the plane.

$\therefore$ $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$

Therefore, the equation we get,

$\left [ (x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0$

$\Rightarrow \left [(x-1)\widehat{i}+(y-4)\widehat{j}+(z-6)\widehat{k}\right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0$

$(x-1)-2(y-4)+(z-6)=0$

$\Rightarrow x-2y+z+1=0$

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

Question:6(a) Find the equations of the planes that passes through three points.

(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)

The equation of the plane which passes through the three points $A(1,1,-1),\ B(6,4,-5),\ and\ C(-4,-2,3)$ is given by;

Determinant method,

$\begin{vmatrix} 1 &1 &-1 \\ 6& 4 & -5\\ -4& -2 &3 \end{vmatrix} = (12-10)-(18-20)-(-12+16)$

Or, $= 2+2-4 = 0$

Here, these three points A, B, C are collinear points.

Hence there will be an infinite number of planes possible which passing through the given points.

Question:6(b) Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

The equation of the plane which passes through the three points $A(1,1,0),\ B(1,2,1),\ and\ C(-2,2,-1)$ is given by;

Determinant method,

$\begin{vmatrix} 1 &1 &0 \\ 1& 2 & 1\\ -2& 2 &-1 \end{vmatrix} = (-2-2)-(2+2)= -8 \neq 0$

As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.

Finding the equation of the plane through the points, $(x_{1},y_{1},z_{1}), (x_{2},y_{2},z_{2})\ and\ (x_{3},y_{3},z_{3})$

$\begin{vmatrix} x-x_{1} &y-y_{1} &z-z_{1} \\ x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ x_{3}-x_{1}&y_{3}-y_{1} &z_{3}-z_{1} \end{vmatrix} = 0$

After substituting the values in the determinant we get,

$\begin{vmatrix} x-1 &y-1 &z \\ 0& 1 &1 \\ -3& 1&-1 \end{vmatrix} = 0$

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

$\Rightarrow-2x+2-3y+3+3z = 0$

$2x+3y-3z = 5$

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

Question:7 Find the intercepts cut off by the plane 2x + y – z = 5.

Given plane $2x + y-z = 5$

We have to find the intercepts that this plane would make so,

Making it look like intercept form first:

By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,

$\frac{2}{5}x+\frac{y}{5}-\frac{z}{5} =1$

$\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5} =1$

So, as we know that from the equation of a plane in intercept form, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1$ where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.

Therefore after comparison, we get the values of a,b, and c.

$a = \frac{5}{2},\ b=5,\ and\ c=-5$ .

Hence the intercepts are $\frac{5}{2},\ 5,\ and\ -5$ .

Question:8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Given that the plane is parallel to the ZOX plane.

So, we have the equation of plane ZOX as $y = 0$ .

And an intercept of 3 on the y-axis $\Rightarrow b =3$

Intercept form of a plane given by;

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1$

So, here the plane would be parallel to the x and z-axes both.

we have any plane parallel to it is of the form, $y=a$ .

Equation of the plane required is $y=3$ .

Question:9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

The equation of any plane through the intersection of the planes,

$3x-y+2z-4=0\ and\ x+y+z-2=0$

Can be written in the form of; $(3x-y+2z-4)\ +\alpha( x+y+z-2)= 0$ , where $\alpha \epsilon R$

So, the plane passes through the point $(2,2,1)$ , will satisfy the above equation.

$(3\times2-2+2\times1-4)+\alpha(2+2+1-2) = 0$

That implies $2+3\alpha= 0$

$\alpha = \frac{-2}{3}$

Now, substituting the value of $\alpha$ in the equation above we get the final equation of the plane;

$(3x-y+2z-4)\ +\alpha( x+y+z-2)= 0$

$(3x-y+2z-4)\ +\frac{-2}{3}( x+y+z-2)= 0$

$\Rightarrow 9x-3y+6z-12\ -2 x-2y-2z+4= 0$

$\Rightarrow 7x-5y+4z-8= 0$ is the required equation of the plane.

Question:10 Find the vector equation of the plane passing through the intersection of the planes $\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})=7$ , $\overrightarrow{r}(2\widehat{i}+5\widehat{j}+3\widehat{k})=9$ and through the point (2, 1, 3).

Here $\vec{n_{1}} =2 \widehat{i}+2\widehat{j}-3\widehat{k}$ and $\vec{n_{2}} = 2\widehat{i}+5\widehat{j}+3\widehat{k}$

and $d_{1} = 7$ and $d_{2} = 9$

Hence, using the relation $\vec{r}.(\vec{n_{1}}+\lambda\vec{n_{2}}) = d_{1}+\lambda d_{2}$ , we get

$\vec{r}.[2\widehat{i}+2\widehat{j}-3\widehat{k}+\lambda(2\widehat{i}+5\widehat{j}+3\widehat{k})] = 7+9\lambda$

or $\vec{r}.[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda$ ..............(1)

where, $\lambda$ is some real number.

Taking $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$ , we get

$(\vec{x\widehat{i}+y\widehat{j}+z\widehat{k}}).[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda$

or $x(2+2\lambda) + y(2+5\lambda) +z(3\lambda-3) = 7+9\lambda$

or $2x+2y-3z-7 + \lambda(2x+5y+3z-9) = 0$ .............(2)

Given that the plane passes through the point $(2,1,3)$ , it must satisfy (2), i.e.,

$(4+2-9-7) + \lambda(4+5+9-9) = 0$

or $\lambda = \frac{10}{9}$

Putting the values of $\lambda$ in (1), we get

$\vec{r}\left [\left ( 2+\frac{20}{9} \right )\widehat{i}+\left ( 2+\frac{50}{9} \right )\widehat{j}+\left ( \frac{10}{3}-3 \right )\widehat{k} \right ] = 7+10$

or $\vec{r}\left ( \frac{38}{9}\widehat{i}+\frac{68}{9}\widehat{j}+\frac{1}{3}\widehat{k} \right ) = 17$

or $\vec{r}.\left ( 38\widehat{i}+68\widehat{j}+3\widehat{k} \right ) = 153$

which is the required vector equation of the plane.

Question:11 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.

The equation of the plane through the intersection of the given two planes, $x+y+z =1$ and $2x+3y+4z =5$ is given in Cartesian form as;

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

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

So, the direction ratios of (1) plane are $a_{1},b_{1},c_{1}$ which are $(1+2\lambda),(1+3\lambda),\ and\ (1+4\lambda)$ .

Then, the plane in equation (1) is perpendicular to $x-y+z= 0$ whose direction ratios $a_{2},b_{2},c_{2}$ are $1,-1,\ and\ 1$ .

As planes are perpendicular then,

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

we get,

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

or $1+3\lambda = 0$

or $\lambda = -\frac{1}{3}$

Then we will substitute the values of $\lambda$ in the equation (1), we get

$\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3} = 0$

or $x-z+2=0$

This is the required equation of the plane.

Question:12 Find the angle between the planes whose vector equations are $\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5$ and $\overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3$ .

Given two vector equations of plane

$\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5$ and $\overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3$ .

Here, $\vec{n_{1}} = 2\widehat{i}+2\widehat{j}-3\widehat{k}$ and $\vec{n_{2}} = 3\widehat{i}-3\widehat{j}+5\widehat{k}$

The formula for finding the angle between two planes,

$\cos A = \left | \frac{\vec{n_{1}}.\vec{n_{2}}}{|\vec{n_{1}}||\vec{n_{2}}|} \right |$ .............................(1)

$\vec{n_{1}}.\vec{n_{2}} = (2\widehat{i}+2\widehat{j}-3\widehat{k})(3\widehat{i}-3\widehat{j}+5\widehat{k}) = 2(3)+2(-3)-3(5) = -15$

$|\vec{n_{1}}| =\sqrt{(2)^2+(2)^2+(-3)^2} =\sqrt{17}$

and $|\vec{n_{2}}| =\sqrt{(3)^2+(-3)^2+(5)^2} =\sqrt{43}$

Now, we can substitute the values in the angle formula (1) to get,

$\cos A = \left | \frac{-15}{\sqrt{17}\sqrt{43}} \right |$

or $\cos A =\frac{15}{\sqrt{731}}$

or $A = \cos^{-1}\left ( \frac{15}{\sqrt{731}} \right )$

Question:13(a) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $7x + 5y + 6z + 30 = 0\ and\ 3x -y - 10z + 4 = 0$

Here,

$a_{1} = 7,b_{1} = 5, c_{1} = 6$ and $a_{2} = 3,b_{2} = -1, c_{2} = -10$

So, applying each condition to check:

Parallel check: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{7}{3}, \frac{b_{1}}{b_{2}}=\frac{5}{-1},\frac{c_{1}}{c_{2}} = \frac{6}{-10}$

Clearly, the given planes are NOT parallel. $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Perpendicular check: $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

$\Rightarrow 7(3)+5(-1)+6(-10) = 21-5-60 = -44 \neq 0$ .

Clearly, the given planes are NOT perpendicular.

Then find the angle between them,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$

$= \cos^{-1}\left | \frac{-44}{\sqrt{7^2+5^2+6^2}.\sqrt{3^2+(-1)^2+(-10)^2}} \right |$

$= \cos^{-1}\left | \frac{-44}{\sqrt{110}.\sqrt{110}} \right |$

$= \cos^{-1}\left ( \frac{44}{110} \right )$

$= \cos^{-1}\left ( \frac{2}{5} \right )$

Question:13(b) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $2x + y + 3z -2 = 0\ and\ x -2y + 5 = 0$

Here,

$a_{1} = 2,b_{1} = 1, c_{1} = 3$ and $a_{2} = 1,b_{2} = -2, c_{2} = 0$

So, applying each condition to check:

Perpendicular check: $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

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

Thus, the given planes are perpendicular to each other.

Question:13(c) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $2x - 2y + 4z + 5 = 0\ and\ 3x -3y +6z -1 = 0$

Here,

$a_{1} = 2,b_{1} = -2, c_{1} = 4$ and $a_{2} = 3,b_{2} = -3, c_{2} = 6$

So, applying each condition to check:

Parallel check: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{-2}{-3}=\frac{2}{3},\ and\ \frac{c_{1}}{c_{2}} = \frac{4}{6}=\frac{2}{3}$

Thus, the given planes are parallel as $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Question:13(d) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $2x - y + 3z -1 = 0\ and\ 2x -y +3z + 3 = 0$

Here,

$a_{1} = 2,b_{1} = -1, c_{1} = 3$ and $a_{2} = 2,b_{2} = -1, c_{2} = 3$

So, applying each condition to check:

Parallel check: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{2}{2}=1, \frac{b_{1}}{b_{2}}=\frac{-1}{-1} =1,\frac{c_{1}}{c_{2}} = \frac{3}{3} = 1$

Therefore $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Thus, the given planes are parallel to each other.

Question:13(e) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

4x + 8y + z – 8 = 0 and y + z – 4 = 0

Two planes

$L_{1}:a_{1}x+b_{1}y+c_{1}z = 0$ whose direction ratios are $a_{1},b_{1},c_{1}$ and $L_{2}:a_{2}x+b_{2}y+c_{2}z = 0$ whose direction ratios are $a_{2},b_{2},c_{2}$ ,

are said to Parallel:

If, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

and Perpendicular:

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

And the angle between $L_{1}\ and\ L_{2}$ is given by the relation,

$A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |$
So, given two planes $4x + 8y + z -8 = 0\ and\ y + z - 4 = 0$

Here,

$a_{1} = 4,b_{1} = 8, c_{1} = 1$ and $a_{2} = 0,b_{2} = 1, c_{2} = 1$

So, applying each condition to check:

Parallel check: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}} =\frac{4}{0}, \frac{b_{1}}{b_{2}}=\frac{8}{1},\frac{c_{1}}{c_{2}} = \frac{1}{1}$

Clearly, the given planes are NOT parallel as $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ .

Perpendicular check: $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

More About NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3

• Fourteen questions in total are given in the exercise 11.3 Class 12 Maths.

• There are sub-questions to certain question numbers.

• All these 14 questions are detailed in the NCERT solutions for Class 12 Maths chapter 11 exercise 11.3

Also Read| Three Dimensional Geometry Class 12th Notes

Significance of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3

• The topic plane covers many concepts and the questions from this part are important for the CBSE Board exam preparation for Class 12.

• Exercise 11.3 is a part of the topic plane and the NCERT solutions for Class 12 Maths chapter 11 exercise 11.3 will be useful to score well in the exam.

Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 11
• NCERT Solutions for Class 12 Maths Chapter 11

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology