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- Three Dimensional Geometry

Edited By Ramraj Saini | Updated on Dec 04, 2023 09:16 AM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 11 Exercise 11.3

NCERT Solutions for Exercise 11.3 Class 12 Maths Chapter 11 Three Dimensional Geometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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.

12th class Maths exercise 11.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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

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.

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.

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.

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.

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:

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:

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:

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.

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 $\left [ \frac{24}{29}, \frac{36}{49}, \frac{48}{29} \right ]$

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 $\left ( 0, \frac{18}{25}, \frac{24}{25} \right )$

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

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

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.

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.

(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.

(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.

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 $\frac{5}{2},\ 5,\ and\ -5$ .

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 $y=3$ .

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.

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.

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.

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

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,

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

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

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

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

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

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.

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

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.

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

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.

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.

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.

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.

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:

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:

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:

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.

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 $\left [ \frac{24}{29}, \frac{36}{49}, \frac{48}{29} \right ]$

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 $\left ( 0, \frac{18}{25}, \frac{24}{25} \right )$

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

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

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.

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.

(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.

(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.

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 $\frac{5}{2},\ 5,\ and\ -5$ .

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 $y=3$ .

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.

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.

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.

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

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,

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

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

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

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

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

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.

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

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.

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
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

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.

## Key Features Of NCERT Solutions for Exercise 11.3 Class 12 Maths Chapter 11

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 11.3 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 11.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 11.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 11.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 11.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 11.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

## Subject Wise NCERT Exemplar Solutions

1. What is the main topic that is to be covered to solve exercise 11.3 Class 12 Maths?

The topic 11.6 plane

2. What is discussed after Class 12 Maths chapter 11 exercise 11.3?

Miscellaneous examples are given after Class 12th Maths chapter 11 exercise 11.3

3. Who solved the NCERT solutions for Class 12 Maths chapter 11 exercise 11.3?

A team of mathematics experts solved exercise 11.3 discussed here

4. Why students should solve Class 12 Maths chapter 11 exercise 11.3?

To understand how much students have grasped the concepts of plane discussed in the NCERT mathematics book, it is good to solve exercise 11.3

5. Is there any supporting NCERT material for more practice questions?

Yes, NCERT exemplars have a good number of practice questions and will be useful in the preparation of the chapter.

6. Are NCERT solutions helpful in the CBSE board examination?

Yes, for the CBSE board there will be a good number of similar questions as discussed in the NCERT book.

7. Is three-dimensional geometry important for JEE Main examination?

Yes. Questions are asked from three-dimensional geometry in the JEE Main papers.

8. Which NCERT Class 12 chapter explains the concepts of vectors?

Chapter 10 of Class 12 NCERT book

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9