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

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

Answer:

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

Answer:

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

Answer:

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 $3\widehat{i}+5\widehat{j}-6\widehat{k}$ .

Answer:

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:

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

Answer:

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:

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

Answer:

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:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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 $\widehat{i}+\widehat{j}-\widehat{k}.$

Answer:

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 $\widehat{i}-2\widehat{j}+\widehat{k}$ .

Answer:

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)

Answer:

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)

Answer:

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.

Answer:

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.

Answer:

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).

Answer:

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).

Answer:

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.

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

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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

Answer:

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.