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, $1,\text{1},and1$ .

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

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

$\frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

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 $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$ and the distance of the plane from the origin is $\frac{1}{\sqrt{3}}$ 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, $2,\text{3},and-1$ .

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, $0,-5,and0$ .

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 $0,-1,and0$ 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 $3\hat{i}+5\hat{j}-6\hat{k}$ .

Answer:

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

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

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

Calculating $\hat{n}$ ;

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

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

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

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

Answer:

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

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,

$\overrightarrow{r}=x\hat{i}+y\hat{j}-z\hat{k}$

Hence we have,

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

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:

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

Answer:

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

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,

$\overrightarrow{r}=x\hat{i}+y\hat{j}-z\hat{k}$

Hence we have,

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

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:

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

Answer:

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

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, $\overrightarrow{r}=x\hat{i}+y\hat{j}-z\hat{k}$

Hence we have,

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

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;

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

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;

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

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

So, now dividing both sides of the equation by $\sqrt{3}$ we will obtain,

$\frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

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;

$(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}})$ or $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ ..

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;

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

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

Answer:

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

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

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

$(\overrightarrow{r}-\overrightarrow{a}).\hat{n}=0$

Or $[\overrightarrow{r}-(\hat{i}-2\hat{k})].(\hat{i}+\hat{j}-\hat{k})=0$

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

$\therefore$ $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Therefore, the equation we get,

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

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

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

$\Rightarrow x+y-z-3=0$ 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 $\hat{i}-2\hat{j}+\hat{k}$ .

Answer:

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

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

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

$(\overrightarrow{r}-\overrightarrow{a}).\hat{n}=0$

Or $[\overrightarrow{r}-(\hat{i}+4\hat{j}+6\hat{k})].(\hat{i}-2\hat{j}+\hat{k})=0$

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

$\therefore$ $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Therefore, the equation we get,

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

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

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

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

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 $A(1,1,-1),B(6,4,-5),andC(-4,-2,3)$ is given by;

Determinant method,

$\left|\begin{array}{ccc}1& 1& -1\\ 6& 4& -5\\ -4& -2& 3\end{array}\right|=(12-10)-(18-20)-(-12+16)$

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 $A(1,1,0),B(1,2,1),andC(-2,2,-1)$ is given by;

Determinant method,

$\left|\begin{array}{ccc}1& 1& 0\\ 1& 2& 1\\ -2& 2& -1\end{array}\right|=(-2-2)-(2+2)=-8\ne 0$

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, $(x_1,y_1,z_1),(x_2,y_2,z_2)and(x_3,y_3,z_3)$

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$

After substituting the values in the determinant we get,

$\left|\begin{array}{ccc}x-1& y-1& z\\ 0& 1& 1\\ -3& 1& -1\end{array}\right|=0$

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

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

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

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.

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

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,

$3x-y+2z-4=0andx+y+z-2=0$

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

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

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

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;

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

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

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

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

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

Answer:

Here $\overrightarrow{n_1}=2\hat{i}+2\hat{j}-3\hat{k}$ and $\overrightarrow{n_2}=2\hat{i}+5\hat{j}+3\hat{k}$

and $d_1=7$ and $d_2=9$

Hence, using the relation $\overrightarrow{r}.(\overrightarrow{n_1}+\lambda \overrightarrow{n_2})=d_1+\lambda d_2$ , we get

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

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

where, $\lambda$ is some real number.

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

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

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

$\overrightarrow{r}[(2+\frac{20}{9})\hat{i}+(2+\frac{50}{9})\hat{j}+(\frac{10}{3}-3)\hat{k}]=7+10$

or $\overrightarrow{r}(\frac{38}{9}\hat{i}+\frac{68}{9}\hat{j}+\frac{1}{3}\hat{k})=17$

or $\overrightarrow{r}.(38\hat{i}+68\hat{j}+3\hat{k})=153$

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 $1,-1,and1$ .

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

Answer:

Given two vector equations of plane

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

Here, $\overrightarrow{n_1}=2\hat{i}+2\hat{j}-3\hat{k}$ and $\overrightarrow{n_2}=3\hat{i}-3\hat{j}+5\hat{k}$

The formula for finding the angle between two planes,

$\cos A=\left|\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}\right|$ .............................(1)

$\overrightarrow{n_1}.\overrightarrow{n_2}=(2\hat{i}+2\hat{j}-3\hat{k})(3\hat{i}-3\hat{j}+5\hat{k})=2(3)+2(-3)-3(5)=-15$

$|\overrightarrow{n_1}|=\sqrt{(2{)}^2+(2{)}^2+(-3{)}^2}=\sqrt{17}$

and $|\overrightarrow{n_2}|=\sqrt{(3{)}^2+(-3{)}^2+(5{)}^2}=\sqrt{43}$

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 $7x+5y+6z+30=0and3x-y-10z+4=0$

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. $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Perpendicular check: $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

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

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 $a_2,b_2,c_2$ ,