Solution

(a) The equation of the plane is $z=2$ or $0 x+0 y+z=2 \ldots$ (1)

The direction ratios of normal are 0,0 , and 1 .

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

Dividing both sides of equation (1) by 1 , we obtain

$0 . x+0 . y+1 . z=2$

This is of the form $I x+m y+n z=d$, 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 are 0,0 , and 1 and the distance of the plane from the origin is 2 units.

(b) $x+y+z=1 \ldots$ (1)

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

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

Dividing both sides of equation (1) by $\sqrt{3}$, we obtain

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

This equation is of the form $l x+m y+n z=d$, where $l, m, n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.

Therefore, the direction cosines of the normal are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$, and $\frac{1}{\sqrt{3}}$ and the distance of normal from the origin is $\frac{1}{\sqrt{3}}$ units.

(c) $2 x+3 y-z=5$

The direction ratios of normal are 2,3 , and -1 .

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

Dividing both sides of equation (1) by $\sqrt{14}$, we obtain

$\frac{2}{\sqrt{14}} x+\frac{3}{\sqrt{14}} y-\frac{1}{\sqrt{14}} z=\frac{5}{\sqrt{14}}$

This equation is of the form $l x+m y+n z=d$, where $l, m, n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are $\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$, and $\frac{-1}{\sqrt{14}}$ and

the distance of normal from the origin is $\frac{5}{\sqrt{14}}$ units.

(d) $5 y+8=0$

$\Rightarrow 0 x-5 y+0 z=8$

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

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

Dividing both sides of equation (1) by 5 , we obtain

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

This equation is of the form $l x+m y+n z=d$, where $l, m, n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are $0,-1$, and 0 and the

distance of normal from the origin is $\frac{8}{5}$ units.

Solution

The normal vector is, $\vec{n}=3 \hat{i}+5 \hat{j}-6 \hat{k}$

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

It is known that the equation of the plane with position vector $\vec{r}$ is given by, $\vec{r} \cdot \hat{n}=d$

$\Rightarrow \hat{r} \cdot(\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{70}})=7$

This is the vector equation of the required plane.

Solution

(a) It is given that equation of the plane is

$\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2$

For any arbitrary point $P(x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r}=x \hat{i}+y \hat{j}-z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

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

$\Rightarrow x+y-z=2$

This is the Cartesian equation of the plane.

(b) $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$

For any arbitrary point $P(x, y, z)$ on the plane, position vector $\vec{r}$ is given by,

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

Substituting the value of $\vec{r}$ in equation (1), we obtain

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

$\Rightarrow 2 x+3 y-4 z=1$

This is the Cartesian equation of the plane.

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

For any arbitrary point $P(x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r}=x \hat{i}+y \hat{j}-z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

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

$\Rightarrow(s-2 t) x+(3-t) y+(2 s+t) z=15$

This is the Cartesian equation of the given plane.

Solution

(a) Let the coordinates of the foot of perpendicular $P$ from the origin to the plane be $(x_1, y_1, z_1)$. $2 x+3 y+4 z-12=0$

$\Rightarrow 2 x+3 y+4 z=12$

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

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

Dividing both sides of equation (1) by $\sqrt{29}$, we obtain

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

This equation is of the form $l x+m y+n z=d$, where $l, m, n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

( $l d, m d, n d)$.

Therefore, the coordinates of the foot of the perpendicular are

$(\frac{2}{\sqrt{29}} \cdot \frac{12}{\sqrt{29}}, \frac{3}{\sqrt{29}} \cdot \frac{12}{\sqrt{29}}, \frac{4}{\sqrt{29}} \cdot \frac{12}{\sqrt{29}})$ i.e., $(\frac{24}{29}, \frac{36}{49}, \frac{48}{29})$.

(b) Let the coordinates of the foot of perpendicular $P$ from the origin to the plane be ( $x_1$, $.y_1, z_1)$.

$3 y+4 z-6=0$

$\Rightarrow 0 x+3 y+4 z=6$

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

$\therefore \sqrt{0+3^{2}+4^{2}}=5$

Dividing both sides of equation (1) by 5 , we obtain

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

This equation is of the form $l x+m y+n z=d$, where $l, m, n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(Id, $m d, n d)$.

Therefore, the coordinates of the foot of the perpendicular are

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

(c) Let the coordinates of the foot of perpendicular $P$ from the origin to the plane be ( $x_1$, $.y_1, z_1)$.

$x+y+z=1$

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

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

Dividing both sides of equation (1) by $\sqrt{3}$, we obtain

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

This equation is of the form $l x+m y+n z=d$, where $l, m, n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

( $l d, m d, n d)$.

Therefore, the coordinates of the foot of the perpendicular are

$(\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}})$ i.e., $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$.

(d) Let the coordinates of the foot of perpendicular $P$ from the origin to the plane be ( $x_1$, $.y_1, z_1)$.

$5 y+8=0$

$\Rightarrow 0 x-5 y+0 z=8$

The direction ratios of the normal are $0,-5$, and 0 . $\therefore \sqrt{0+(-5)^{2}+0}=5$

Dividing both sides of equation (1) by 5 , we obtain

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

This equation is of the form $l x+m y+n z=d$, where $l, m, n$ are the direction cosines of normal to the plane and $d$ is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(Id, $m d, n d)$.

Therefore, the coordinates of the foot of the perpendicular are

$ (0,-1(\frac{8}{5}), 0) \text{ i.e., }(0,-\frac{8}{5}, 0) \text{. } $

Solution

(a) The position vector of point $(1,0,-2)$ is $\vec{a}=\hat{i}-2 \hat{k}$

The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N}=\hat{i}+\hat{j}-\hat{k}$

The vector equation of the plane is given by, $(\vec{r}-\vec{a}) \cdot \vec{N}=0$

$\Rightarrow[\vec{r}-(\hat{i}-2 \hat{k})] \cdot(\hat{i}+\hat{j}-\hat{k})=0$

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

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

Therefore, equation (1) becomes

$ \begin{aligned} & {[(x \hat{i}+y \hat{j}+z \hat{k})-(\hat{i}-2 \hat{k})] \cdot(\hat{i}+\hat{j}-\hat{k})=0} \\ & \Rightarrow[(x-1) \hat{i}+y \hat{j}+(z+2) \hat{k}] \cdot(\hat{i}+\hat{j}-\hat{k})=0 \\ & \Rightarrow(x-1)+y-(z+2)=0 \\ & \Rightarrow x+y-z-3=0 \\ & \Rightarrow x+y-z=3 \end{aligned} $

This is the Cartesian equation of the required plane.

(b) The position vector of the point $(1,4,6)$ is $\vec{a}=\hat{i}+4 \hat{j}+6 \hat{k}$

The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N}=\hat{i}-2 \hat{j}+\hat{k}$

The vector equation of the plane is given by, $(\vec{r}-\vec{a}) \cdot \vec{N}=0$

$\Rightarrow[\vec{r}-(\hat{i}+4 \hat{j}+6 \hat{k})] \cdot(\hat{i}-2 \hat{j}+\hat{k})=0$

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

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

Therefore, equation (1) becomes

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

This is the Cartesian equation of the required plane.

Solution

(a) The given points are $A(1,1,-1), B(6,4,-5)$, and $C(-4,-2,3)$.

$ \begin{aligned} \begin{vmatrix} 1 & 1 & -1 \\ 6 & 4 & -5 \\ -4 & -2 & 3 \end{vmatrix} & =(12-10)-(18-20)-(-12+16) \\ & =2+2-4 \\ & =0 \end{aligned} $

Since A, B, C are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are A $(1,1,0), B(1,2,1)$, and $C(-2,2,-1)$.

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

Therefore, a plane will pass through the points $A, B$, and $C$.

It is known that the equation of the plane through the points, $(x_1, y_1, z_1),(x_2, y_2, z_2)$, and

$ \begin{aligned} & (x_3, y_3, z_3), \text{ is } \\ & \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 \\ & \Rightarrow \begin{vmatrix} x-1 & y-1 & z \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} =0 \\ & \Rightarrow(-2)(x-1)-3(y-1)+3 z=0 \\ & \Rightarrow-2 x-3 y+3 z+2+3=0 \\ & \Rightarrow-2 x-3 y+3 z=-5 \\ & \Rightarrow 2 x+3 y-3 z=5 \end{aligned} $

This is the Cartesian equation of the required plane.

Solution

$2 x+y-z=5$

Dividing both sides of equation (1) by 5 , we obtain

$\begin{aligned} & \frac{2}{5} x+\frac{y}{5}-\frac{z}{5}=1 \\ & \Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1 \quad …(2)\end{aligned} $

It is known that the equation of a plane in intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where $a, b, c$ are the intercepts cut off by the plane at $x, y$, and $z$ axes respectively.

Therefore, for the given equation,

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

Thus, the intercepts cut off by the plane are $\frac{5}{2}, 5$, and -5 .

Solution

The equation of the plane $Z O X$ is

$y=0$

Any plane parallel to it is of the form, $y=a$

Since the $y$-intercept of the plane is 3 ,

$\therefore a=3$

Thus, the equation of the required plane is $y=3$

Solution

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

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

$(3 x-y+2 z-4)+\alpha(x+y+z-2)=0$, where $\alpha \in R$

The plane passes through the point $(2,2,1)$. Therefore, this point will satisfy equation (1).

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

$\Rightarrow 2+3 \alpha=0$

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

Substituting $\alpha=-\frac{2}{3}$ in equation (1), we obtain

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

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

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

$\Rightarrow 7 x-5 y+4 z-8=0$

This is the required equation of the plane.

Solution

The equations of the planes are $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$

$\Rightarrow \vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})-7=0$

$\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9=0$

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

$[\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})-7]+\lambda[\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9]=0$, where $\lambda \in R$

$$ \begin{align*} & \vec{r} \cdot[(2 \hat{i}+2 \hat{j}-3 \hat{k})+\lambda(2 \hat{i}+5 \hat{j}+3 \hat{k})]=9 \lambda+7 \\ & \vec{r} \cdot[(2+2 \lambda) \hat{i}+(2+5 \lambda) \hat{j}+(3 \lambda-3) \hat{k}]=9 \lambda+7 \tag{3} \end{align*} $$

The plane passes through the point $(2,1,3)$. Therefore, its position vector is given by, $\vec{r}=2 \hat{i}+2 \hat{j}+3 \hat{k}$

Substituting in equation (3), we obtain

$ \begin{aligned} & (2 \hat{i}+\hat{j}-3 \hat{k}) \cdot[(2+2 \lambda) \hat{i}+(2+5 \lambda) \hat{j}+(3 \lambda-3) \hat{k}]=9 \lambda+7 \\ & \Rightarrow(2+2 \lambda)+(2+5 \lambda)+(3 \lambda-3)=9 \lambda+7 \\ & \Rightarrow 18 \lambda-3=9 \lambda+7 \\ & \Rightarrow 9 \lambda=10 \\ & \Rightarrow \lambda=\frac{10}{9} \end{aligned} $

Substituting $\lambda=\frac{10}{9}$ in equation (3), we obtain

$\vec{r} \cdot(\frac{38}{9} \hat{i}+\frac{68}{9} \hat{j}+\frac{3}{9} \hat{k})=17$

$\Rightarrow \vec{r} \cdot(38 \hat{i}+68 \hat{j}+3 \hat{k})=153$

This is the vector equation of the required plane.

Solution

The equation of the plane through the intersection of the planes, $x+y+z=1$ and

$$ \begin{align*} & 2 x+3 y+4 z=5, \text{ is } \\ & (x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \\ & \Rightarrow(2 \lambda+1) x+(3 \lambda+1) y+(4 \lambda+1) z-(5 \lambda+1)=0 \tag{1} \end{align*} $$

The direction ratios, $a_1, b_1, c_1$, of this plane are $(2 \lambda+1),(3 \lambda+1)$, and $(4 \lambda+1)$.

The plane in equation (1) is perpendicular to $x-y+z=0$

Its direction ratios, $a_2, b_2, c_2$, are $1,-1$, and 1 .

Since the planes are perpendicular,

$a_1 a_2+b_1 b_2+c_1 c_2=0$

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

$\Rightarrow 3 \lambda+1=0$

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

Substituting $\lambda=-\frac{1}{3}$ in equation (1), we obtain

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

$\Rightarrow x-z+2=0$

This is the required equation of the plane.

Solution

The equations of the given planes are $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})=3$

It is known that if $ \vec{n} _1$ and $ \vec{n} _2$ are normal to the planes, $\vec{r} \cdot \vec{n} _1=d_1$ and $\vec{r} \cdot \vec{n} _2=d_2$, then the angle between them, $Q$, is given by,

$\cos Q=|\frac{ \vec{n} _1 \cdot \vec{n} _2}{| \vec{n} _1|| \vec{n} _2|}|$

Here, $ \vec{n} _1=2 \hat{i}+2 \hat{j}-3 \hat{k}$ and $ \vec{n} _2=3 \hat{i}-3 \hat{j}+5 \hat{k}$

$\therefore \vec{n} _1 \cdot \vec{n} _2=(2 \hat{i}+2 \hat{j}-3 \hat{k})(3 \hat{i}-3 \hat{j}+5 \hat{k})=2.3+2 \cdot(-3)+(-3) \cdot 5=-15$

$| \vec{n} _1|=\sqrt{(2)^{2}+(2)^{2}+(-3)^{2}}=\sqrt{17}$

$| \vec{n} _2|=\sqrt{(3)^{2}+(-3)^{2}+(5)^{2}}=\sqrt{43}$

Substituting the value of $\vec{n} \cdot \vec{n} _2,| \vec{n} _1|$ and $| \vec{n} _2|$ in equation (1), we obtain

$\cos Q=|\frac{-15}{\sqrt{17} \cdot \sqrt{43}}|$

$\Rightarrow \cos Q=\frac{15}{\sqrt{731}}$

$\Rightarrow \cos Q^{-1}=(\frac{15}{\sqrt{731}})$

Solution

The direction ratios of normal to the plane, $L_1: a_1 x+b_1 y+c_1 z=0$, are $a_1, b_1, c_1$ and $L_2: a_1 x+b_2 y+c_2 z=0$ are $a_2, b_2, c_2$.

$L_1 | L_2$, if $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$L_1 \perp L_2$, if $a_1 a_2+b_1 b_2+c_1 c_2=0$

The angle between $L_1$ and $L_2$ is given by,

$Q=\cos ^{-1}|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2} \cdot \sqrt{a_2^{2}+b_2^{2}+c_2^{2}}}}|$

(a) The equations of the planes are $7 x+5 y+6 z+30=0$ and

$3 x-y-10 z+4=0$

Here, $a_1=7, b_1=5, c_1=6$ $a_2=3, b_2=-1, c_2=-10$

$a_1 a_2+b_1 b_2+c_1 c_2=7 \times 3+5 \times(-1)+6 \times(-10)=-44 \neq 0$

Therefore, the given planes are not perpendicular.

$\frac{a_1}{a_2}=\frac{7}{3}, \frac{b_1}{b_2}=\frac{5}{-1}=-5, \frac{c_1}{c_2}=\frac{6}{-10}=\frac{-3}{5}$

It can be seen that, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq, \frac{c_1}{c_2}$

Therefore, the given planes are not parallel.

The angle between them is given by,

$ \begin{aligned} Q & =\cos ^{-1}|\frac{7 \times 3+5 \times(-1)+6 \times(-10)}{\sqrt{(7)^{2}+(5)^{2}+(6)^{2}} \times \sqrt{(3)^{2}+(-1)^{2}+(-10)^{2}}}| \\ & =\cos ^{-1}|\frac{21-5-60}{\sqrt{110} \times \sqrt{110}}| \\ & =\cos ^{-1} \frac{44}{110} \\ & =\cos ^{-1} \frac{2}{5} \end{aligned} $

(b) The equations of the planes are $2 x+y+3 z-2=0$ and $x-2 y+5=0$

Here, $a_1=2, b_1=1, c_1=3$ and $a_2=1, b_2=-2, c_2=0$

$\therefore a_1 a_2+b_1 b_2+c_1 c_2=2 \times 1+1 \times(-2)+3 \times 0=0$

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x-2 y+4 z+5=0$ and $3 x-3 y+6 z-1=0$

Here, $a_1=2, b_1-2, c_1=4$ and

$a_2=3, b_2=-3, c_2=6 \quad a_1 a_2+b_1 b_2+c_1 c_2=2 \times 3+(-2)(-3)+4 \times 6=6+6+24=36 \neq 0$

Thus, the given planes are not perpendicular to each other.

$\frac{a_1}{a_2}=\frac{2}{3}, \frac{b_1}{b_2}=\frac{-2}{-3}=\frac{2}{3}$ and $\frac{c_1}{c_2}=\frac{4}{6}=\frac{2}{3}$ $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Thus, the given planes are parallel to each other.

(d) The equations of the planes are $2 x-y+3 z-1=0$ and $2 x-y+3 z+3=0$

Here, $a_1=2, b_1=-1, c_1=3$ and $a_2=2, b_2=-1, c_2=3$

$\frac{a_1}{a_2}=\frac{2}{2}=1, \frac{b_1}{b_2}=\frac{-1}{-1}=1$ and $\frac{c_1}{c_2}=\frac{3}{3}=1$

$\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Thus, the given lines are parallel to each other.

(e) The equations of the given planes are $4 x+8 y+z-8=0$ and $y+z-4=0$

Here, $a_1=4, b_1=8, c_1=1$ and $a_2=0, b_2=1, c_2=1$

$a_1 a_2+b_1 b_2+c_1 c_2=4 \times 0+8 \times 1+1=9 \neq 0$

Therefore, the given lines are not perpendicular to each other.

$\frac{a_1}{a_2}=\frac{4}{0}, \frac{b_1}{b_2}=\frac{8}{1}=8, \frac{c_1}{c_2}=\frac{1}{1}=1$

$\therefore \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq, \frac{c_1}{c_2}$

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

$Q=\cos ^{-1}|\frac{4 \times 0+8 \times 1+1 \times 1}{\sqrt{4^{2}+8^{2}+1^{2}} \times \sqrt{0^{2}+1^{2}+1^{2}}}|=\cos ^{-1}|\frac{9}{9 \times \sqrt{2}}|=\cos ^{-1}(\frac{1}{\sqrt{2}})=45^{\circ}$

Solution

It is known that the distance between a point, $p(x_1, y_1, z_1)$, and a plane, $A x+B y+C z=$ $D$, is given by,

$d=|\frac{A x_1+B y_1+C z_1-D}{\sqrt{A^{2}+B^{2}+C^{2}}}|$

(a) The given point is $(0,0,0)$ and the plane is $3 x-4 y+12 z=3$

$\therefore d=|\frac{3 \times 0-4 \times 0+12 \times 0-3}{\sqrt{(3)^{2}+(-4)^{2}+(12)^{2}}}|=\frac{3}{\sqrt{169}}=\frac{3}{13}$

(b) The given point is $(3,-2,1)$ and the plane is $2 x-y+2 z+3=0$

$\therefore=|\frac{2 \times 3-(-2)+2 \times 1+3}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}}|=|\frac{13}{3}|=\frac{13}{3}$

(c) The given point is $(2,3,-5)$ and the plane is $x+2 y-2 z=9$

$\therefore d=|\frac{2+2 \times 3-2(-5)-9}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}|=\frac{9}{3}=3$

(d) The given point is $(-6,0,0)$ and the plane is $2 x-3 y+6 z-2=0$

$ d=|\frac{2(-6)-3 \times 0+6 \times 0-2}{\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}}|=|\frac{-14}{\sqrt{49}}|=\frac{14}{7}=2 $