RD Sharma Class 12 Exercise VSA The Plane Solutions Maths

RD Sharma Class 12 Exercise VSA The Plane Solutions Maths - Download PDF Free Online

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RD Sharma Class 12th VSAQ is an exercise from the chapter ‘The Plane,’ containing 24 VSAQs. RD Sharma solutions Here students will learn about concepts like finding the equation of plane passing through points, equation of plane in scalar product form, angle, and vertices parallel and perpendicular to other planes. This exercise will help students learn critical concepts of vector algebra.

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RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise
The Plane Exercise: VSA
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

The Plane Exercise: VSA

The Plane exercise very short answer type question 1

Answer:
Z = 5
Hint:
Equation of the XY plane ⇒ Z = 0
Given:
plane parallel to XOY plane and passing through the point (2,-3, 5).
Solution:
Equation of the XY plane ⇒ Z = 0
Equation of the plane parallel to XOY plane ⇒Z=k …(i)
If Z = k Passing through (2, -3, 5) then 5 = k …(ii)
Equating (i) and (ii) ⇒ Z = 5

The Plane exercise very short answer type question 2

Answer:
X = - 4
Hint:
Equation of YZ plane ⇒ X = 0
Given:
plane parallel to YOZ plane and passing through (-4, 1, 0)
Solution:
Equation of YZ plane ⇒ X = 0
Equation of the plane parallel to YOZ plane ⇒ X = k …(i)
If X=k Passing through (-4, 1, 0) then -4 = k …(ii)
Equating (i) and (ii) ⇒ X = -4

The Plane exercise very short answer type question 3

Answer:
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Hint:
use formula
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Given:
plane passing through (a, o, o), (o, b, o) and (o, o, c)
Solution:
plane will cut x- axis at a
y- axis at b
z- axis at c
General eqn of plane
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

The Plane exercise very short answer type question 4

Answer:
by + cz + d = 0
Hint:
plane lie in yz plane.
Given:
plane parallel to x-axis
Solution:
Let the general eqn of plane is ax + by + cz + d = 0 …(i)
This plane is parallel to x-axis.
It means that this plane will pass through
a.0 + b.y + c.z + d = 0
⇒ by + cz + d = 0

The Plane exercise very short answer type question 5

Answer:
k = -8
Hint:
If two vectors are perpendicular then their dot product is zero
$\overrightarrow{n_1}.\overrightarrow{n_2}=0$
Given:
Equation of plane x - 2y + kz = 4 and 2x + 5y - z = 9
Solution:
Equation of first plane x - 2y + kz = 4 … (i)
Normal vector of plane (i) is
$\overrightarrow{n_1}=\widehat{i}-2\widehat{j}+k\widehat{k}$
Equation of second plane 2x + 5y - z = 9 …(ii)
Normal vector of plane (ii) is
$\overrightarrow{n_2}=2\widehat{i}+5\widehat{j}-\widehat{k}$
(i) and (ii) are perpendicular to each other
$\begin{aligned} &\Rightarrow \overrightarrow{n_1}.\overrightarrow{n_2}=0\\ &\Rightarrow (\widehat{i}-2\widehat{j}+k\widehat{k}).(2\widehat{i}+5\widehat{j}-\widehat{k})=0\\ &\Rightarrow 2-10-k\\ &\Rightarrow k = -8 \end{aligned}$

The Plane exercise very short answer type question 6

Answer:
x- axis at 6, y- axis at -4, z-axis at 3
Hint:
Intercept form of plane
$\frac{x}{a}+\frac{y}{b}+\frac{x}{c}=1$
Given:
Equation of plane 2x - 3y + 4z = 12
Solution:
2x - 3y + 4z = 12
$\begin{aligned} &\frac{x}{6}-\frac{y}{4}+\frac{x}{3}=1\\ &\frac{x}{6}+\frac{y}{-4}+\frac{x}{3}=1 \end{aligned}$
Plane intercept x- axis at 6
y- axis at -4
z- axis at 3

The Plane exercise very short answer type question 7

Answer:
2 : 1
Hint:
$C = \left ( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n} \right )$
Given:
4x + 5y - 3z = 8 divides the line segment joining the points (-2, 1, 5) & (3, 3, 2)
Solution:
Let the coordinate of A (-2, 1, 5) and B(3, 3, 2)
Let 4x + 5y - 3z = 8 …(i) divides the line segment AB at C in K : 1
C = internal section of AB
Co- ordinate of
$C = \left ( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n} \right )$
$= \left ( \frac{3k-2}{k+1}, \frac{3k+1}{k+1}, \frac{2k+5}{k+1} \right )$
Put the coordinate of C in (i)
$\begin{aligned} &4 \left ( \frac{3k-2}{k+1} \right )+5 \left ( \frac{3k+1}{k+1} \right )-3 \left ( \frac{2k+5}{k+1} \right )=8\\ &\Rightarrow 12k -8+15k+5-6k-15=8k+8\\ &\Rightarrow 13k=26\\ &\Rightarrow k=2 \end{aligned}$
Ratio K: 1 will be 2:1

The Plane exercise very short answer type question 8

Answer:
$\sqrt{14}\: \: units$
Hint:
Distance =
$\frac{\left | D_1-D_2 \right |}{\sqrt{a^2+b^2+c^2}}$
Given:
parallel planes are 2x - y + 3z = 4 and 2x - y + 3z = 18
Solution:
2x - y + 3z = 4
2x - y + 3z = 18
Distance =
$\frac{\left | D_1-D_2 \right |}{\sqrt{a^2+b^2+c^2}}$
$\sqrt{a^2+b^2+c^2}=\sqrt{(2)^2+(-1)^2+(3)^2}=\sqrt{4+1+9}=\sqrt{14}\: units$
$D_1-D_2=18-4=14$
$\therefore \text { Distance }=\frac{14}{\sqrt{14}}=\sqrt{14}\: units$

The Plane exercise very short answer type question 9

Answer:
$\overrightarrow{r}.\left ( \frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k} \right )=2$
Hint:
$\overrightarrow{r}.\frac{\overrightarrow{n}}{\left | \overrightarrow{n} \right |}=\frac{d}{\left | \overrightarrow{n} \right |}$
Given:
$\overrightarrow{r}.(2\widehat{i}+3\widehat{j}-6\widehat{k})=14$
Solution:
$\begin{aligned} &\overrightarrow{n}=2\widehat{i}+3\widehat{j}-6\widehat{k}\\ &\left |\overrightarrow{n} \right |=\sqrt{2^2+3^2+(-6)^2}\\ &=\sqrt{4+9+36}\\ &=\sqrt{49} = 7 \end{aligned}$
Plane in normal form,
$\overrightarrow{r}.\left ( \frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k} \right )=2$

The Plane exercise very short answer type question 10

Answer:
4 units
Hint:
$d= \left | \frac{Ax+By+Cz+D}{\sqrt{a^2+b^2+c^2}} \right |$
Given:
$\overrightarrow{r}.(2\widehat{i}-\widehat{j}+2\widehat{k})=12$
Solution:
$\begin{aligned} &\overrightarrow{r}.(2\widehat{i}-\widehat{j}+2\widehat{k})=12\\ &(x\widehat{i}+y\widehat{j}+z\widehat{k}).(2\widehat{i}-\widehat{j}+2\widehat{k})=12\\ &\Rightarrow 2x-y+2z=12\\ &\Rightarrow 2x-y+2z-12=0 \end{aligned}$
Point (0, 0, 0)
$\begin{aligned} &d= \left | \frac{Ax+By+Cz+D}{\sqrt{a^2+b^2+c^2}} \right |\\ &= \left | \frac{2(0)+(-1)(0)+2(0)-12}{\sqrt{2^2+(-1)^2+2^2}} \right |\\ &=\left | \frac{12}{\sqrt{9}} \right |\\ &=\left |\frac{-12}{3} \right |\\ &=4 \: \: units \end{aligned}$

The Plane exercise very short answer type question 11

Answer:
$(\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{b}\times \overrightarrow{c})$
Hint:
Equation of plane in scalar product form
$(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$
Given:
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}+\mu \overrightarrow{c}$
Solution:
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}+\mu \overrightarrow{c}$
Plane passes through $\overrightarrow{a}$ and parallel to $\overrightarrow{b}$ & $\overrightarrow{c}$ .
Normal vector will be ⊥ to $\overrightarrow{b}$ & $\overrightarrow{c}$ .
$\overrightarrow{n}=\overrightarrow{b} \times \overrightarrow{c}$
Putting in
$(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$
$(\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{b}\times \overrightarrow{c})=0$

The Plane exercise very short answer type question 12

Answer:
$\overrightarrow{n}=\overrightarrow{b} \times \overrightarrow{c}$
Hint:
Plane parallel to
$\overrightarrow{b} \text { and } \overrightarrow{c}$
Given:
$\overrightarrow{r}=l\overrightarrow{b}+m\overrightarrow{c}$
Solution:
$\overrightarrow{r}=l\overrightarrow{b}+m\overrightarrow{c} \qquad \qquad \dots(i)$
Plane (i) parallel to
$\overrightarrow{b} \text { and } \overrightarrow{c}$
Normal vector of (i) ⊥ to be both
$\overrightarrow{b} \times \overrightarrow{c}$
$\overrightarrow{n}=\overrightarrow{b} \times \overrightarrow{c}$

The Plane exercise very short answer type question 13

Answer:
3x + 2y - z = 3
Hint:
Plane parallel to 3x + 2y - z = 7 is 3x + 2y - z = λ
Given:
Plane passing through (2,-1,1) and ? to 3x + 2y - z = 7
Solution:
3x + 2y - z = 7 ...(i)
∴ Eqn of parallel to (i)
3x + 2y - z = λ ...(ii)
(ii) Passes through (2,-1,1)
3(2) + 2(-1) - (1) = λ
λ = 3
∴ Eqn. (ii) Become 3x + 2y - z = 3

The Plane exercise very short answer type question 14

Answer:
$(\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{b}\times \overrightarrow{c})$
Hint:
Equation of plane in scalar product form
$(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$
Given:
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \text { and }\overrightarrow{r}=\overrightarrow{a}+\mu \overrightarrow{c}$
Solution:
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \text { and }\overrightarrow{r}=\overrightarrow{a}+\mu \overrightarrow{c}$
Plane will pass through $\overrightarrow{a}$ .
Normal of plane will be ⊥ to be both $\overrightarrow{b}$ & $\overrightarrow{c}$
$\overrightarrow{n}=\overrightarrow{b}\times \overrightarrow{c}$
∴ General Eqn of the plane
$(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$
$\Rightarrow (\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{b}\times \overrightarrow{c})=0$

The Plane exercise very short answer type question 15

Answer:
$\overrightarrow{r}=\overrightarrow{a}+\left ( \frac{\overrightarrow{a}-\overrightarrow{n}}{\overrightarrow{b}-\overrightarrow{n}} \right ).\overrightarrow{b}$
Hint:
Put Eqn (i) in Eqn (ii)
Given:
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \text { meets the plane }\overrightarrow{r}.\overrightarrow{n}=0$
Solution:
$\begin{aligned} &\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \qquad \qquad \dots(i)\\ &\overrightarrow{r}.\overrightarrow{n}=0 \qquad \qquad \dots(ii) \end{aligned}$
Put (i) in(ii).
$\begin{aligned} &(\overrightarrow{a}+\lambda \overrightarrow{b}).\overrightarrow{n}=0\\ &\Rightarrow \overrightarrow{a}.\overrightarrow{n}+\lambda \overrightarrow{b}.\overrightarrow{n}=0\\ &\Rightarrow \lambda \overrightarrow{b}.\overrightarrow{n}=-\overrightarrow{a}.\overrightarrow{n}\\ &\Rightarrow \lambda =-\frac{\overrightarrow{a}.\overrightarrow{n}}{\overrightarrow{b}.\overrightarrow{n}} \end{aligned}$
Putting the value of λ in eqn (i), we get
$\vec{r}=\vec{a}-(\frac{\vec{a}\cdot\vec{n}}{\vec{b}\cdot\vec{n}})\vec{b}$

The Plane exercise very short answer type question 16

Answer:
$-\frac{13}{4}$
Hint:
Use this formula if two planes are perpendicular a₁a₂+b₁b₂+c₁c₂=0
Given:
$\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{k} \perp \text { to normal of plane } \overrightarrow{r}.(2\widehat{i}+3\widehat{j}+4\widehat{k})=4$
Solution:
$\overrightarrow{r}.(2\widehat{i}+3\widehat{j}+4\widehat{k})=4 \qquad \qquad \dots(i)$
So the vector normal to the plane as,
$\overrightarrow{n}=2\widehat{i}+3\widehat{j}+4\widehat{k}$
a₁= 2, b₁= 3, c₁ = 4
and
$\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{k} \qquad \qquad \dots (ii)$
a₂= 2, b₂= 3, c₂ = k
Since (i) and (ii) ⊥ to each other.
$\begin{aligned} &a_1a_2+b_1b_2+c_1C_2=0\\ &2.2+3.3+4.k=0\\ &\Rightarrow 4+9+4k=0\\ &\Rightarrow 4k=-13\\ &\Rightarrow k=-\frac{13}{4} \end{aligned}$

The Plane exercise very short answer type question 17

Answer:
45°
Hint:
$cos(90-\theta )=\frac{\left | \overrightarrow{n}.\overrightarrow{b} \right |}{\left | \overrightarrow{n} \right |.\left | \overrightarrow{b} \right |}$
Given:
$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2} \text { and } x+y+4=0$
Solution:
$x+y+4=0 \qquad \qquad \qquad \dots(i)$
So the vector normal to the plane (i) is
$\overrightarrow{n}=\overrightarrow{i}+\overrightarrow{j}$
Cartesian equation of plane is
$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2} \qquad \qquad \qquad \dots(ii)$
$\overrightarrow{b}=2\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$
Since
$cos(90-\theta )=\frac{\left | \overrightarrow{n}.\overrightarrow{b} \right |}{\left | \overrightarrow{n} \right |.\left | \overrightarrow{b} \right |}$
$\begin{aligned} &cos(90-\theta )=\frac{2+1}{\sqrt{1+1}.\sqrt{4+1+4}}\\ &cos(90-\theta )=\frac{3}{\sqrt{2}.3}\\ &cos(90-\theta )=\frac{1}{\sqrt{2}}\\ &\Rightarrow sin\theta =\frac{1}{\sqrt{2}}\\ &\Rightarrow \theta =45^{\circ} \end{aligned}$

The Plane exercise very short answer type question 18

Answer:
$\frac{5}{2}$
Hint:
Intercept form of plane
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Given:
2x + y - z = 5
Solution:
2x + y - z = 5 .....(i)
$\begin{aligned} &\Rightarrow \frac{2x}{5}+\frac{y}{5}-\frac{z}{5}=1\\ &\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}-\frac{z}{5}=1\\ &\Rightarrow a=\frac{5}{2} \end{aligned}$
Plane (i) cut x-axis at
$\begin{aligned} &\frac{5}{2} \end{aligned}$

The Plane exercise very short answer type question 19

Answer:
3 units
Hint:
$d=\left | \frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \right |$
Given:
2x - 3y + 6z + 21 = 0
Solution:
2x - 3y + 6z + 21 = 0 ......(i)
Point of origin (0,0,0)
Distance of plane (i) from (0,0,0)
$\begin{aligned} &d=\left | \frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \right |\\ &d=\left | \frac{2\times 0+3\times 0+6\times 0+21}{\sqrt{2^2+(-3)^2+6^2}} \right |\\ &d=\left | \frac{21}{\sqrt{49}} \right |\\ &d=\left | \frac{21}{7} \right |\\ &d=3 \end{aligned}$

The Plane exercise very short answer type question 20

Answer:
$(\widehat{i}-2\widehat{j}-3\widehat{k})+(2\widehat{i}+\widehat{j}+2\widehat{k})$
Hint:
we will use vector equation of line as
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
Given:
Line passing through (1,-2,-3) and normal the plane
$\overrightarrow{r}.(2\widehat{i}+\widehat{j}+2\widehat{k})=5$
Solution:
Position vector of point (1,-2,-3).
$\overrightarrow{a}=(\widehat{i}-2\widehat{j}-3\widehat{k})$
Given equation of plane
$\overrightarrow{r}.(2\widehat{i}+\widehat{j}+2\widehat{k})=5$
So the vector, normal to the plane is
$\overrightarrow{b}=(2\widehat{i}+\widehat{j}+2\widehat{k})$
∴Eqn. of line
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
$\overrightarrow{r}=(\widehat{i}-2\widehat{j}-3\widehat{k})+\lambda (2\widehat{i}+\widehat{j}+2\widehat{k})$
Where λ is parameter.

The Plane exercise very short answer type question 21

Answer:
$\left \{ \overrightarrow{r}-(a\widehat{i}+b\widehat{j}+c\widehat{k}) \right \}(\widehat{i}+\widehat{j}+\widehat{k})=0$
Hint:
we will use equation of plane as
$\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{d}$
Given:
Plane passing through (a,b,c) and parallel to plane
$\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2$
Solution:
$\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2 \qquad \qquad \dots(i)$
So the vector, normal to the plane is
$\overrightarrow{n}=\widehat{i}+\widehat{j}+\widehat{k}$
∴Eqn. of plane to (i) is
$\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{d}$
$\Rightarrow \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=\overrightarrow{d} \qquad \qquad \dots(ii)$
Plane (ii) passing through (a,b,c)
$\Rightarrow (a\widehat{i}+b\widehat{j}+c\widehat{k}) .(\widehat{i}+\widehat{j}+\widehat{k})=\overrightarrow{d} \qquad \qquad \dots(iii)$
Putting (iii) in (ii), we get
$\begin{aligned} &\Rightarrow \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=(a\widehat{i}+b\widehat{j}+c\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})\\ \end{aligned}$
$\Rightarrow \left \{ \overrightarrow{r}-(a\widehat{i}+b\widehat{j}+c\widehat{k}) \right \}(\widehat{i}+\widehat{j}+\widehat{k})=0$

The Plane exercise very short answer type question 22

Answer:
$\overrightarrow{r}.(2\widehat{i}-3\widehat{j}+6\widehat{k})=35$
Hint:
we will use equation of plane as
$\overrightarrow{r}.\widehat{n}=\overrightarrow{d}$
Given:
Normal vector of plane is
$2\widehat{i}-3\widehat{j}+6\widehat{k}$
Solution:
$\overrightarrow{n}=2\widehat{i}-3\widehat{j}+6\widehat{k}$
and distance from the origin =5 units
We know that
$\overrightarrow{r}.\widehat{n}=\overrightarrow{d}$
$\begin{aligned} &\Rightarrow \overrightarrow{r}.\frac{(2\widehat{i}-3\widehat{j}+6\widehat{k})}{\sqrt{4+9+36}}=5\\ &\Rightarrow \overrightarrow{r}.(2\widehat{i}-3\widehat{j}+6\widehat{k})=5\times \sqrt{49}\\ &\Rightarrow \overrightarrow{r}.(2\widehat{i}-3\widehat{j}+6\widehat{k})=5\times 7\\ &\Rightarrow \overrightarrow{r}.(2\widehat{i}-3\widehat{j}+6\widehat{k})=35 \end{aligned}$

The Plane exercise very short answer type question 23

Answer:
x + y + z = 15
Hint:
we will use equation of plane as lx + my + nz = d
Given:
Distance from the origin $5\sqrt{3}$
Solution:
Since the normal of plane equation inclined to co-ordinate axes
$\therefore cos \: \alpha =cos\: \beta =cos\: \gamma$
We know that
$\begin{aligned} &cos^2 \: \alpha +cos^2\: \beta +cos^2\: \gamma =1\\ &\Rightarrow 3cos^2\alpha =1\\ &\Rightarrow cos\: \alpha =\frac{1}{\sqrt{3}}\\ &cos \: \alpha =cos\: \beta =cos\: \gamma =\frac{1}{\sqrt{3}}\\ &l=\frac{1}{\sqrt{3}},\; m=\frac{1}{\sqrt{3}},\;n= \frac{1}{\sqrt{3}} \end{aligned}$
∴Equation of plane lx + my + nz = d, d = distance from origin $\begin{aligned} &(5\sqrt{3}) \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=5\sqrt{3}\\ &\Rightarrow x + y+z =15 \end{aligned}$

The Plane exercise very short answer type question 24

Answer:
$cos^{-1}\frac{11}{21}$
Hint:
$cos\theta =\frac{\left | \overrightarrow{n_1}\overrightarrow{n_2} \right |}{\left | \overrightarrow{n_1} \right |.\left | \overrightarrow{n_2} \right |}$
Given:
$\overrightarrow{r}.(\widehat{i}-2\widehat{j}-2\widehat{k})=1 \text { and }\overrightarrow{r}.(3\widehat{i}-6\widehat{j}+2\widehat{k})=0$
Solution:
$\overrightarrow{r}.(\widehat{i}-2\widehat{j}-2\widehat{k})=1 \qquad \qquad \dots(i)$
So the vector, normal to the plane is
$\overrightarrow{n_1}=(\widehat{i}-2\widehat{j}-2\widehat{k})$
$\overrightarrow{r}.(3\widehat{i}-6\widehat{j}+2\widehat{k})=0 \qquad \qquad \dots (ii)$
So the vector, normal to the plane is
$\overrightarrow{n_2}=(3\widehat{i}-6\widehat{j}+2\widehat{k})$
Angle between (i) & (ii)
$\begin{aligned} &\Rightarrow cos\theta =\frac{\left | \overrightarrow{n_1}\overrightarrow{n_2} \right |}{\left | \overrightarrow{n_1} \right |.\left | \overrightarrow{n_2} \right |}\\ &\Rightarrow cos\theta = \frac{\left | (\widehat{i}-2\widehat{j}-2\widehat{k}).(3\widehat{i}-6\widehat{j}+2\widehat{k}) \right |}{\left | (\widehat{i}-2\widehat{j}-2\widehat{k}) \right |\left | (3\widehat{i}-6\widehat{j}+2\widehat{k}) \right |}\\ &\Rightarrow cos\theta =\frac{3+12-4}{\sqrt{1+4+4}.\sqrt{9+36+4}}\\ &\Rightarrow cos\theta =\frac{11}{3\times 7}\\ &\Rightarrow cos\theta =\frac{11}{21}\\ &\Rightarrow \theta =cos^{-1}\frac{11}{21} \end{aligned}$

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