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

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

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:52 AM IST

When it comes to preparing for board exams, RD Sharma books are the best. For many students, board exams have always been a ruse. Still, when they refer to this book in combination with Rd Sharma Class 12th Exercise MCQ solutions, they find it much easier to relax and understand the concept in a much better manner. RD Sharma solutions This exercise contains 24 questions which are based on the Equation of the plane containing two lines, Equation of the plane in scalar product form, Reflection of point, Vector and Cartesian form, equation in normal form and distance between the plane.

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RD Sharma Class 12 Solutions Chapter28 MCQ The Plane - Other Exercise

The Plane Excercise: MCQ

The Plane exercise multiple choice question 1

Answer:
Option (a).
Hint:
Simplify the given equation.
Given:
$2x-(1+\lambda )y+3\lambda z=0$
Solution:
The given equation plane is,
$2x-(1+\lambda )y+3\lambda z=0$
We can rewrite the equation of the given plane as,
$2x-(1+\lambda )y+3\lambda z=0$
$2x-y-\lambda (y-3z)=0$

So, the given plane passes through the intersection of the planes

$2x-y=0 \text { and }y-3z=0$ .

The Plane exercise multiple choice question 2

Answer:
Option(b)
Hint:
Angle between two planes is 0.
Given:
$2x-y+z=6 \text { and } x+y+2z=3$
Solution:
We know, the angle between two planes
$a_1x+b_1y+c_1z+d_1=0 \text { and } a_2x+b_2y+c_2z+d_2=0 \text { is }$
$\theta =cos^{-1}\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1\, ^2+b_1\, ^2+c_1\, ^2}.\sqrt{a_2\, ^2+b_2\, ^2+c_2\, ^2}}$
Here a₁ = 2; b₁ = -1; c₁ = 1; d₁ = -6 and a₂ = 1; b₂ = 1; c₂ = 2; d₂ = -3
So, the acute angle between the planes
$2x-y+z=6 \text { and } x+y+2z=3$ is
$\begin{aligned} &\theta =cos^{-1}\frac{((2.1)+(-1.1)+(1.2))}{\sqrt{2^2+(-1)^2+1^2}.\sqrt{1^2+1^2+2^2}}\\ &=cos^{-1}\frac{(2-1+2)}{\sqrt{4+1+1}.\sqrt{1+1+4}}\\ &=cos^{-1}\frac{3}{\sqrt{6}.\sqrt{6}}\\ &=cos^{-1}\frac{3}{6}\\ &\theta =60^{o} \end{aligned}$
Therefore, the acute angle between the planes
$2x-y+z=6 \text { and } x+y+2z=3 \text { is } 60^o$

The Plane exercise multiple choice question 3

Answer:
Option (d)
Hint:
$\lambda$ is a scalar quantity.
Given:
$x+2y+3z=4 \text { and }2x+y-z=-5$
are perpendicular to the plane
$5x+3y+6z+8=0$
Solution:
The equation of the plane through the intersection of the planes
$x+2y+3z=4 \text { or } x+2y+3x-4=0 \text { and }2x+y-z=-5 \text { or } 2x+y-z+5=0$
is given as,
$\begin{aligned} &(x+2y+3z-4)-\lambda (2x+y-z+5)=0\\ &x(1+2\lambda )+y(2+\lambda )+z(3-\lambda )-4+5\lambda =0, \end{aligned}$
where $\begin{aligned} &\lambda \end{aligned}$ is a scalar.
Given, that the required plane is perpendicular to the plane
$\begin{aligned} &5x+3y+6z+8=0 \end{aligned}$
so we have,
$\begin{aligned} &5(1+2\lambda )+3(2+\lambda )+6(3-\lambda )=0\\ &5+10\lambda +6+3\lambda +18-6\lambda =0\\ &29+7\lambda =0\\ &\lambda =-\frac{29}{7} \end{aligned}$
Therefore, the equation of the required plane is,
$\begin{aligned} &(x+2y+3z-4)-\frac{29}{7}(2x+y-z+5)=0\\ &7(x+2y+3z-4)-29(2x+y-z+5)=0\\ &7x+14y+21z-28-58x-29y+29z-145=0\\ &-51x-15y+50z-173=0\\ &51x+15y-50z+173=0 \end{aligned}$

The Plane exercise multiple choice question 4

Answer:
Option (c).
Hint:
Distance between two parallel planes is 0.
Given:
$2x+2y-z=0 \text { and } 4x+4y-2z+5=0$
Solution:
We know that, the distance between two parallel planes:
$Ax+By+Cz+D_1=0 \qquad \qquad \dots(1)$
and
$Ax+By+Cz+D_2=0 \qquad \qquad \dots(2)$
is given by
$D=\frac{\left | D_2-D_{1} \right |}{\sqrt{A^2+B^2+C^2}}$
Here, the two parallel planes are given as:
$2x+2y-z+2=0 \qquad \qquad \dots (3)$
and
$4x+4y-2z+5=0$
i.e.
$2x+2y-z+\frac{5}{2}=0 \qquad \qquad \dots (4)$
Compiling equation (3) with equation (1), and equation (4) with equation (2), we get;
$A = 2,B=2,C=-1,D_1=2,D_2=\frac{5}{2}.$
So, the distance between the given two parallel planes are,
$\begin{aligned} &D=\frac{\left | \frac{5}{2}-2 \right |}{\sqrt{2^2+2^2+(-1)^2}}\\ &D=\frac{\frac{1}{2}}{\sqrt{4+4+1}}\\ &D=\frac{1}{2\sqrt{9}}\\ &D=\frac{1}{2\times 3}\\ &D=\frac{1}{6} \end{aligned}$
Hence, the distance between the parallel planes
$2x+2y-z=0 \text { and } 4x+4y-2z+5=0 \text { is } \frac{1}{6}.$

The Plane exercise multiple choice question 5

Answer:
Option (b)
Hint:
Solve the equation simultaneously.
Given:
(1, 3, 4) in the plane 2x - y + z + 3 = 0
Solution:
We know, if the image of a point P(x₀, y₀, z₀) on a plane Ax + By + Cz + D = 0 . . . (1) is Q(x₁, y₁, z₁) then

The given plane is 2x - y + z + 3 = 0 . . . . .(2)
By comparing equations (2) and (3)
A = 2, B = -1, C = 1, D = 3
And here x_0 = 1, y_0 = 3, z_0 = 4
So
$\begin{aligned} &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2[(2\times 1)+((-1)\times 3)+(1\times 4)]}{2^2+(-1)^2+1^2}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2(2-3+4+3)}{4+1+1}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=\frac{-2\times 6}{6}\\ &\frac{x_1-1}{2}=\frac{y_1-3}{-1}=\frac{z_1-4}{1}=-2 \end{aligned}$
Therefor,
x₁ = -4 + 1
=-3
y₁ = 2 + 3
=5
z₁ = -2 + 4
=2
So, the image of the (1, 3, 4), in the plane 2x - y + 3 = 0 , is (-3, 5, 2).

The Plane exercise multiple choice question 6

Answer:
Option (d).
Hint:
See for co-planer.
Given:
$\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-0}{3} \text { and } \frac{x-0}{-2}=\frac{y-2}{-3}=\frac{z+1}{-1}$
Solution:
We know, the two lines given as,
$\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-0}{3} \text { and } \frac{x-0}{-2}=\frac{y-2}{-3}=\frac{z+1}{-1}$
$\begin{aligned} &\vec{b_1}=2\hat{i}-\hat{j}+3\hat{k}\\ &\vec{b_2}=-\hat{i}-3\hat{j}-\hat{k}\\ &=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 2 &-1 &3 \\ -1 &-3 &-1 \end{vmatrix}\\ &=-8\hat{i}-\hat{j}+5\hat{k} \end{aligned}$
Lines passes through (1,-1,0)
$\begin{aligned} &(x-x_1)a+(y-y_1)b+(z-z_1)c=0\\ &(x-1)(-8)+(y+1)(-1)+(z-0)5=0\\ &-8x+8-y-1+5z=0\\ &-8x-y+5z+7=0\\ &8x+y-5z-7=0 \end{aligned}$
Hence, option a is correct.

The Plane exercise multiple choice question 7

Answer:
Option (a)
Hint:
Use cross product of vector.
Given:
$\vec{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})$
Solution:
The given plane is
$\vec{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})$
So, it is clear from the given equation of plane, that the plane passing through a point $(\hat{i}-\hat{j})$ and parallel to the two vectors $(\hat{i}+\hat{j}+\hat{k}) \text { and } (\hat{i}-2\hat{j}+3\hat{k}).$
Therefore, the equation of the vector normal to the plane is given as
$\begin{aligned} &\widehat{n}=[(\hat{i}+\hat{j}+\hat{k}) \times (\hat{i}-2\hat{j}+3\hat{k})]\\ &=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 &1 &1 \\ 1 &-2 &3 \end{vmatrix}\\ &=\left [ \left \{ (1\times 3)-(-2\times 1) \right \}\widehat{i}-\left \{ (1\times 3)-(1\times 1) \right \}\widehat{j}+\left \{ (1\times -2)-(1\times 1)\widehat{k} \right \} \right ]\\ &=5\widehat{i}-2\widehat{j}-3\widehat{k} \end{aligned}$
So, in scalar product form the vector equation of the plane is given as,
$\begin{aligned} &(\overrightarrow{r}-( \hat{i}-\hat{j})).\widehat{n}=0\\ &(\overrightarrow{r}-( \hat{i}-\hat{j})).(5\widehat{i}-2\widehat{j}-3\widehat{k})=0\\ &\overrightarrow{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=5+2\\ &\overrightarrow{r}.(5\widehat{i}-2\widehat{j}-3\widehat{k})=7 \end{aligned}$

The Plane exercise multiple choice question 8

Answer:
Option (b)
Hint:
Use scalar product of vector
Given:
$\overrightarrow{r}=2\widehat{i}-2\widehat{j}+3\widehat{k}+\lambda (\widehat{i}+\widehat{j}+4\widehat{k})$
from the plane
$\overrightarrow{r}.(\widehat{i}+5\widehat{j}+\widehat{k})=5$
Solution:
We have the straight line given as,
$\overrightarrow{r}=2\widehat{i}-2\widehat{j}+3\widehat{k}+\lambda (\widehat{i}+\widehat{j}+4\widehat{k})$
And the plane as,
$\begin{aligned} &\overrightarrow{r}.(\widehat{i}+5\widehat{j}+\widehat{k})=5\\ &x-5y+2=5\\ &\text { i.e. }x-5y+2-5=0 \end{aligned}$
the normal vector of the plane given as,
$\overrightarrow{n}=(\widehat{i}-5\widehat{j}+\widehat{k})$
if the straight line and the plane are parallel,
scalar product will be zero.
$\begin{aligned} &(\widehat{i}+\widehat{j}+4\widehat{k}).(\widehat{i}-5\widehat{j}+\widehat{k})=1+[1\times (-5)]+(4\times 1)\\ &=1-5+4\\ &=0 \end{aligned}$
Hence, the point (2, -2, 3) is on the straight line. Distance from point (2, -2, 3) to the plane will be equal to the distance of the line from the plane. We know, that the distance of a point (x0, y0, z0) from plane Ax + By + Cz + D = 0 ...(2) is
$\begin{aligned} &=\frac{\left | Ax_0+By_0+Cz_0+D \right |}{\sqrt{A^2+B^2+C^2}} \end{aligned}$
On comparing, equation (1), i.e., x - 5y + z - 5 = 0 with equation (2), we get,
A = 1, B = -5, C = 1, D = -5
So, the distance is,
$\begin{aligned} &=\frac{\left [ (1\times 2)+(-5\times -2)+(1\times 3)+(-5) \right ]}{\sqrt{1^2+(-5)^2+1^2}}\\ &=\frac{\left | 2+10+3-5 \right |}{\sqrt{1+25+1}}\\ &=\frac{10}{\sqrt{27}}\\ &=\frac{10}{3\sqrt{3}} \end{aligned}$

The Plane exercise multiple choice question 9

Answer:
Option (a)
Hint:
Solve the given equations simultaneously.
Given:
$x+y+z+3=0,\: 2x-y+3z+1=0 \text { and } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
Solution:
Equation of line passing through the line
$x+y+z+3=0,\: 2x-y+3z+1=0$
is given by,
$\begin{aligned} &x+y+z+3+k(2x-y+3z+1)=0 \qquad \qquad \dots (1)\\ &x(1+2k)+y(1-k)+z(1+3k)+3+k=0 \end{aligned}$
where k is constant
Again, the required plane is parallel to the line
$\begin{aligned} &\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \end{aligned}$
So, we have
$\begin{aligned} &\left [ 1\times (1+2k) \right ]+\left [ 2\times (1-k) \right ]+\left [ 3\times (1+3k) \right ]=0\\ &1+2k+2-2k+3+9k=0\\ &9k=-6\\ &k=-\frac{2}{3} \end{aligned}$
Put k in eq. (1)
$\begin{aligned} &(x+y+z+3)-\frac{-2}{3}(2x-y+3z+1)=0\\ &3(x+y+z+3)-2(2x-y+3z+1)=0\\ &3x+3y+3z+9-4x+2y-6z-2=0\\ &x-5y+3z=7 \end{aligned}$
Therefore, the equation of the plane through the line x + y + z + 3 = 0, 2x - y + 3z + 1 = 0 and the parallel to the line
$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
x - 5y + 3z = 7

The Plane exercise multiple choice question 10

Answer:
Option (a)
Hint:
Put $\lambda = 1$
Given:
$\overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})$
Solution:
The plane contains the line
$\overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})$
so, the plane contains the point
$(\widehat{i}+2\widehat{j}+3\widehat{k})$
Put $\lambda = 1$
$-2\widehat{i}+3\widehat{j}+4\widehat{k}+3\widehat{i}-2\widehat{j}-\widehat{k} \text { i.e } (\widehat{i}-5\widehat{j}+3\widehat{k})$
so, we got the plane, they are:
$(\widehat{i}+2\widehat{j}+3\widehat{k}); (-2\widehat{i}-3\widehat{j}+4\widehat{k}); \text { and } (\widehat{i}-5\widehat{j}+3\widehat{k})$
Let,
$\begin{aligned} &\overrightarrow{a}=(\widehat{i}-5\widehat{j}+3\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k}) \text { and }\\ &\overrightarrow{b}=(\widehat{i}-5\widehat{j}+3\widehat{k})-(-2\widehat{i}-3\widehat{j}+4\widehat{k})\\ &\text { So, }\overrightarrow{a}=-7\widehat{j}\text { and }\overrightarrow{b}=3\widehat{i}-2\widehat{j}-\widehat{k} \end{aligned}$

The normal of two vectors $\begin{aligned} &\overrightarrow{a} \end{aligned}$ and $\begin{aligned} &\overrightarrow{b} \end{aligned}$

$\begin{aligned} &\overrightarrow{a}=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 0 &-7 &0 \\ 3 &-2 &1 \end{vmatrix}\\ &=\left [ \left \{ (-7)\times (-1)-((-2)\times 0) \right \}\widehat{i}+\left \{ (0\times (-1))-(0\times (-2)) \right \}\widehat{j}+\left \{ (0\times (-2))-((-7)\times 3) \right \}\widehat{k} \right ]\\ &=7\widehat{i}+21\widehat{k} \end{aligned}$
The general equation of plane is,
$\begin{aligned} &\left [ (x-1)\widehat{i}+(y-2)\widehat{j}+(z-3)\widehat{k} \right ].(7\widehat{i}+21\widehat{k})=0\\ &7(x-1)+21(z-3)=0\\ &7x-7+21z-63=0\\ &7x+21z=70\\ &x+3z=10\\ &\text { or }\\ &\overrightarrow{r}.(\widehat{i}+3\widehat{k})=10, \qquad \left [ \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k} \right ] \end{aligned}$
Hence, the vector equation of the plane containing the line
$\overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})$
and of the point
$(\widehat{i}+2\widehat{j}+3\widehat{k})$
is
$\begin{aligned}&\overrightarrow{r}.(\widehat{i}+3\widehat{k})=10 \end{aligned}$

The Plane exercise multiple choice question 11

Answer:
Option (c)
Hint:
Use the centroid formula.
Given:
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=k$
Solution:
A plane meets the co-ordinate area of (A, B, C) such that the centroid of $\Delta ABC$ is the point (a,b,c).
Let the co-ordinate of the point
$A(\alpha , 0, 0),\: B(0,\beta ,0),\: C(0,0,\gamma )$
By the centroid formula,
$\begin{aligned} &a =\frac{\alpha +0+0}{3}\\ &\Rightarrow \alpha =3a\\ &b=\frac{0+\beta +0}{3}\\ &\Rightarrow \beta =3b\\ &c=\frac{0+0+\gamma }{3}\\ &\Rightarrow \gamma =3c \end{aligned}$
The Intercept form of the plane–
$\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$
If the plane makes intercepts at (p, 0, 0), (0, q, 0) and (0, 0, r) with the axes x, y, z respectively.
Here,
$p=\alpha =3a,\: q=\beta =3b,\: r=\gamma =3c$
So, the equation of the plane is,
$\begin{aligned} &\frac{x}{3a}+\frac{y}{3b}+\frac{z}{3c}=1\\ &\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3 \qquad \qquad \qquad \dots(1) \end{aligned}$
By comparing the equation (1) with given equation of the plane
$\begin{aligned} &\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=k \end{aligned}$ ,
we get k = 3.

The Plane exercise multiple choice question 12

Answer:
Option (c)
Hint:
The plane, must satisfy both the equation.
Given:
$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$
Solution:
Let, the point of intersection of the line
$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$
And the plane x + y + z = 17 be (x_0,y₀, z₀) So, (x_0,y₀, z₀) satisfy the both equations,
$\begin{aligned} &\frac{x_0-3}{1}=\frac{y_0-4}{2}=\frac{z_0-5}{2}=k, \text { (Let) }\\ &\text { i.e. }x_0=k+3\\ &y_0=2k+4\\ &z_0=2k+5 \end{aligned}$
Put these values in the equation of plane x + y + z = 17
$\begin{aligned} &(k+3)+(2k+4)+(2k+5)=17\\ &5k+12=17\\ &k=1\\ &\therefore x_0=1+3\\ &=4\\ &y_0=2\times 1+4\\ &=6\\ &z_0=2\times 1+5\\ &=7\\ \end{aligned}$
Hence, the distance between point (4, 6, 7) and (3, 4, 5) is,
$\begin{aligned} &=\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}\\ &=\sqrt{1^2+2^2+2^2}\\ &=\sqrt{1+4+4}\\ &=3 \end{aligned}$

The Plane exercise multiple choice question 13

Answer:
$-2\widehat{i}+7\widehat{j}+13\widehat{k}$
Hint:
Solve linear equations in two variables
Given:
$\overrightarrow{r}.(3\widehat{i}-\widehat{j}+\widehat{k})=1 \text { and } \overrightarrow{r}.(\widehat{i}-4\widehat{j}-2\widehat{k})=2$
Solution:
The two planes are
$\overrightarrow{r}.(3\widehat{i}-\widehat{j}+\widehat{k})=1 \text { and } \overrightarrow{r}.(\widehat{i}-4\widehat{j}-2\widehat{k})=2$
The line of intersection of planes
$\overrightarrow{r}.(3\widehat{i}-\widehat{j}+\widehat{k})=1 \text { and } \overrightarrow{r}.(\widehat{i}-4\widehat{j}-2\widehat{k})=2$
Can be written as 3x - y + z = 1 ……………… (i)
x + 4y - 2z = 2 …………… (ii)
Solving eqn (i) & (ii)
$\frac{x}{-2}=\frac{y}{7}=\frac{z}{13}$
Vector equation is
$-2\widehat{i}+7\widehat{j}+13\widehat{k}$

The Plane exercise multiple choice question 14

Answer:
Option (c)
Hint:
Calculate the perpendicular distance of the plane from the origin.
Given:
$\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$
Solution:
Let, the equation of the plane be Ax + By + Cz + D = 0, as the plane is perpendicular to, so
We have,
A = 3, B = 0 and C = 4 as the plane passes through (1, 1, 1), we have,
$\begin{aligned} &(A\times 1)+(B\times 1)+(C\times 1)+D=0\\ &A+B+C+D=0\\ &3+0+4+D=0\\ &D=-7 \end{aligned}$
So, the equation of the plane becomes
3x + 4z - 7 = 0
Now, the perpendicular distance of the plane from the origin is
$\begin{aligned} &\frac{\left | Ax_0+By_0+Cz_0+D \right |}{\sqrt{A^2+B^2+C^2}}=\frac{\left | (3\times 0)+(0\times 0)+(4\times 0)-7 \right |}{\sqrt{3^2+0^2+4^2}}\\ &\left [ \because (x_0, y_0,z_0)\approx (0,0,0) \right ]\\ &=\frac{\left | 0-7 \right |}{\sqrt{9+16}}\\ &=\frac{\left | -7 \right |}{\sqrt{25}}\\ &=\frac{7}{5} \end{aligned}$

The Plane exercise multiple choice question 15

Answer:
Option (a)
Hint:
Use straight line & vector cross product.
Given:
x - 1 = 2y - 5 = 2z and 3x = 4y -11 = 3z - 4
Solution:
The required plane is parallel to the lines
x - 1 = 2y - 5 = 2z and 3x = 4y - 11 = 3z - 4
Equation of the lines can be written as,
$\frac{x-1}{1}=\frac{2y-5}{1}=\frac{2z}{1}$
or
$\frac{x-1}{1}=\frac{y-\frac{5}{2}}{\frac{1}{2}}=\frac{z}{\frac{1}{2}}$
and,
$\frac{3x}{1}=\frac{4y-11}{1}=\frac{3z-4}{1}$
or,
$\frac{x}{\frac{1}{3}}=\frac{y-\frac{11}{4}}{\frac{1}{4}}=\frac{z-\frac{4}{3}}{\frac{1}{3}}=\mu , \text { Let }$
So, we know the straight lines as
$\begin{aligned} &(1+\lambda )\widehat{i}+\left ( \frac{5}{2}+\frac{1}{2}\lambda \right )\widehat{j}+\frac{1}{2}\widehat{k}=0 \text { or, }\\ &\widehat{i}+\frac{5}{2}\widehat{j}+\lambda \left ( \widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k} \right )=0 \end{aligned}$
And,
$\begin{aligned} &\frac{1}{3}\mu \widehat{i}+\left ( \frac{11}{4}+\frac{1}{4}\mu \right )\widehat{j}+\left ( \frac{4}{3}+\frac{1}{3}\mu \right )\widehat{k}=0, \text { or }\\ &\frac{11}{4}\widehat{j}+\frac{4}{3}\widehat{k}+\mu \left ( \frac{1}{3}\widehat{i}+\frac{1}{4}\widehat{j}+\frac{1}{3}\widehat{k} \right )=0 \end{aligned}$
We have the normal vector of the plane as,
$\begin{aligned} &\overrightarrow{n}=\left (\widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k} \right )\times \left ( \frac{1}{3}\widehat{i}+\frac{1}{4}\widehat{j}+\frac{1}{3}\widehat{k} \right )\\ &=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ \\ 1 &\frac{1}{2} &\frac{1}{2} \\ \\ \frac{1}{3} &\frac{1}{4} &\frac{1}{3} \end{vmatrix}\\ &=\left ( \left ( \frac{1}{2}\times \frac{1}{3} \right )-\left ( \frac{1}{2}\times \frac{1}{4} \right ) \right )\widehat{i}-\left ( \left ( 1\times \frac{1}{3} \right )-\left ( \frac{1}{2}\times \frac{1}{3} \right ) \right )\widehat{j}+\left ( \left ( 1\times \frac{1}{4} \right )-\left (\frac{1}{2}\times \frac{1}{3} \right ) \right )\widehat{k}\\ &=\left ( \frac{1}{6}-\frac{1}{8} \right )\widehat{i}-\left ( \frac{1}{3}-\frac{1}{6} \right )\widehat{j}+\left ( \frac{1}{4}-\frac{1}{6} \right )\widehat{k}\\ &=\left (\frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right ) \end{aligned}$
So, the equation of plane is
$\begin{aligned} &(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0\\ &\text { Where } \overrightarrow{a}=2\widehat{i}+3\widehat{j}+3\widehat{k},\\ &\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}\\ &\left ( x\widehat{i}+y\widehat{j}+z\widehat{k} \right ).\overrightarrow{n}=(2\widehat{i}+3\widehat{j}+3\widehat{k}).\overrightarrow{n}\\ &\left ( x\widehat{i}+y\widehat{j}+z\widehat{k} \right ).\left ( \frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right )=(2\widehat{i}+3\widehat{j}+3\widehat{k}).\left ( \frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right ) \end{aligned}$
x - 4y + 2z + 4 = 0, (by solving further)

The Plane exercise multiple choice question 16

Answer:
Option (b)
Hint:
Use simultaneous equation.
Given:
(-1, -5, -10)
Solution:
Let the point of intersection of the line
$\overrightarrow{r}.(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k}) -1, -5$
And the plane
$\overrightarrow{r}.(\widehat{i}-\widehat{j}+\widehat{k})=5 \text { be } \left (x_0,y_0,z_0 \right )$
As x₀, y₀, z₀ is the point of intersection of the line and the plane, so the position vector of this point i.e.,
$\overrightarrow{r_0}.(x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k})$
Must satisfy both equations of line and the equation of plane.
Substituting $\overrightarrow{r_0}$ in plane of $\overrightarrow{r}$ in both the equations, we get,
$(x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k})=(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k})$
and
$(x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k}).(\widehat{i}-\widehat{j}+\widehat{k})=5 \qquad \qquad \dots (2)$
Substituting, these values in equation (2)
$\begin{aligned} &\left [ ((2+3\lambda )\times 1)-(1\times (-1+4\lambda )) \right ]+\left [ 1\times (2+12\lambda ) \right ]=5\\ &2+3\lambda +1-4\lambda +2+12\lambda =5\\ &11\lambda =0\\ &\lambda =0 \end{aligned}$
Therefore,
$\begin{aligned} &x_0=2+3\lambda \\ &=2,\\ &y_0=-1+4\lambda \\ &=-1\\ &z_0=2+12\lambda \\ &=2 \end{aligned}$
Hence, the point of intersection is (2, -1, 2)
Now, the distance between the point (-1, -5, -10) and (2, -1, 2) is
$\begin{aligned} &=\sqrt{[2-(-1)]^2+[(-1)-(-5)]^2+[2-(-10)]^2}\\ &=\sqrt{3^2+4^2+12^2}\\ &=\sqrt{169}\\ &=13 \end{aligned}$
Hence, the required distance between the point (-1, -5, -10) and the point where the line
$\begin{aligned} &\overrightarrow{r}.(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k}) \end{aligned}$
And the plane
$\begin{aligned} &\overrightarrow{r}.(\widehat{i}-\widehat{j}+\widehat{k})=5 \end{aligned}$
intersects is 13 units

The Plane exercise multiple choice question 17

Answer:
Option (a)
Hint:
$\lambda$ is a scalar.
Given:
ax + by + cz + d = 0
Solution:
The equation of the plane through the intersection of the planes
ax + by + cz + d = 0 and lx + my + nz + p = 0
Is given by
(ax + by + cz + d) + (lx + my + nz + p) = 0, [ where $\lambda$ is a scalar ]
x (a + l $\lambda$ ) + y (b + m $\lambda$ ) + z (c + n $\lambda$ ) + d + p $\lambda$ = 0
Given that the required plane is parallel to the lines y = 0, z = 0, i.e., x-axes so, we have
1 (a + l $\lambda$ ) + 0 ( b + m $\lambda$ ) + 0 (c + n $\lambda$ ) = 0
a + l $\lambda$ = 0
$\lambda =-\frac{a}{l}$
(a + l $\lambda$ )x + (b + m $\lambda$ ) y + (c + n $\lambda$ ) z + (d + pλ ) = 0
$\begin{aligned} &\left ( a+l\times \frac{-a}{l} \right )x +\left ( b+m\times \frac{-a}{l} \right )y+\left ( c+n\times \frac{-a}{l} \right )z+\left ( d+p\times \frac{-a}{l} \right )=0 \end{aligned}$
(b l - a m) y + (c l - a n) z + d l - a p = 0
Hence, option a is correct.

The Plane exercise multiple choice question 18

Answer:
Option (a)
Hint:
Use simultaneous equation.
Given:
The equation of the plane which cuts equal intercepts of unit length of the co-ordinate axes.
Solution:
We know that the general equation of a planes
$Ax+By+Cz+D=0, \text { where } D\neq 0 \qquad \qquad \dots(1)$
Here, A, B, C are the coordinate of a normal vector to the plane, while (x, y, z) are the coordinates of any point through which the plane passes.
Again, we know the intercept form of plane
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \qquad \qquad \dots(2), \text { where }A=\frac{D}{a},B=-\frac{D}{b},C=-\frac{D}{c}$ A
And the plane makes intercepts at (a, 0, 0), (0, b, 0) and (0, 0, c) with the axes x, y, z respectively.
Here a = b = c = 1
Put in the equation (2)
$\frac{x}{1}+\frac{y}{1}+\frac{z}{1}=1$
Hence the equation is x + y + z = 1

The Plane exercise multiple choice question 19

Answer:
Option (d)
Hint:
Solve the equation to get the values of x, y and z
Given:
xy + yz = 0
Solution:
We have
xy + yz = 0
x (y + z) = 0
therefore x = 0, y + z = 0
Above one equation of planes normal to the plane x = 0 is $\widehat{i}$
y + z = 0 is
$\widehat{j}+\widehat{k},$
And normal to the plane
$\widehat{i}.(\widehat{j}+\widehat{k})=0$
So, planes are perpendicular.

The Plane exercise multiple choice question 20

Answer:
Option (d)
Hint:
See the reflection with the graph
Given:
$(\alpha ,\beta ,\gamma )$
Solution:
With the help of graph,
In xy-plane, the reflection of the point $(\alpha ,\beta ,\gamma )$ is $(\alpha ,\beta ,-\gamma )$

The Plane exercise multiple choice question 21

Answer:
Option (c)
Hint:
Use vector dot product for direction.
Given:
2x - 3y + 6z - 11 = 0
Solution:
We have equation of plane are 2x - 3y + 6z - 11 = 0
Normal to the plane is
$\overrightarrow{n} = 2\widehat{i}-3\widehat{j}+6\widehat{k}$
Also x-axis is along the vector
$\overrightarrow{a} = \widehat{i}+0\widehat{j}+0\widehat{k}$
According to the question
$\begin{aligned} &sin \alpha =\frac{\left | \overrightarrow{a}.\overrightarrow{n} \right |}{\left | \overrightarrow{a} \right |.\left | \overrightarrow{n} \right |}\\ &=\frac{\left | \widehat{i}.(2\widehat{i}-3\widehat{j}+6\widehat{k}) \right |}{\sqrt{1}.\sqrt{4+9+36}}\\ &=\frac{2}{7} \end{aligned}$

The Plane exercise multiple choice question 22

Answer:
Option (d)
Hint:
Find the angle with vector dot product.
Given:
$\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$
Solution:
Given line
$\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$
is parallel to the vector
$\overrightarrow{b}=3\widehat{i}+4\widehat{j}+5\widehat{k}$
Equation to the plane is
2x - 2y + z = 5
Normal to the plane is
$\overrightarrow{n}=2\widehat{i}-2\widehat{j}+\widehat{k}$
If angle between line and plane is $\theta$
Then
$\begin{aligned} &sin\theta =\frac{\left | \overrightarrow{b}.\overrightarrow{n} \right |}{\left | \overrightarrow{b} \right |.\left | \overrightarrow{n} \right |}\\ &=\frac{\left | (3\widehat{i}+4\widehat{j}+5\widehat{k}).(2\widehat{i}-2\widehat{j}+\widehat{k}) \right |}{\sqrt{3^2+4^2+5^2.\sqrt{4+4+1}}}\\ &=\frac{\left | 6-8+5 \right |}{\sqrt{50}.\sqrt{9}}\\ &=\frac{3}{15\sqrt{3}}\\ &=\frac{1}{5\sqrt{3}}\\ &\therefore sin\theta =\frac{\sqrt{2}}{10} \end{aligned}$

The Plane exercise multiple choice question 23

Answer:
Option (a)
Hint:
Find magnitude of n vector.
Given:
$\overrightarrow{r}.\left ( \frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k} \right )=1$
Solution:
$\overrightarrow{r}.\left ( \frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k} \right )=1$
Let,
$\begin{aligned} &\overrightarrow{n}=\left ( \frac{2}{7}\widehat{i}+\frac{3}{7}\widehat{j}-\frac{6}{7}\widehat{k} \right )\\ &\left | \overrightarrow{n} \right |=\sqrt{\left ( \frac{2}{7} \right )^2+\left ( \frac{3}{7} \right )^2+\left ( -\frac{6}{7} \right )^2}\\ &=1 \end{aligned}$
N is a unit vector.

The Plane exercise multiple choice question 24

Answer:
Option (b)
Hint:
Use vector addition.
Given:
(-1, 5, 4)
Solution:

Line passes through (-1, 5, 4) and perpendicular to the plane z = 0

Equation of plane

0. x + 0. y + 1. z = 0
Normal of plane (0, 0, 1)
Equation of line
r = (-1, 5, 4) + $\lambda$ (0, 0, 1)
$\therefore \overrightarrow{r}=-\widehat{i}+5\widehat{j}+4\widehat{k}+\lambda \widehat{k}$

Rd Sharma Class 12 Chapter 28 Exercise MCQ is organised in a stepwise manner with hints for complex problems so that a student can find the best path to solve them. The advantages of these solutions are as follows:

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A group of experts creates RD Sharma Class 12 The Plane Ex. MCQ solutions and they provide the most precise answers possible. When students refer to these solutions, they can find various ways to solve a single problem and pick the one that is easiest for them.

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RD Sharma Chapter-wise Solutions

Frequently Asked Question (FAQs)

1. What is the definition of a Plane Figure?

Any geometric figure with no thickness is referred to as a plane figure.

2. Is the RD Sharma textbook a good choice for class 12?

The CBSE recommends RD Sharma as a reference book. RD Sharma's class 12 solutions include numerous solved illustrations, pictorial representations, and more than one method for problem-solving. It also teaches students mathematical tricks and concepts that help them solve problems quickly and accurately. It is an excellent choice for practice questions.

3. Is R S Agrawaal superior to RD Sharma?

RD Sharma explains the concepts more clearly. It also includes a large number of practise questions of various types. As a result, it prepares students to perform better in exams.

4. What is the main concept of the chapter ‘The Plane’?

This chapter explains the plane concept of geometry.

5. What is the best website for studying RD Sharma Solutions for Class 12 Maths?

On the Career360 website, you can find Rd Sharma Class 12th Exercise MCQ solutions, as well as step-by-step answers to all of the questions in the RD Sharma textbook. As a result, in order to understand the important topics, students in Class 12 should learn all of the concepts covered in the syllabus.

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