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.

## The Plane Excercise: MCQ

The Plane exercise multiple choice question 1

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

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

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

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

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

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

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

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

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

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

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

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 (x0, y0, z0) So, (x0, 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

$-2\widehat{i}+7\widehat{j}+13\widehat{k}$
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

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

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

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

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

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

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

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

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

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

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

Option (b)
Hint:
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:

1. Experts create RD Sharma Class 12th Exercise MCQ Solutions:

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.

1. Covers the entire syllabus

RD Sharma Class 12th Exercise MCQ material provided by Career360 contains step-by-step answers to all the questions from the syllabus. As teachers can't cover all the questions through their lectures, students can refer to this material to stay in line with their class and prepare accordingly. Moreover, as Class 12 RD Sharma Chapter 28 Exercise MCQ solutions material complies with the CBSE syllabus, students can refer to it without worrying about the difference in concepts.

1. Best source for preparation and revision:

Because of the simple language and clarity of concept, RD Sharma Class 12 Solutions Chapter 28 ex MCQ are an excellent choice for preparation. Both of these characteristics make it an excellent choice for studying and revising.

1. NCERT-based questions:

Teachers typically assign questions from NCERT, and our subject experts are well aware of this fact, as they work on these questions while keeping NCERT and CBSE guidelines in mind.

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## RD Sharma Chapter-wise Solutions

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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##### Visual Communication Designer

Individuals who want to opt for a career as a Visual Communication Designer will work in the graphic design and arts industry. Every sector in the modern age is using visuals to connect with people, clients, or customers. This career involves art and technology and candidates who want to pursue their career as visual communication designer has a great scope of career opportunity.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available

Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

2 Jobs Available
##### Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available
##### QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
##### Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available