RD Sharma Class 12 Exercise 28.15 The Plane Solutions Maths - Download PDF Free Online

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Also Read - RD Sharma Solutions For Class 9 to 12 Maths

This chapter, 'The Plane,' discusses 2D surfaces having infinite dimensions with no thickness such that when two lines join them, both of them lie on the surface entirely and find the shortest distance between them.

RD Sharma Class 12 Solutions Chapter28 The Plane - Other Exercise

• Chapter 28- The Plane - Ex-28.1

• Chapter 28- The Plane - Ex-28.2

• Chapter 28- The Plane - Ex-28.3

• Chapter 28- The Plane - Ex-28.4

• Chapter 28- The Plane - Ex-28.5

• Chapter 28- The Plane - Ex-28.6

• Chapter 28- The Plane - Ex-28.7

• Chapter 28- The Plane - Ex-28.8

• Chapter 28- The Plane - Ex-28.9

• Chapter 28- The Plane - Ex-28.10

• Chapter 28- The Plane - Ex-28.11

• Chapter 28- The Plane - Ex-28.12

• Chapter 28- The Plane - Ex-28.13

• Chapter 28- The Plane - Ex-28.14

• Chapter 28- The Plane - Ex-FBQ

• Chapter 28- The Plane - Ex-MCQ

• Chapter 28- The Plane - Ex-VSA

The Plane Excercise: 28.15

The Plane exercise 28.15 question 1

Answer: $(-\frac{6}{61},\frac{-8}{61},\frac{12}{61})$
Hint:
Suppose point $P(0,0,0)$, Let$PM\perp \text{ Plane }$. Direction of $PM÷3,4,-6$
Equation of line$=\frac{x-x_1}{\alpha }=\frac{y-y_1}{\beta }=\frac{z-z_1}{\gamma }$
$P(0,0,0)$


Given:

Find the image of the point $(0,0,b)$ in the plane $3x+4y-6z+1=0$
Solution:
$\begin{array}{rl}& 3x+4y-6z+1=0\\ \\ & \text{ Let }{x}_1=0=y_1=31\end{array}$
Equation of line $=\frac{x-0}{3}=\frac{y-0}{4}=\frac{z-0}{-6}$
$\begin{array}{rl}& \text{ Say }\lambda ,\\ & x=3\lambda \Rightarrow x=3\times \left(\frac{-1}{61}\right)=\frac{-3}{61}\\ & y=4\lambda \Rightarrow y=4\times \left(\frac{-1}{61}\right)=\frac{-4}{61}\\ & z=-6\lambda \Rightarrow y=-6\times \left(\frac{-1}{61}\right)=\frac{6}{61}\end{array}$
$\begin{array}{rl}& \therefore x=\frac{-3}{61},y=\frac{-4}{61},z=\frac{6}{61}\\ \\ & \frac{A}{z}=\frac{-3}{61},\frac{B}{z}=\frac{-4}{61},\frac{C}{z}=\frac{6}{61}\end{array}$

The Plane exercise 28.15 question 2

Answer: $(\frac{73}{25},-\frac{6}{5},\frac{39}{25})$
Hint:
Suppose point $P(1,2,-1)$, Let $PM\perp \text{ Plane. }$
Direction $=(3,-5,4)$
$P(1,2,-1)$


Given:

Find the reflection of the point $(1,2,-1)$ in the plane $3x-5y+4z=5$

Solution:
Direction of line $=(3,-5,4)$$=(3,-5,4)$$=(3,-5,4)$$=(3,-5,4)$
Equation of line $\frac{x-1}{3}=\frac{y-2}{5}=\frac{3+1}{4}=x$

$x=3\lambda +1,y=-5x+2,z=4\lambda -1$

Put in $(3x-5y+4z=5)$

$3(3\lambda +1)-5(-5\lambda +2)+4(4\lambda -1)=5$
$\begin{array}{rl}& 9\lambda +3+25\lambda -10+16\lambda -4=5\\ \\ & 50\lambda =16\\ \\ & \lambda =\frac{16}{50}=\frac{8}{25}\end{array}$

So that
$\begin{array}{rl}& x=3\left(\frac{8}{25}\right)+1=\frac{73}{25}\\ \\ & y=-5\left(\frac{8}{25}\right)+2=\frac{6}{5}\\ \\ & z=4\left(\frac{8}{25}\right)-1=\frac{39}{25}\end{array}$
coordinate $(\frac{73}{25},\frac{-6}{5},\frac{39}{25})$ Ans.

The Plane exercise 28.15 question 3

Answer: $(1,6,0),2\sqrt{6}$
Hint:
Suppose $\frac{x+1}{2}=\frac{y-3}{3}=\frac{3-1}{-1}=\lambda$
$PQ=\sqrt{{(x_2-x_1)}^2{(y_2-y_1)}^2{(z_2-z_1)}^2}$
Given:
Find the coordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ to the line
$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$ hence or otherwise deduce the length of the perpendicular.
$\frac{x+1}{2}=\frac{y-3}{3}=\frac{3-1}{-1}$
$P(5,4,2)$

Solution:
$\begin{array}{rl}& x=2\lambda +1\\ \\ & y=3\lambda +3\\ \\ & z=-\lambda +1\end{array}$
Direction ratio of $P$ as $=(2\lambda -1-5,3\lambda +z-4,-\lambda +1-2)$
$\begin{array}{rl}& =(2\lambda -6,3\lambda -1,-\lambda -1)\\ & a_1{b}_1{c}_1\\ \\ & a_1{a}_2+b_1{b}_2+c_1{c}_2=0\end{array}$
$\begin{array}{rl}& =2(2\lambda -6)+3(3\lambda -1)-1(-\lambda -1)=0\\ \\ & =4\lambda -12+9\lambda -3+\lambda +1=0\\ \\ & =14\lambda -14=0\\ \\ & \lambda =1\end{array}$
Putting the value of x in eq
$\begin{array}{rl}& x=2\lambda +1,\\ \\ & y=3\lambda +3,\\ \\ & z=-\lambda +1\end{array}$
$\begin{array}{rl}& x=3\\ & y=6\\ & z=0\end{array}$
co-ordinates $(1,6,0)$
$\begin{array}{rl}& PQ=\sqrt{(5-1{)}^2+(4-8{)}^2+(2-0{)}^2}\\ \\ & =\sqrt{16+14+4}\\ \\ & =\sqrt{24}\\ \\ & =2\sqrt{6}\mathrm{Ans}\end{array}$

The Plane exercise 28.15 question 4

Answer: $(1,2,1),2i+\frac{3}{2}\hat{ȷ}+\frac{3}{2}\hat{k}$
$\overrightarrow{r}=3i+\hat{ȷ}+2\hat{k}+\lambda (2\hat{ı}-\hat{ȷ}+\hat{k})$
Hint:
Point $(3,1,2)$
Suppose $\overrightarrow{r}=x\hat{ı}+y\hat{ȷ}+z\hat{k}$
Given:
Find the image of the point with position vector $3i+2\hat{ȷ}+\hat{k}$ in the plane $\overrightarrow{r}\cdot (2\hat{ı}+\hat{j}+\hat{k})=\mathrm{Y}.$. Also find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through $3i+2\hat{ȷ}+\hat{k}$ $P(3,1,2)$

$Q(1,2,1)$
Solution:
$\begin{array}{rl}& \overrightarrow{r}=x\hat{ı}+y\hat{ȷ}+z\hat{k}\\ \\ & 2x-y+z=4\\ \\ & \frac{x-3}{2}=\frac{y-1}{-1}=\frac{z-2}{1}\end{array}$
$\begin{array}{rl}& =\frac{-2(6-1+2-4)}{(4+1+1)}\\ \\ & =\frac{-2(3)}{6}\\ \\ & =-1\end{array}$
$\begin{array}{rl}& \frac{x-3}{2}=-1\\ \\ & \frac{y-1}{-1}=-1\\ \\ & \frac{z-2}{1}=1\end{array}$
$\begin{array}{rl}& x=1\\ & y=2\\ & z=1\end{array}$
Point$(1,2,1)$
Passing $(3,1,2),$ the point.

direction $\to 2,-1,1$
$\begin{array}{rl}& \frac{x-3}{2}=\frac{y-1}{-1}=\frac{z-2}{1}\\ \\ & \Rightarrow \overrightarrow{r}=3\hat{ı}+\hat{ȷ}+2\hat{k}\\ \\ & (2\hat{ı}+-\hat{ȷ}+\hat{k})\end{array}$
point $(3,1,2)$
point $q=(1,2,1)$
So the middle point $=2+\frac{3}{2}+\frac{3}{2}$

i.e, $2i+\frac{3}{2}\hat{ȷ}+\frac{3}{2}k$

The Plane exercise 28.15 question 5

Answer: $(-\frac{1}{12},\frac{25}{12},-\frac{1}{6}),\frac{13}{\sqrt{24}}$
Hint:
Let point

directional ratio are perpendicular to the line
Given:
Find the coordinate of the foot of the perpendicular from the point $(1,1,2)$ to the plane $2x-2y+4z+5=0$ . Also find the length of the perpendicular
Solution:
Direction ratio of plane $\overrightarrow{n}=2,-2,4$

equation of line $\frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{4}=\lambda (\text{ say })$
$\begin{array}{rl}& x=2\lambda +1\\ \\ & y=-2\lambda +1,\\ \\ & z=4\lambda +2\end{array}$
Put in $2x-2y+4z+5=0$
$2(2\lambda +1)-2(-2\lambda +1)+4(4\lambda +2)+5=0$
$\begin{array}{rl}& 4\lambda +2+4\lambda -2+16\lambda +8+5=0\\ \\ & 24\lambda +13=0\\ \\ & \lambda =\frac{-13}{24}\end{array}$
Putting the value of λ iner
$\begin{array}{rl}& x=2\lambda +1\\ \\ & y=-2\lambda +1\\ \\ & z=4\lambda +2\end{array}$
$\begin{array}{rl}& x=2(-\frac{13}{24})+1\\ \\ & y=-2\left(\frac{-13}{24}\right)+1\\ \\ & z=4\left(\frac{-13}{24}\right)+2\end{array}$
$\begin{array}{rl}& x=\frac{-13+12}{12}\\ \\ & y=\frac{13+12}{12}\\ \\ & z=\frac{-13+12}{6}\end{array}$
Distance $=(-\frac{1}{12},\frac{25}{12},-\frac{1}{6})$
Given point $(1,1,2)$
Distance $=\sqrt{{(1+\frac{1}{12})}^2+{(1-\frac{25}{12})}^2+{(2+\frac{1}{6})}^2}$
$\begin{array}{rl}& =\sqrt{\frac{169}{144}+\frac{169}{144}+\frac{169}{36}}\\ \\ & =13\sqrt{\frac{1}{144}+\frac{1}{144}+\frac{1}{36}}\end{array}$
$\begin{array}{rl}& =\frac{13}{12}\sqrt{1+1+4}\\ \\ & =\frac{13\sqrt{6}}{12}\end{array}$

The Plane exercise 28.15 question 8

Answer: $(-3,5,2)$
Hint:
Formula ,
$\begin{array}{rl}& \frac{x-x_1}{a}=\frac{y+y_1}{b}=\frac{z-z_1}{c}\\ \\ & =\frac{-2(a{x}_1+b{y}_1+c{z}_1+d)}{(a^2+b^2+c^2)}\\ \\ & \hat{k}\mathrm{P}(1,3,4)\end{array}$


$2x-y+z+3=0$
Given:
Find the image of the point with $(1,3,4)$ in the plane $2x-y+z+3=0$
Solution:
$\begin{array}{rl}& (x_1,y_1,z_1)\\ \\ & ax+by+z+d=0\\ \\ & \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{3-31}{c}=\frac{-2(a{x}_1+b{y}_1+c_1+d)}{(a^2+b^2+c^2)}\end{array}$
$\begin{array}{rl}& \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{-1}=\frac{-2(2-3+4+3)}{4+1+1}=-2\\ \\ & x=-4+1\\ \\ & y=2+3\\ \\ & z=-2+4\end{array}$
point $=(1,3,4)$
$\begin{array}{rl}& x=-3\\ & y=5,\\ & z=2\end{array}$
Mirror image $(-3,5,2)$

The Plane exercise 28.15 question 10

Answer: $\frac{13}{12}\sqrt{6},(-\frac{1}{12},\frac{25}{12},-\frac{2}{12})$
Hint:
For P as distance put
$\begin{array}{rl}& \overrightarrow{r}=i+\hat{ȷ}+2\hat{k}\text{ in }\\ \\ & \overrightarrow{r}\cdot (\hat{ı}-2\hat{ȷ}+4\hat{k})+5=0P(1,1,2)\end{array}$

Given:
Find the length and foot of the perpendicular from the point $(1,1,2)$ to the plane $\overrightarrow{r}\cdot (\hat{ı}-2\hat{ȷ}+4\hat{k})+5=0$
Solution:
Distance $=\frac{(\hat{ı}+\hat{ȷ}+2\hat{k})\cdot (\hat{ı}-2\hat{ȷ}+4\hat{i})+5}{\sqrt{1^2+(-2{)}^2+4^2}}$
$\begin{array}{rl}& =\frac{1-2+8+5}{\sqrt{1+4+16}}\\ \\ & =\frac{12}{\sqrt{21}}\end{array}$
Direction of $\overrightarrow{n}=(1,-2,4)$

equation of line $\frac{x-1}{1}=\frac{y-1}{-2}=\frac{3-2}{4}=\lambda (\text{ say })$
$\begin{array}{rl}& x=\lambda +1\\ \\ & y=2\lambda +1,\\ \\ & z=4\lambda +2\end{array}$
$Put\overrightarrow{r}=(\lambda +1)i+(-2\lambda +1)\hat{ȷ}+(4\lambda +2)\hat{k}\text{ in }\overrightarrow{r}\cdot (\hat{ı}-\hat{y}+4\hat{k})+5=0$
$\begin{array}{rl}& [(\lambda +1)i+(-2\lambda +1)\hat{ȷ}+(4\lambda +2)\hat{k}]\cdot [\hat{l}-\hat{y}+4\hat{k}]+5=0\\ \\ & \lambda +1+4\lambda -2+16\lambda +8+5=0\\ \\ & 21\lambda +12=0\end{array}$
$\begin{array}{rl}& \lambda =-\frac{12}{21}\\ \\ & x=\frac{-4}{7}+1\\ \\ & =\frac{3}{7}\end{array}$
$\begin{array}{rl}& y=\frac{8}{7}+1\\ \\ & =\frac{15}{7}\\ \\ & z=\frac{-16}{7}+2\end{array}$
$\begin{array}{rl}& =-\frac{2}{7}\\ \\ & (\frac{3}{7},\frac{15}{7},\frac{-2}{7})\text{ Ans. }\end{array}$

The Plane exercise 28.15 question 11

Answer:$(7,0,3)(-1,4,-1),\sqrt{6}$
Hint:
$\hat{N}(3,2,1)$

If $M$ is lying on the plane and the normal vector as well $\hat{N}$
$(ax+by+cz+d),\hat{N}=\frac{(a)}{\sqrt{a^2+b^2+i^2}}\hat{ı}+\frac{b}{\sqrt{a^2+b^2+c^2}}\hat{ȷ}+\frac{c}{\sqrt{a^2+b^2}{t}^2}\hat{k}$
Given:
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point $P(3,2,1)$ from the plane $2x-y+z+1=0$. Find also the image of the point in the plane
Solution:
$\begin{array}{rl}& \hat{N}=\frac{2}{\sqrt{6}}\hat{ı}-\frac{\hat{ȷ}}{\sqrt{6}}+\frac{\hat{k}}{\sqrt{6}}\\ & \hat{N}\end{array}$

$\begin{array}{rl}& \therefore \frac{x-3}{\frac{2}{\sqrt{6}}}=\frac{y-2}{(-\frac{1}{\sqrt{6}})}=\frac{3-1}{\left(\frac{1}{\sqrt{6}}\right)}=\lambda \\ \\ & x=\frac{3+2\lambda }{\sqrt{6}},y=\frac{2-\lambda }{\sqrt{6}},z=\frac{1+\lambda }{\sqrt{6}}\end{array}$
$\begin{array}{rl}& =2(3+\frac{2\lambda }{\sqrt{6}})-(2-\frac{\lambda }{\sqrt{6}})+(1+\frac{\lambda }{\sqrt{6}})+1=0\\ \\ & =6+\frac{4\lambda }{\sqrt{6}}-2+\frac{\lambda }{\sqrt{6}}+2+\frac{\lambda }{\sqrt{6}}=0\end{array}$
$\begin{array}{rl}& 6+\frac{6}{\sqrt{6}}=0\\ \\ & \Rightarrow \lambda =-\frac{6\sqrt{6}}{6}=-\sqrt{6}\\ \\ & M[3-2,2+1,1-1]\end{array}$
$\begin{array}{rl}& \therefore M(1,3,0)\\ & \to Q=\sqrt{2^2+1+1}=\sqrt{6}\\ & P(3,2,1)\end{array}$

$\begin{array}{rl}& \Rightarrow \frac{(3-a)}{(2/\sqrt{6})}=\frac{(2-b)}{(-1/\sqrt{6})}=\frac{(1-c)}{1/\sqrt{6}}=2\sqrt{6}\\ \\ & \Rightarrow (3-a)=4\text{ or }-4\\ \\ & \Rightarrow -10r+7\end{array}$
$\begin{array}{rl}& \Rightarrow (2-b)=-2\text{ or }2\\ \\ & \Rightarrow b=-4\text{ or }0\\ \\ & \Rightarrow (1-c)=2\text{ or }-2\end{array}$
$\begin{array}{rl}& \Rightarrow c=-1\\ \\ & p!(-1,4,-1)\text{ or }\\ \\ & p=7,0,3\end{array}$

The Plane exercise 28.15 question 14

Answer:$(0,\frac{5}{2},0);\sqrt{6}$
Hint:
Direction ratio on $(2,-2,4)$
Equation of line $=\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

$2x-2y+4z-6=0$
Given:
Find the length and foot of the perpendicular from the point $(1,3/2,2)$ to the plane
$2x-2y+4z-6=0$
Solution:
$2x-2y+4z+5=0$
$\begin{array}{rl}& D{R}^{\prime }\mathrm{\Delta }=(2,-2,4)\\ \\ & ABD{R}^{\prime }S=(2,-2,4)\end{array}$
$\begin{array}{rl}& \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{3-3}{c}\\ \\ & \frac{x-1}{2}=\frac{y-3/2}{-2}=\frac{2-2}{4}=1\end{array}$
$\begin{array}{rl}& x=2\lambda +1\\ \\ & y=-2\lambda +\frac{3}{2}\\ \\ & z=4\lambda +2\end{array}$
$\begin{array}{rl}& \Rightarrow 2(2\lambda +1)-2(-2\lambda +\frac{3}{2})+4(4\lambda +2)+5=0\\ \\ & \Rightarrow 4\lambda +2+4\lambda -3+16\lambda +8+5=0\end{array}$
$\begin{array}{rl}& 24\lambda =-12\\ \\ & \lambda =-\frac{12}{24}\\ \\ & =-\frac{1}{2}\end{array}$
Foot of the perpendicular,
$\begin{array}{rl}& x=2\lambda +1\\ \\ & =2\left(\frac{-1}{2}\right)+1=0\\ \\ & y=-2\left(\frac{-1}{2}\right)+\frac{3}{2}=\frac{5}{2}\\ \\ & z=4\left(\frac{-1}{2}\right)+2=0\end{array}$
Foot of the perpendicular $(0,\frac{5}{2},0)$

$AB=\sqrt{1+1+4}=\sqrt{6}$

The Plane exercise 28.15 question 15

Answer:$(3,\frac{7}{2},\frac{11}{2}),(4,4,7)$
Hint:
Distance $\to \sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2+{(z_1-z_2)}^2}$
$\overrightarrow{d}=\overrightarrow{r}\cdot \overrightarrow{n}$

Given:
Find the position vector of the foot of the perpendicular and the perpendicular distance from the point P with position vector $2\tilde{i}+3\hat{ȷ}+4\hat{k}$ to the plane $\overrightarrow{r}\cdot (\hat{ı}+\hat{ȷ}+3\hat{k})-26=0$ . Also find the image of $P$ in the plane
Solution:
$\begin{array}{rl}& \frac{x-2}{2}=\frac{y-3}{1}=\frac{z-4}{3}=t\\ \\ & x=2+2t\\ \\ & y=3+t\\ \\ & z=4+3t\end{array}$
$\begin{array}{rl}& \Rightarrow [(2+2t)\hat{i}+(3+t)\hat{j}+(4+3t)\hat{k}][2\hat{i}+\hat{j}+3\hat{k}]=26\\ \\ & \Rightarrow 19+14t=26\\ \\ & t=\frac{1}{2}\end{array}$
Putting the value of t in equation 1
i.e,
$\begin{array}{rl}& x=3,\\ \\ & y=\frac{7}{2},\\ \\ & z=\frac{11}{2}\end{array}$
$\mathrm{PQ}\Rightarrow (3-2{)}^2+{(\frac{7}{2}-3)}^2+{(\frac{11}{2}-2)}^2$
$\begin{array}{rl}& =\sqrt{1+{\left(\frac{1}{2}\right)}^2+{\left(\frac{11}{2}\right)}^2},\\ \\ & =\sqrt{1+\frac{1}{4}+\frac{9}{4}}\end{array}$
$\begin{array}{rl}& =\sqrt{\frac{14}{4}}\\ \\ & =\frac{\sqrt{14}}{2}\end{array}$
Image ,
$x=2+2(1)=4$
$\begin{array}{rl}& y=3+1=4\\ \\ & z=4+3(1)=7\end{array}$
Image $(4,4,7)$ Ans
Foot $(3,\frac{7}{2},\frac{11}{2})$ Ans Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Chapter-wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable