Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 14

Answers (1)

Answer: Therefore the equation of the plane is $9 x-8 y+7 z-21=0$ and distance is $\mathrm{D}=\frac{13}{\sqrt{194}}$

Hint: Use formula $a x+b y+c z+d=0$ and $D=\frac{a x_{1}+b y_{1}+c z_{1}+d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}}$

Given: Planes are $x-2 y+z=1 \& 2 x+y+z=8$ and point $\left ( 1,1,1 \right )$ ratios proportional is $1, 2, 1.$

Solution: We know that equation of plane passing through the intersection of planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by

$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$

So, equation of plane passing through the intersection of planes $x-2 y+z-1=0 \& 2 x+y+z-8=0$ is

$\begin{aligned} &\Rightarrow(x-2 y+z-1)+k(2 x+y+z-8)=0 \\ & \end{aligned}$ ………………. (1)

$\Rightarrow x(1+2 k)+y(-2+k)+z(1+k)+(-1-8 k)=0$

We know that line

$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$

Is parallel to plane $a_{2} x+b_{2} y+c_{2} z+d_{2}=0 \; \text { if }\; a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

Given the plane is parallel to line with direction ratios 1, 2, 1

$\begin{aligned} &1 \times(1+2 k)+2 \times(-2+k)+1 \times(1+k)=0 \\ & \end{aligned}$

$\Rightarrow 1+2 k-4+2 k+1+k=0 \\$

$\Rightarrow k=\frac{2}{5}$

Putting the value of k in equation (1) we get

$\begin{aligned} &\Rightarrow(x-2 y+z-1)+\frac{2}{5}(2 x+y+z-8)=0 \\ & \end{aligned}$

$\Rightarrow x+\frac{4}{5} x-2 y+\frac{2}{5} y+z+\frac{2}{5} z-1-\frac{16}{5}=0 \\$

$\Rightarrow \frac{9 x}{5}-\frac{8 y}{5}+\frac{7 z}{5}-\frac{21}{5}=0 \\$

$\Rightarrow 9 x-8 y+7 z-21=0$

We know that the distance (D) of point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ from plane $a x+b y+c x-d=0$ is given by

$D=\frac{a_{1} x+b_{1} y+c_{1} z+d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}}$

So, distance of Point $\left ( 1,1,1 \right )$ from plane is

$\begin{aligned} D &=\frac{9 \times 1+(-8) \times 1+7 \times 1-21}{\sqrt{9^{2}+(-8)^{2}+7^{2}}} \\ \end{aligned}$

$=\frac{9-8+7-21}{\sqrt{81+64+49}} \\$

$=\frac{-13}{\sqrt{194}}$

Taking the mod value we have $D=\frac{13}{\sqrt{194}} \text { unit }$

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon

anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers

anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question