Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter The Plane exercise 28.3 question 9 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.3  question 9 maths textbook solution

Answers (1)

Answer:

The answer of given question is 2x+3y-z=14

Hint:

By using formula \vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}

Given:

P is the point (2,3,-1) and the require plane is passing through P at right angle to OP.

Solution:

As per the given criteria, it means that the plane is passing through P and OP is the vector normal to the plane.

Let the position vector of the point P be.

\vec{n}=2 \hat{\imath}+3 \hat{\jmath}-\hat{k} \ldots \ldots(i)

And it is also given the planes is normal to the line joining the points O(0,0,0) and P(2,-3,1).

Then \vec{n}=\overrightarrow{O P}

? \vec{n} =position vector of \vec{P}-  Position vector of \vec{n}
\begin{aligned} &\vec{n}=(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-(0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k}) \\ &\vec{n}=2 \hat{\imath}+3 \hat{\jmath}-\hat{k} \ldots \ldots .(i i) \end{aligned}

We know that

(\vec{r}-\vec{a}) \cdot \vec{n}=0

Substituting the values from equ (i) and equ (ii) in the above equ we get

\begin{aligned} &{[\vec{r}-(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})] \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-(2 \hat{\imath}+3 \hat{\jmath}-\hat{k}) \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=0 \\ &\Rightarrow \vec{r}-(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-[(2)(2)+(3)(3)+(-1)(-1)]=0 \end{aligned}

By multiplying the two vectors using the formula

\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-[4+9+1]=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-14=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=14 \end{aligned}

Then, the above vector equation of the plane becomes

(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(2 \hat{\imath}+3 \hat{\jmath}-3 \hat{k})=14

By multiplying the two vectors using the formula

\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(2)+(y)(3)+(z)(-1)=14 \\ &\Rightarrow 2 x+3 y-z=14 \end{aligned}

This is the cartesian form of equation of the required plane.

Posted by

infoexpert26

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