Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 4 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 4 maths textbook solution

Answers (1)

Answer: Vectors are co-planar

Hint: Express four points as the linear combination of one vector into two vectors

Given: Let,

\begin{aligned} &P=6 \hat{i}-7 \hat{j} \\ &Q=16 \hat{i}-19 \hat{j}-4 \hat{k} \\ &R=3 \hat{j}-6 \hat{k} \\ &S=2 \hat{i}-5 \hat{j}+10 \hat{k} \end{aligned}

\RightarrowNow, Expressing one vector as a linear combination of other two.

\begin{aligned} &\overrightarrow{P Q}=(16 \hat{i}-19 \hat{j}-4 \hat{k})-(6 \hat{i}-7 \hat{j}+0 \hat{k})\\ &=(16-6) \hat{i}+(-19+7) \hat{j}+(-4-0) \hat{k}\\ &=10 \hat{i}-12 \hat{j}-4 \hat{k}\\ &\overrightarrow{P R}=(0 \hat{i}+3 \hat{j}-6 \hat{k})-(6 \hat{i}-7 \hat{j}+0 \hat{k})\\ &=(0-6) \hat{i}+(3+7) \hat{j}+(-6-0) \hat{k}\\ &=-6 \hat{i}+10 \hat{j}-6 \hat{k}\\ &\overrightarrow{P S}=(2 \hat{i}-5 \hat{j}+10 \hat{k})-(6 \hat{i}-7 \hat{j}+0 \hat{k})\\ &=(2-6) \hat{i}+(-5+7) \hat{j}+(0-0) \hat{k} \end{aligned}

 \begin{aligned} &=-4 \hat{i}+2 \hat{j}+10 \hat{k} \\ &\overrightarrow{P Q}=x \overrightarrow{P R}+y \overrightarrow{P S} \\ &10 \hat{i}-12 \hat{j}-4 \hat{k}=x(-6 \hat{i}+10 \hat{j}-6 \hat{k})+y(-4 \hat{i}+2 \hat{j}+10 \hat{k}) \\ &10 \hat{i}-12 \hat{j}-4 \hat{k}=\hat{i}(-6 x-4 y)+\hat{j}(10 x+2 y)+\hat{k}(-6 x+10) \end{aligned}

Comparing coefficients of i, j and k on both sides

\begin{aligned} &-6 x-4 y=10 \Rightarrow-2(3 x+2 y)=10 \Rightarrow 3 x+2 y=-5 \ldots \ldots .(1)\\ &10 x+2 y=-12 \Rightarrow 2(5 x+y)=-12 \Rightarrow 5 x+y=-6 \ldots \ldots .\left ( 2 \right )\\ &-6 x+10 y=-4 \Rightarrow-2(3 x-5 y)=-4 \Rightarrow 3 x-5 y=2 \ldots \ldots .(3) \end{aligned}

Solving (1) and (3) by elimination

\begin{aligned} &\begin{array}{c} 3 x+2 y=-5 \\ 3 x-5 y=2 \\ \frac{7 y}{y}=-7 \\ y=\frac{-7}{7} \Rightarrow y=-1 \\ 3 x-5(-1)=2 \\ 3 x+5=2 \\ 3 x=2-5 \\ x=\frac{-3}{3}=-1 \end{array} \end{aligned}

Now, Put x=-1 & y=-1  in equation (2)

\begin{aligned} &5 x+y=-6 \\ &\text { L.H } S=5(-1)+(-1) \\ &=-5-1 \\ &=-6 \\ &=R \cdot H . S \end{aligned}

As x=-1 & y=-1  satisfies all the equation we can say that

All four points are coplanar.

Posted by

Infoexpert

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