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  • Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 2 sub question 3 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 2 sub question 3 maths textbook solution

Answers (1)

Answer: A,B and C are collinear

Hint: Represent one vector as a scalar product of another.

\Rightarrow Given the points,

\begin{aligned} &A=\hat{i}+2 \hat{j}+7 \hat{k} \\ &B=2 \hat{i}+6 \hat{j}+3 \hat{k} \\ &C=3 \hat{i}+10 \hat{j}-\hat{k} \end{aligned}

\overrightarrow{AB} = Position of vector B – Position of vector A

\begin{aligned} &=(2 \hat{i}+6 \hat{j}+3 \hat{k})-(\hat{i}+2 \hat{j}+7 \hat{k}) \\ &=(2-1) \hat{i}+(6-2) \hat{j}+(3-7) \hat{k} \\ &=\hat{i}+4 \hat{j}+(-4) \hat{k} \\ &=\hat{i}+4 \hat{j}-4 \hat{k} \end{aligned}                                             ..........(1)

 Similarly, \overrightarrow{BC} = Position of vector C – Position of vector B

\begin{aligned} &=(3 \hat{i}+10 \hat{j}-\hat{k})-(2 \hat{i}+6 \hat{j}+3 \hat{k})\\ &=(3-2) \hat{i}+(10-6) \hat{j}+(-1-3) \hat{k}\\ &=\hat{i}+4 \hat{j}-4 \hat{k}\\ \end{aligned}                                      .........(2)

\text { From }(1) \&(2)\\

\overrightarrow{A B}=\overrightarrow{B C}

So, vectors are parallel. But B is a point common to them

Hence, the given points A,B and C are collinear

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