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  • Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 2 sub question 1 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 2 sub question 1 maths textbook solution

Answers (1)

Answer: A,B and C are collinear

Hint: Use formula \overrightarrow{AB}=\lambda \overrightarrow{BC} i.e. representing one vector as a scalar product of another vector

\Rightarrow Given, the points,

\begin{aligned} &A=6 \hat{i}-7 \hat{j}-\hat{k} \\ &B=2 \hat{i}-3 \hat{j}+\hat{k} \\ &C=4 \hat{i}-5 \hat{j} \end{aligned}

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

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

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

\begin{aligned} &=(4 \hat{i}-5 \hat{j}+0 \hat{k})-(2 \hat{i}-3 \hat{j}+\hat{k}) \\ &=(4-2) \hat{i}+(-5+3) \hat{j}+(0-1) \hat{k} \\ &=2 \hat{i}-2 \hat{j}-\hat{k} \end{aligned}

From \left ( 1 \right )

\overrightarrow{AB}=-2\overrightarrow{BC}

 Thus, both vectors are parallel to each other and as B is the common point

A\left ( 6,-7,-1 \right ),B\left ( 2,-3,1 \right )&C\left ( 4,-5,0 \right ) are collinear

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