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  • Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 2 sub question 4 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 22 Algebra of vectors exercise 22.8 question 2 sub question 4 maths textbook solution

Answers (1)

Answer: Vectors 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=-3 \hat{i}-2 \hat{j}-5 \hat{k} \\ &B=\hat{i}+2 \hat{j}+3 \hat{k} \\ &C=3 \hat{i}+4 \hat{j}+7 \hat{k} \end{aligned}

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

\begin{aligned} &=(\hat{i}+2 \hat{j}+3 \hat{k})-(-3 \hat{i}-2 \hat{j}-5 \hat{k})\\ &=(1+3) \hat{i}+(2+2) \hat{j}+(3+5) \hat{k}\\ &=4 \hat{i}+4 \hat{j}+8 \hat{k}\\ &=4(\hat{i}+\hat{j}+2 \hat{k}) \end{aligned}                                       .....(1)

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

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

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

\overrightarrow{A B}=2 \overrightarrow{B C}

 Thus both vectors are parallel to each other and B being the common point

Hence, vectors are collinear

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