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Please solve RD Sharma class 12 chapter 22 Algebra of Vectors exercise Fill in the blanks question 5 maths textbook solution

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Answer:- 9

Hint:- To solve this question we use  \vec{b}=\lambda \vec{a}   where \lambda  is constant.

Given:- If the vector \vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \text { and } \vec{b}=3 \hat{i}-6 \hat{j}+m \hat{k} are collinear then m= ----

 \begin{aligned} &\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \\\\ &\vec{b}=3 \hat{i}-6 \hat{j}+m \hat{k} \end{aligned}

Using \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

 

Then, \frac{1}{3}=\frac{-2}{-6}=\frac{3}{m}

 

\frac{1}{3}=\frac{3}{m}

Therefore, m = 9.

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