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Explain Solution R.D.Sharma Class 12 Chapter 22 Algebra of Vectors  Exercise Very Short Answer Type Question 40 Maths Textbook Solution.

Answers (1)

Answer: -4

Hint: You must know the rules of vector functions

Given: For what value of ‘a’ the vectors 2\hat{i}-3\hat{j}+4\hat{k}&a\hat{i}+6\hat{j}-8\hat{k}are collinear

Solution: Two vectors are

\begin{aligned} &\vec{p}=2 \hat{i}-3 \hat{j}+4 \hat{k}, q=a \hat{i}+6 \hat{j}-8 \hat{k} \\ \end{aligned}

Vectors are collinear,

\begin{aligned} &\vec{p}=\lambda \vec{q} \\ \end{aligned}

\begin{aligned} &2 \hat{i}-3 \hat{j}+4 \hat{k}=\lambda(\hat{a i}+6 \hat{j}-8 \hat{k}) \\ \end{aligned}

\begin{aligned} &2 \hat{i}-3 \hat{j}+4 \hat{k}=\lambda a \hat{i}+6 \lambda \hat{j}-8 \lambda \hat{k} \\ \end{aligned}

By comparing

\begin{aligned} &-8 \lambda=-3 \Rightarrow \lambda=\frac{-1}{2} \\ \end{aligned}

\begin{aligned} &-6 \lambda=-3 \Rightarrow \lambda=\frac{-1}{2} \\ \end{aligned}

\begin{aligned} &\lambda a=2 \Rightarrow \frac{-1}{2} \times a=2 \Rightarrow a=-4 \end{aligned}

Hence a=-4

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