The vectors $\mathbf{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$ and $\mathbf{b}$ are collinear and $|\mathbf{b}|=2…

The vectors $\mathbf{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$ and $\mathbf{b}$ are collinear and $|\mathbf{b}|=2 \mathrm{l}$, then $\mathbf{b}=$
  1. $\pm(2 \hat{i}+3 \hat{j}+6 \hat{k})$
  2. $\pm(6 \hat{i}+9 \hat{j}+18 \hat{k})$
  3. $\frac{21}{\sqrt{3}}(\hat{i}+. \hat{j}+\hat{k})$
  4. $\pm 21(2 \hat{i}+3 \hat{j}+6 \hat{k})$

Solution

If $\mathbf{a}$ and $\mathbf{b}$ are collinear, then $ \begin{aligned} & \mathbf{b}=\lambda \mathbf{a} \\ & |\mathbf{b}|=|\lambda||\mathbf{a}| \\ & \Rightarrow \quad 21=|\lambda| \times \sqrt{4+9+36} \\ & \Rightarrow|\lambda|=\frac{21}{7} \Rightarrow|\lambda|=3 \\ & \lambda= \pm 3 \\ & \therefore \quad \mathbf{b}= \pm 3(2 \hat{i}+3 \hat{j}+6 \hat{k})= \pm(6 \hat{i}+9 \hat{j}+18 \hat{k}) \\ & \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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