Q75 / 160 · AP EAMCET 2005 · Mathematics

If the vector $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $\mathbf{b}$ are collinear and $|\mathbf{b}|=21$, then $\mathbf{b}$ equal to:

  • A
    $\pm(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})$
  • Correct answer
    B
    $\pm 3(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})$
  • C
    $(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$
  • D
    $\pm 21(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})$

Solution

Given that $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $|\mathbf{b}|=21$ Now, taking option (b) Let $\begin{aligned} \mathbf{b} & = \pm 3(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\ |\mathbf{b}| & =3 \sqrt{4+9+36}=21 \\ \mathbf{b} & = \pm 3 \mathbf{a} \end{aligned}$ and $\therefore \mathbf{a}$ and $\mathbf{b}$ are collinear and magnitude of $\mathbf{b}$ is 21 .

Hence, the correct answer is option: 'B'.

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