Q63 / 180 · COMEDK 2018 · Mathematics

Unit vector perpendicular to $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and lying in the plane containing $\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is

  • A
    $8 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$
  • B
    $8 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-11 \hat{\mathbf{k}}$
  • C
    $8 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-11 \hat{\mathbf{k}}$
  • Correct answer
    D
    $\frac{1}{\sqrt{234}}(8 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-11 \hat{\mathbf{k}})$

Solution

Let $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{c}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ Now, $\mathbf{b} \times \mathbf{c}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & -2 \\ -1 & 2 & 1\end{array}\right|=5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ Now, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & 2 \\ 5 & 1 & 3\end{array}\right|=-8 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$ Now, any vector perpendicular to a and lying in the plane containing $\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is $\pm(a \times(b \times c))$ $$ \begin{aligned} \therefore \text { Required unit vector } &=\pm \frac{(-8 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+11 \hat{\mathbf{k}})}{\sqrt{(-8)^{2}+(7)^{2}+(11)^{2}}} \\ &=\pm \frac{(-8 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+11 \hat{\mathbf{k}})}{\sqrt{234}} \end{aligned} $$

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

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