Q198 / 270 · NDA 2017 (Phase 2) · Mathematics

For any vector $\vec{a}|\vec{a} \times \hat{i}|^2+|\vec{a} \times \hat{j}|^2+|\vec{a} \times \hat{k}|^2$ is equal to

  • A
    $|\vec{a}|^2$
  • Correct answer
    B
    $2|\vec{a}|^2$
  • C
    $3|\overrightarrow{\mathrm{a}}|^2$
  • D
    $4|\vec{a}|^2$

Solution

$\begin{aligned} & \text { Let } \vec{a}=x \hat{i}+y \hat{j}+z \hat{k} \\ & |\vec{a} \times \hat{i}|^2=y^2+z^2\end{aligned}$ $\begin{aligned} & |\overrightarrow{\mathrm{a}} \times \hat{\mathrm{j}}|^2=\mathrm{x}^2+\mathrm{z}^2 \\ & |\overrightarrow{\mathrm{a}} \times \hat{\mathrm{k}}|^2=\mathrm{x}^2+\mathrm{y}^2\end{aligned}$ $\begin{aligned} & \therefore|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{i}}|^2+|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{j}}|^2+|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{k}}|^2 \\ & =y^2+z^2+x^2+z^2+x^2+y^2\end{aligned}$ $\begin{aligned} & =2 x^2+2 y^2+2 z^2 \\ & =2\left(x^2+y^2+z^2\right)=2|\vec{a}|^2\end{aligned}$

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

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