Q215 / 270 · NDA 2015 (Phase 2) · Mathematics

If $|\vec{a}|=2$ and $|\vec{b}|=3$, then $|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2$ is equal to

  • A
    72
  • B
    64
  • C
    48
  • Correct answer
    D
    36

Solution

$\begin{aligned} & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta \hat{\mathrm{n}} \\ & \therefore|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta=6 \sin \theta \\ & \&|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=6 \cos \theta \\ & |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|^2=(6 \sin \theta)^2+(6 \cos \theta)^2 \\ & =36\left(\sin ^2 \theta+\cos ^2 \theta\right) \\ & =36\end{aligned}$

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

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