Q242 / 270 · NDA 2023 (Phase 2) · Mathematics

If $(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=144$ and $|\vec{b}|=4$, then what is the value of $|\vec{a}|$ ?

  • Correct answer
    A
    3
  • B
    4
  • C
    5
  • D
    6

Solution

$\begin{aligned} & \text { Given, }(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})^2+(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})^2=144 \\ & \Rightarrow|\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{~b}}|^2 \rightarrow \sin ^2 \theta+|\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{~b}}|^2 \cos ^2 \theta=144 \\ & \Rightarrow|\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{~b}}|^2=144 \\ & |\overrightarrow{\mathrm{a}}|^2 \times 16=144 \Rightarrow|\overrightarrow{\mathrm{a}}|=3\end{aligned}$

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

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