If $|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=144$ and $|\bar{a}|=4$, then $|\bar{b}|=$

If $|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=144$ and $|\bar{a}|=4$, then $|\bar{b}|=$
  1. 8
  2. 12
  3. 3
  4. 16

Solution

$\begin{aligned} & |\overline{\mathrm{a}} \times \overline{\mathrm{b}}|^2+(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2=144 \text { and }|\overline{\mathrm{a}}|=4 \\ & \therefore\left(|\overline{\mathrm{a}}|^2 \cdot|\overline{\mathrm{b}}|^2 \cdot \sin ^2 \theta\right)+\left(|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2 \cos ^2 \theta\right)=144 \\ & \therefore|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2\left(\cos ^2 \theta+\sin ^2 \theta\right)=144 \\ & \therefore(4)^2|\overline{\mathrm{b}}|^2=144 \Rightarrow|\overline{\mathrm{b}}|^2=9 \Rightarrow|\overline{\mathrm{b}}|=3\end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 2)

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