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$|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=144$ and $|\mathbf{a}|=4$, then $|\mathbf{b}|$ is equal to
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The correct answer is:
$3$
Given, $|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=144$ and $|\mathbf{a}|=4$
$\begin{aligned} & \Rightarrow|\mathbf{a}|^2|\mathbf{b}|^2 \sin ^2 \theta+|\mathbf{a} \| \mathbf{b}|^2 \cos ^2 \theta=144 \\ & \Rightarrow \quad|\mathbf{a}|^2|\mathbf{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=144\end{aligned}$
$\begin{array}{ll}\Rightarrow \quad & (4)^2 \times|b|^2=144 \\ \Rightarrow \quad & |b|^2=\frac{144}{16}=9 \\ & |b|=3\end{array}$
$\begin{aligned} & \Rightarrow|\mathbf{a}|^2|\mathbf{b}|^2 \sin ^2 \theta+|\mathbf{a} \| \mathbf{b}|^2 \cos ^2 \theta=144 \\ & \Rightarrow \quad|\mathbf{a}|^2|\mathbf{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=144\end{aligned}$
$\begin{array}{ll}\Rightarrow \quad & (4)^2 \times|b|^2=144 \\ \Rightarrow \quad & |b|^2=\frac{144}{16}=9 \\ & |b|=3\end{array}$
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