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If $(\mathbf{a} \times \mathbf{b})^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=144$ and $|\mathbf{a}|=4$, then $|\mathbf{b}|$ is equal to
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Given, $(\mathbf{a} \times \mathbf{b})^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=144$
$\Rightarrow \quad\left(a^{2} b^{2} \cdot 1 \cdot \sin ^{2} \theta\right)+a^{2} b^{2} \cos ^{2} \theta=144$
$\Rightarrow \quad \mathrm{a}^{2} \mathrm{~b}^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=144$
$\Rightarrow \quad a^{2} b^{2}=144$
$\Rightarrow \quad 16 \mathrm{~b}^{2}=144 \quad(\because|\mathrm{a}|=4)$
$\Rightarrow \quad \mathrm{b}^{2}=9$
$\Rightarrow \quad \mathrm{b}=3$
or $|\mathbf{b}|=3$
$\Rightarrow \quad\left(a^{2} b^{2} \cdot 1 \cdot \sin ^{2} \theta\right)+a^{2} b^{2} \cos ^{2} \theta=144$
$\Rightarrow \quad \mathrm{a}^{2} \mathrm{~b}^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=144$
$\Rightarrow \quad a^{2} b^{2}=144$
$\Rightarrow \quad 16 \mathrm{~b}^{2}=144 \quad(\because|\mathrm{a}|=4)$
$\Rightarrow \quad \mathrm{b}^{2}=9$
$\Rightarrow \quad \mathrm{b}=3$
or $|\mathbf{b}|=3$
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