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If \( |\bar{a} \times \bar{b}|^{2}+|\bar{a} \cdot \bar{b}|^{2}=144 \) and \( |\bar{a}|=4 \), then the value of \( |\bar{b}| \) is
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The correct answer is:
\( 03 \)
Given that, $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$
We know,
$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2} \cdot|\vec{b}|^{2} \cos ^{2} \theta$
$=|\vec{a}|^{2} \cdot|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=|\vec{a}|^{2}|\vec{b}|^{2}$
$\Rightarrow|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}$
So, $(144)=16|\vec{b}|^{2}$
$\Rightarrow 9=|\vec{b}|^{2} \Rightarrow|\vec{b}|=3$
We know,
$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2} \cdot|\vec{b}|^{2} \cos ^{2} \theta$
$=|\vec{a}|^{2} \cdot|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=|\vec{a}|^{2}|\vec{b}|^{2}$
$\Rightarrow|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}$
So, $(144)=16|\vec{b}|^{2}$
$\Rightarrow 9=|\vec{b}|^{2} \Rightarrow|\vec{b}|=3$
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