If $|\vec{a}|=4,|\vec{b}|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi / 6$ then $(\vec{a} imes \vec{b})^2=2$ is equal to

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If $|\vec{a}|=4,|\vec{b}|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi / 6$ then $(\vec{a} 	imes \vec{b})^2=2$ is equal to

Aicharya · Accepted Answer

We have, $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \frac{\pi}{6}=4 	imes 2 	imes \frac{\sqrt{3}}{2}=4 \sqrt{3}$ Now, $(\vec{a} 	imes \vec{b})^2+(\vec{a} \cdot \vec{b})^2=a^2 b^2 \Rightarrow(\vec{a} 	imes \vec{b})^2+48=16 	imes 4 \Rightarrow(\vec{a} 	imes \vec{b})^2=16$

If $|\vec{a}|=4,|\vec{b}|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi / 6$ then $(\vec{a}…

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