If theta is the angle between the vectors 4 h | vedclass

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(C) Let $\vec{a} = 4 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = \hat{i}+3 \hat{j}-2 \hat{k}$.The dot product is $\vec{a} \cdot \vec{b} = (4)(1) + (-1)(

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$-\frac{\sqrt{285}}{49}$

Vedclass · Answer

(C) Let $\vec{a} = 4 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = \hat{i}+3 \hat{j}-2 \hat{k}$.The dot product is $\vec{a} \cdot \vec{b} = (4)(1) + (-1)(3) + (2)(-2) = 4 - 3 - 4 = -3$.The magnitudes are $|\vec{a}| = \sqrt{4^2 + (-1)^2 + 2^2} = \sqrt{16+1+4} = \sqrt{21}$ and $|\vec{b}| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{1+9+4} = \sqrt{14}$.$\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-3}{\sqrt{21} \sqrt{14}} = \frac{-3}{\sqrt{294}} = \frac{-3}{7 \sqrt{6}}$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{9}{49 	imes 6} = 1 - \frac{9}{294} = \frac{285}{294}$,we have $\sin 	heta = \sqrt{\frac{285}{294}} = \frac{\sqrt{285}}{7 \sqrt{6}}$.$\sin 2 	heta = 2 \sin 	heta \cos 	heta = 2 \left( \frac{\sqrt{285}}{7 \sqrt{6}} ight) \left( \frac{-3}{7 \sqrt{6}} ight) = \frac{-6 \sqrt{285}}{49 	imes 6} = -\frac{\sqrt{285}}{49}$.

If $\theta$ is the angle between the vectors $4 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+3 \hat{j}-2 \hat{k}$,then $\sin 2 \theta=$

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