If overrightarrow{A} = 3hat{i} + hat{j} + 2ha | vedclass

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(A) Given vectors are $\overrightarrow{A} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\overrightarrow{B} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.First,calculate t

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$\frac{2}{\sqrt{7}}$

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(A) Given vectors are $\overrightarrow{A} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\overrightarrow{B} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.First,calculate the dot product $\overrightarrow{A} \cdot \overrightarrow{B} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12$.Next,calculate the magnitudes $|\overrightarrow{A}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$ and $|\overrightarrow{B}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.Using the formula $\cos 	heta = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|}$,we get $\cos 	heta = \frac{12}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{6}{\sqrt{84}} = \frac{6}{2\sqrt{21}} = \frac{3}{\sqrt{21}} = \sqrt{\frac{9}{21}} = \sqrt{\frac{3}{7}}$.Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we have $\sin^2 	heta = 1 - \frac{3}{7} = \frac{4}{7}$.Therefore,$\sin 	heta = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$.

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

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