If theta is the angle between the vectors vec | vedclass

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(A) The formula for the cosine of the angle $	heta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by $\cos 	heta = \frac{\vec{a} \cdot \vec{b

Vedclass · Accepted Answer

$\cos 	heta = \frac{4}{21}$

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(A) The formula for the cosine of the angle $	heta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.First,calculate the dot product $\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4$.Next,calculate the magnitudes of the vectors:$|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.$|\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.Finally,substitute these values into the formula:$\cos 	heta = \frac{4}{3 	imes 7} = \frac{4}{21}$.

If $\theta$ is the angle between the vectors $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then:

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