If vec{a} = hat{i} + 2hat{j} + 2hat{k},|vec{b | vedclass

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(C) Given $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$.First,calculate the magnitude of $\vec{a}$:$|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4}

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$\frac{15}{4}$

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(C) Given $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$.First,calculate the magnitude of $\vec{a}$:$|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.We are given $|\vec{b}| = 5$ and the angle $	heta = \frac{\pi}{6}$.The area of a triangle formed by two vectors $\vec{a}$ and $\vec{b}$ as adjacent sides is given by the formula:$	ext{Area} = \frac{1}{2} |\vec{a}| |\vec{b}| \sin(	heta)$.Substituting the values:$	ext{Area} = \frac{1}{2} 	imes 3 	imes 5 	imes \sin\left(\frac{\pi}{6}ight)$.Since $\sin\left(\frac{\pi}{6}ight) = \frac{1}{2}$,we get:$	ext{Area} = \frac{1}{2} 	imes 15 	imes \frac{1}{2} = \frac{15}{4}$.

If $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$,$|\vec{b}| = 5$,and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then the area of the triangle formed by these two vectors as two sides is

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