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

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(D) The area of a triangle with sides $\vec{a}$ and $\vec{b}$ is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{a} 	imes \vec{b}|$. Since $|\

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$\frac{9}{2}$ sq. units

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(D) The area of a triangle with sides $\vec{a}$ and $\vec{b}$ is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{a} 	imes \vec{b}|$. Since $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta$, where $	heta$ is the angle between the vectors: First, calculate the magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$. Given $|\vec{b}| = 6$ and $	heta = \frac{\pi}{6}$. Substitute these values into the area formula: $	ext{Area} = \frac{1}{2} 	imes |\vec{a}| 	imes |\vec{b}| 	imes \sin \left(\frac{\pi}{6}ight)$. $	ext{Area} = \frac{1}{2} 	imes 3 	imes 6 	imes \sin \left(\frac{\pi}{6}ight)$. Since $\sin \left(\frac{\pi}{6}ight) = \frac{1}{2}$, we have: $	ext{Area} = \frac{1}{2} 	imes 3 	imes 6 	imes \frac{1}{2} = \frac{18}{4} = \frac{9}{2}$ sq. units.

If $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}$, $|\vec{b}| = 6$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$, then the area of the triangle (in square units) with $\vec{a}$ and $\vec{b}$ as two of its sides is

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