The angle between two consecutive sides vec{a | vedclass

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(A) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the formula: $	ext{Area} = |\vec{a}| |\vec{b}| \sin(	heta)$.

Vedclass · Accepted Answer

$9$

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

The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.

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