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(B) The area of a parallelogram defined by two adjacent vectors $\vec{A}$ and $\vec{B}$ is given by the magnitude of their cross product,i.e.,$	ext{A

Vedclass · Accepted Answer

$8\sqrt{3}$

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(B) The area of a parallelogram defined by two adjacent vectors $\vec{A}$ and $\vec{B}$ is given by the magnitude of their cross product,i.e.,$	ext{Area} = |\vec{A} 	imes \vec{B}|$.Let $\vec{A} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{B} = 3\hat{i} - 2\hat{j} + \hat{k}$.The cross product is calculated as:$\vec{A} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 3 & -2 & 1 \end{vmatrix}$$= \hat{i}(2(1) - 3(-2)) - \hat{j}(1(1) - 3(3)) + \hat{k}(1(-2) - 2(3))$$= \hat{i}(2 + 6) - \hat{j}(1 - 9) + \hat{k}(-2 - 6)$$= 8\hat{i} + 8\hat{j} - 8\hat{k}$The magnitude is:$|\vec{A} 	imes \vec{B}| = \sqrt{8^2 + 8^2 + (-8)^2} = \sqrt{64 + 64 + 64} = \sqrt{192} = \sqrt{64 	imes 3} = 8\sqrt{3}$ square units.

Two adjacent sides of a parallelogram are represented by the two vectors $\hat{i} + 2\hat{j} + 3\hat{k}$ and $3\hat{i} - 2\hat{j} + \hat{k}$. What is the area of the parallelogram?

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