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(B) When $\vec{A}$ and $\vec{B}$ are the diagonals of a parallelogram,the area of the parallelogram is given by the formula: $	ext{Area} = \frac{1}{2

Vedclass · Accepted Answer

$5$

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(B) When $\vec{A}$ and $\vec{B}$ are the diagonals of a parallelogram,the area of the parallelogram is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{A} 	imes \vec{B}|$.First,calculate the cross product $\vec{A} 	imes \vec{B}$:$\vec{A} 	imes \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & -4 & 3 \ 3 & -2 & -1 \end{vmatrix}$$= \hat{i}((-4)(-1) - (3)(-2)) - \hat{j}((5)(-1) - (3)(3)) + \hat{k}((5)(-2) - (-4)(3))$$= \hat{i}(4 + 6) - \hat{j}(-5 - 9) + \hat{k}(-10 + 12)$$= 10\hat{i} + 14\hat{j} + 2\hat{k}$.Next,calculate the magnitude of the cross product:$|\vec{A} 	imes \vec{B}| = \sqrt{10^2 + 14^2 + 2^2} = \sqrt{100 + 196 + 4} = \sqrt{300} = 10\sqrt{3}$.Finally,calculate the area:$	ext{Area} = \frac{1}{2} |\vec{A} 	imes \vec{B}| = \frac{1}{2} 	imes 10\sqrt{3} = 5\sqrt{3}$.

The diagonals of a parallelogram are represented by vectors $\vec{A} = 5\hat{i} - 4\hat{j} + 3\hat{k}$ and $\vec{B} = 3\hat{i} - 2\hat{j} - \hat{k}$. What is the area of the parallelogram (in $\sqrt{3}$)?

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