Let A(2, 3, 5) and C(-3, 4, -2) be opposite v | vedclass

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(B) The area of a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ is given by $	ext{Area} = \frac{1}{2} |\overrightarr

Vedclass · Accepted Answer

$\frac{1}{2} \sqrt{474}$

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(B) The area of a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ is given by $	ext{Area} = \frac{1}{2} |\overrightarrow{d_1} 	imes \overrightarrow{d_2}|$. Here,$\overrightarrow{d_1} = \overrightarrow{AC} = \vec{C} - \vec{A} = (-3-2)\hat{i} + (4-3)\hat{j} + (-2-5)\hat{k} = -5\hat{i} + \hat{j} - 7\hat{k}$. The second diagonal is $\overrightarrow{d_2} = \overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$. Now,calculate the cross product $\overrightarrow{AC} 	imes \overrightarrow{BD}$: $\overrightarrow{AC} 	imes \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -5 & 1 & -7 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3 - (-14)) - \hat{j}(-15 - (-7)) + \hat{k}(-10 - 1) = 17\hat{i} + 8\hat{j} - 11\hat{k}$. The magnitude is $|17\hat{i} + 8\hat{j} - 11\hat{k}| = \sqrt{17^2 + 8^2 + (-11)^2} = \sqrt{289 + 64 + 121} = \sqrt{474}$. Thus,the area is $\frac{1}{2} \sqrt{474}$.

Let $A(2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$,then the area of the parallelogram is equal to:

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