If diagonals of a parallelogram are (5hat i - | vedclass

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(A) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{

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$\sqrt{171} 	ext{ unit}^2$

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(A) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{d_1} 	imes \vec{d_2}|$.Given $\vec{d_1} = 5\hat{i} - 4\hat{j} + 3\hat{k}$ and $\vec{d_2} = 3\hat{i} + 2\hat{j} - \hat{k}$.First,calculate the cross product $\vec{d_1} 	imes \vec{d_2}$:$\vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & -4 & 3 \ 3 & 2 & -1 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(-5 - 9) + \hat{k}(10 - (-12)) = -2\hat{i} + 14\hat{j} + 22\hat{k}$.Now,find the magnitude of the cross product:$|\vec{d_1} 	imes \vec{d_2}| = \sqrt{(-2)^2 + 14^2 + 22^2} = \sqrt{4 + 196 + 484} = \sqrt{684}$.Simplify the square root: $\sqrt{684} = \sqrt{36 	imes 19} = 6\sqrt{19}$.Finally,the area is $\frac{1}{2} 	imes 6\sqrt{19} = 3\sqrt{19} = \sqrt{9 	imes 19} = \sqrt{171} 	ext{ unit}^2$.

If diagonals of a parallelogram are $(5\hat i - 4\hat j + 3\hat k)$ and $(3\hat i + 2\hat j - \hat k)$,then its area is

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