If 2 hat{i}+3 hat{j}-4 hat{k} and -hat{i}+2 h | vedclass

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(C) Let the diagonals of the parallelogram be $\overrightarrow{d}_1 = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $\overrightarrow{d}_2 = -\hat{i} + 2 \hat

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$\frac{1}{2} \sqrt{174}$

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(C) Let the diagonals of the parallelogram be $\overrightarrow{d}_1 = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $\overrightarrow{d}_2 = -\hat{i} + 2 \hat{j} + \hat{k}$.The area of a parallelogram with diagonals $\overrightarrow{d}_1$ and $\overrightarrow{d}_2$ is given by the formula: $	ext{Area} = \frac{1}{2} |\overrightarrow{d}_1 	imes \overrightarrow{d}_2|$.First, calculate the cross product $\overrightarrow{d}_1 	imes \overrightarrow{d}_2$:$\overrightarrow{d}_1 	imes \overrightarrow{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -4 \ -1 & 2 & 1 \end{vmatrix}$$= \hat{i}(3(1) - (-4)(2)) - \hat{j}(2(1) - (-4)(-1)) + \hat{k}(2(2) - 3(-1))$$= \hat{i}(3 + 8) - \hat{j}(2 - 4) + \hat{k}(4 + 3)$$= 11 \hat{i} + 2 \hat{j} + 7 \hat{k}$.Now, calculate the magnitude of the cross product:$|\overrightarrow{d}_1 	imes \overrightarrow{d}_2| = \sqrt{11^2 + 2^2 + 7^2} = \sqrt{121 + 4 + 49} = \sqrt{174}$.Finally, the area is $\frac{1}{2} \sqrt{174}$.

If $2 \hat{i}+3 \hat{j}-4 \hat{k}$ and $-\hat{i}+2 \hat{j}+\hat{k}$ are the two diagonals of a parallelogram, then the area of the parallelogram in square units is

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