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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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The correct answer is:
$\sqrt{\frac{87}{2}}$
Let $\overrightarrow{\mathrm{d}}_1=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}}_2=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
Now Area is $=\frac{1}{2} \overline{\mathrm{d}}_1 \times \overline{\mathrm{d}}_2$
$=\frac{1}{2}(2 \hat{i}+3 \hat{j}-4 \hat{k}) \times(-\hat{i}+2 \hat{j}+\hat{k})=\sqrt{\frac{87}{2}}$
Now Area is $=\frac{1}{2} \overline{\mathrm{d}}_1 \times \overline{\mathrm{d}}_2$
$=\frac{1}{2}(2 \hat{i}+3 \hat{j}-4 \hat{k}) \times(-\hat{i}+2 \hat{j}+\hat{k})=\sqrt{\frac{87}{2}}$
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