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If the vectors $\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}},-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ represents
the diagonals of a parallelogram, then its area will be
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the diagonals of a parallelogram, then its area will be
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Verified Answer
The correct answer is:
$\frac{\sqrt{21}}{2}$
Let
$\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$
Now, $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{rrr}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \mathbf{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0\end{array}\right|$
$=-4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
Hence, required area $=\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
$=\frac{1}{2} \sqrt{16+4+1}=\frac{\sqrt{21}}{2}$
$\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$
Now, $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{rrr}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \mathbf{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0\end{array}\right|$
$=-4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
Hence, required area $=\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
$=\frac{1}{2} \sqrt{16+4+1}=\frac{\sqrt{21}}{2}$
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