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If $\mathrm{A}(3,2,-1), \mathrm{B}(-2,2,-3)$ and $\mathrm{D}(-2,5,-4)$ are the vertices of a parallelogram, then the area of the parallelogram is
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Verified Answer
The correct answer is:
$\sqrt{286}$ sq. units
We have $\mathrm{A} \equiv(3,2,-1), \mathrm{B} \equiv(-2,2,-3)$ and $\mathrm{D} \equiv(-2,5,-4)$
$$
\therefore \overline{\mathrm{AB}}=-5 \hat{\mathrm{i}}-2 \hat{\mathrm{k}} \text { and } \overline{\mathrm{AD}}=-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}
$$
Area of parallelogram $=|\overline{\mathrm{AB}} \times \overline{\mathrm{AD}}|$
Now
$$
\begin{aligned}
& \overline{\mathrm{AB}} \times \overline{\mathrm{AD}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-5 & 0 & -2 \\
-5 & 3 & -3
\end{array}\right|=\hat{\mathrm{i}}(6)-\hat{\mathrm{j}}(5)+\hat{\mathrm{k}}(-15)=6 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-15 \hat{\mathrm{k}} \\
& \therefore \text { Area }=\sqrt{(6)^2+(-5)^2+(-15)^2}
\end{aligned}
$$
$$
\therefore \overline{\mathrm{AB}}=-5 \hat{\mathrm{i}}-2 \hat{\mathrm{k}} \text { and } \overline{\mathrm{AD}}=-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}
$$
Area of parallelogram $=|\overline{\mathrm{AB}} \times \overline{\mathrm{AD}}|$
Now
$$
\begin{aligned}
& \overline{\mathrm{AB}} \times \overline{\mathrm{AD}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-5 & 0 & -2 \\
-5 & 3 & -3
\end{array}\right|=\hat{\mathrm{i}}(6)-\hat{\mathrm{j}}(5)+\hat{\mathrm{k}}(-15)=6 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-15 \hat{\mathrm{k}} \\
& \therefore \text { Area }=\sqrt{(6)^2+(-5)^2+(-15)^2}
\end{aligned}
$$
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