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If $\mathrm{OA}=6 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}, \mathrm{OB}=2 \hat{\mathbf{j}}+\hat{\mathrm{k}}$, OC $=5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are the coterminous edges of a parallelopiped, then the height of the parallelopiped drawn from the vertex $A$ is
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Verified Answer
The correct answer is:
$\frac{17}{\sqrt{6}}$
We have,
$\begin{aligned}
\mathbf{O A} & =6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\
\mathbf{O B} & =2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\mathbf{O C} & =5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\mathbf{O A} \cdot(\mathbf{O B} \times \mathbf{O C}) & =\left|\begin{array}{ccc}
6 & 3 & -4 \\
0 & 2 & 1 \\
5 & -1 & 2
\end{array}\right| \\
& =6(4+1)-3(-5)-4(-10) \\
\mathbf{O A} \cdot(\mathbf{O B} \times \mathbf{O C}) & =30+15+40=85 \\
\mathbf{O B} \times \mathbf{O C} & =\left|\begin{array}{lll}
1 & 5 & k \\
0 & 2 & 1 \\
5 & -1 & 2
\end{array}\right|=5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-10 \hat{\mathbf{k}} \\
|\mathbf{O B} \times \mathbf{O C}| & =\sqrt{25+25+100}=5 \sqrt{6}
\end{aligned}$
Height of parallelopiped drawn from $A$
$=\frac{\mathrm{OA} \cdot(\mathrm{OB} \times \mathrm{OC})}{|\mathrm{OB} \times \mathrm{OC}|}=\frac{85}{5 \sqrt{6}}=\frac{17}{\sqrt{6}}$
$\begin{aligned}
\mathbf{O A} & =6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\
\mathbf{O B} & =2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\mathbf{O C} & =5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\mathbf{O A} \cdot(\mathbf{O B} \times \mathbf{O C}) & =\left|\begin{array}{ccc}
6 & 3 & -4 \\
0 & 2 & 1 \\
5 & -1 & 2
\end{array}\right| \\
& =6(4+1)-3(-5)-4(-10) \\
\mathbf{O A} \cdot(\mathbf{O B} \times \mathbf{O C}) & =30+15+40=85 \\
\mathbf{O B} \times \mathbf{O C} & =\left|\begin{array}{lll}
1 & 5 & k \\
0 & 2 & 1 \\
5 & -1 & 2
\end{array}\right|=5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-10 \hat{\mathbf{k}} \\
|\mathbf{O B} \times \mathbf{O C}| & =\sqrt{25+25+100}=5 \sqrt{6}
\end{aligned}$
Height of parallelopiped drawn from $A$
$=\frac{\mathrm{OA} \cdot(\mathrm{OB} \times \mathrm{OC})}{|\mathrm{OB} \times \mathrm{OC}|}=\frac{85}{5 \sqrt{6}}=\frac{17}{\sqrt{6}}$
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