If the position vectors of the vertices of a \(\triangle A B C\) are \(\mathbf{O A}=3…
If the position vectors of the vertices of a \(\triangle A B C\) are \(\mathbf{O A}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{O B}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) and \(\mathbf{O C}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\), then the length of the altitude of \(\triangle A B C\) drawn from \(A\) is
\(\sqrt{\frac{3}{2}}\)
\(\frac{3}{\sqrt{2}}\)
\(\frac{\sqrt{3}}{2}\)
\(\frac{3}{2}\)
Solution
Since, length of altitude of \(\triangle A B C\) drawn from \(A\) is
\(\begin{aligned}
& h=\frac{(\text { Area of } \triangle A B C)}{\frac{1}{2}|\mathbf{B C}|}=\frac{\frac{1}{2}|\mathbf{A B} \times \mathbf{A C}|}{\frac{1}{2}|\mathbf{B C}|} \\
& \because \quad \mathbf{A B}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& \mathbf{A C}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\
& \text { and } \quad \mathbf{B C}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}
\end{aligned}\)
\(\begin{aligned}
& \text {So, } \mathbf{A B} \times \mathbf{A C}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
-2 & 1 & 1 \\
-1 & 2 & -1
\end{array}\right| \\
& =\hat{\mathbf{i}}(-\mathbf{l}-2-\mathbf{j}(2+1)+\hat{\mathbf{k}}(-4+1)=-3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
& \therefore|\mathbf{A B} \times \mathbf{A C}|=3 \sqrt{3} \text { and }|\mathbf{B C}|=\sqrt{6} \\
& \therefore h=\frac{3 \sqrt{3}}{\sqrt{2} \sqrt{3}}=\frac{3}{\sqrt{2}}
\end{aligned}\)
Hence, option (2) is correct.