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

Question

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, $\sqrt{\frac{3}{2}}$, $\frac{3}{\sqrt{2}}$, $\frac{\sqrt{3}}{2}$, $\frac{3}{2}$

MARKS App · Accepted Answer

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.

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