Question

If the position vectors of the vertices of a $\Delta ABC$ are $OA=3\hat{i}+\hat{j}+2\hat{k},OB=\hat{i}+2\hat{j}+3\hat{k}$ and $OC=2\hat{i}+3\hat{j}+\hat{k}$, then the length of the altitude of $\Delta ABC$ drawn from $A$ is

• A. $\sqrt{\frac{3}{2}}$
• B. $\frac{3}{\sqrt{2}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

Since, length of altitude of $\Delta ABC$ drawn from $A$ is
$h=\frac{(\text{ Area of }\Delta ABC)}{\frac{1}{2}|BC|}=\frac{\frac{1}{2}|AB\times AC|}{\frac{1}{2}|BC|}$
$\because AB=-2\hat{i}+\hat{j}+\hat{k}$
$AC=-\hat{i}+2\hat{j}-\hat{k}$
and $BC=\hat{i}+\hat{j}-2\hat{k}$
So, $AB\times AC=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 1& 1\\ -1& 2& -1\end{array}\right|$
$=\hat{i}(-1-2)-j(2+1)+\hat{k}(-4+1)=-3\hat{i}-3\hat{j}-3\hat{k}$
$\therefore |AB\times AC|=3\sqrt{3}$ and $|BC|=\sqrt{6}$
$\therefore h=\frac{3\sqrt{3}}{\sqrt{2}\sqrt{3}}=\frac{3}{\sqrt{2}}$