Question

If the position vectors of the vertices of a $\Delta A B C$ are $O A =3 \hat{ i }+\hat{ j }+2 \hat{ k }, O B =\hat{ i }+2 \hat{ j }+3 \hat{ k }$ and $O C =2 \hat{ i }+3 \hat{ j }+\hat{ k }$, then the length of the altitude of $\Delta A B C$ 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 $\Delta A B C$ drawn from $A$ is
$h=\frac{(\text { Area of } \Delta A B C)}{\frac{1}{2}| B C |}=\frac{\frac{1}{2}| A B \times A C |}{\frac{1}{2}| B C |}$
$\because A B =-2 \hat{ i }+\hat{ j }+\hat{ k }$
$A C =-\hat{ i }+2 \hat{ j }-\hat{ k }$
and $B C =\hat{ i }+\hat{ j }-2 \hat{ k }$
So, $A B \times A C =\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & 1 & 1 \\ -1 & 2 & -1\end{vmatrix}$
$=\hat{ i }(-1-2)- j (2+1)+\hat{ k }(-4+1)=-3 \hat{ i }-3 \hat{ j }-3 \hat{ k }$
$\therefore | A B \times A C |=3 \sqrt{3}$ and $| B C |=\sqrt{6}$
$\therefore h=\frac{3 \sqrt{3}}{\sqrt{2} \sqrt{3}}=\frac{3}{\sqrt{2}}$