Question

The perpendicular distance of $A(1,4,-2)$ from $BC$, where coordinates of $B$ and $C$ are respectively $(2,1,-2)$ and $(0,-5,1)$ is

• A. $\frac{3}{7}$
• B. $\frac{\sqrt{26}}{7}$
• C. $\frac{3\sqrt{26}}{7}$
• D. $\sqrt{26}$

Answer 1

Answer: (C)

$AD=AB\sin \theta =AB\cdot \frac{|\overrightarrow{BC}\times \overrightarrow{BA}|}{|\overrightarrow{BC}|\cdot |\overrightarrow{BA}|}=\frac{|\overrightarrow{BC}\times \overrightarrow{BA}|}{|\overrightarrow{BC}|}$
$[\because |\overrightarrow{BA}|=BA=AB]$
Now $\overrightarrow{BC}=-2\hat{i}-6\hat{j}+3\hat{k}$ and $\overrightarrow{BA}=-\hat{i}+3\hat{j}$
$\therefore \overrightarrow{BC}\times \overrightarrow{BA}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& -6& 3\\ -1& 3& 0\end{array}\right|=-9\hat{i}-3\hat{j}-12\hat{k}$
$|\overrightarrow{BC}\times \overrightarrow{BA}|=\sqrt{9^2+3^2+(12{)}^2}=3\sqrt{26}$ and
$|\overrightarrow{BC}|=\sqrt{4+36+9}=7$
$\therefore AD=\frac{3\sqrt{26}}{7}$