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Q.
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
Vector Algebra
Solution:
$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 }|} $
$[\because|\overrightarrow{B A}|=B A=A B]$
Now $\overrightarrow{ BC }=-2 \hat{ i }-6 \hat{ j }+3 \hat{ k }$ and $\overrightarrow{ BA }=-\hat{ i }+3 \hat{ j }$
$\therefore \overrightarrow{ BC } \times \overrightarrow{ BA }=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & -6 & 3 \\ -1 & 3 & 0\end{vmatrix}=-9 \hat{ i }-3 \hat{ j }-12 \hat{ k }$
$|\overrightarrow{ BC } \times \overrightarrow{ BA }|=\sqrt{9^{2}+3^{2}+(12)^{2}}=3 \sqrt{26}$ and
$|\overrightarrow{ BC }|=\sqrt{4+36+9}=7 $
$ \therefore AD =\frac{3 \sqrt{26}}{7}$
