Q. The distance of the point (1, 2, 1) from the line $\frac{x-1}{2} = \frac{ y-2}{1}=\frac{ z-3}{2} $ is
Solution:
$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}=k$
any point on the line = (2k + 1, k + 2, 2k + 3)
drs of AB : 2k, k, 2k + 2
drs of line : 2, 1, 2
$AB \bot line \Rightarrow 4k+k+4k+4=0$
$9k=-4$
$k=\frac{-4}{9}$
$Distance=\sqrt{\frac{64}{81}+\frac{16}{81}+\frac{100}{81}}=\frac{\sqrt{180}}{9}=\frac{2\sqrt{5}}{3}$
