Q. The length of the perpendicular distance of the point $ (-1,\,4,\,0) $ from the line $ \frac{x}{1}=\frac{y}{3}=\frac{z}{1} $ is equal to
J & K CETJ & K CET 2011Three Dimensional Geometry
Solution:
Let L be the foot of the perpendicular drawn from the point
$ P(-1,4,0) $ to the given line.
The coordinated of a general point on $ \frac{x-0}{1}=\frac{y-0}{3}=\frac{z-0}{1} $ are given by
$ \frac{x}{1}=\frac{y}{3}=\frac{z}{1}=r $
i.e., $ x=r,\,\,y=3r,\,\,z=r $
Let the coordinate of L be $ (r,\,3r,r) $ .. (i)
Direction ratios of PL are $ =r+1,\,3r-4,\,r $
Direction ratios of the given line are $ 1,\,\,3,\,\,1 $
Since, PL perpendicular to the given, line,
$ \therefore $ $ (r+1)+(3r-4).2+r.1=0 $
$ \Rightarrow $ $ r+1.1+(3r-4).2+r.1=0 $
$ \Rightarrow $ $ 11r=11 $
$ \Rightarrow $ $ r=1 $
So, the coordinate of L is
$ (1,3,1) $
$ \therefore $ $ PL=\sqrt{{{(1+1)}^{2}}+{{(3-4)}^{2}}+{{(1-0)}^{2}}} $
$ =\sqrt{4+1+1}=\sqrt{6} $
