Question

Find the distance of a point (2,4,-1) from the line. $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$.

Answer 1

We have, equation of the line as $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda$.
$\Rightarrow x=\lambda -5,y=4\lambda -3,z=6-9\lambda$
Let the coordinates of L be $(\lambda -5,4\lambda -3,6-9\lambda )$, then Dr's of PL are $(\lambda -7,4\lambda -7,7-9\lambda )$.
Also, the direction ratios of given line are proportional to 1,4,-9.
Since, PL is perpendicular to the given line.
$\therefore (\lambda -7).1+(4\lambda -7).4+(7-9\lambda ).(-9)=0$
$\Rightarrow \lambda -7+16\lambda -28+81\lambda -63=0$
$\Rightarrow 98\lambda =98\Rightarrow \lambda =1$
So, the coordinates of L are (-4,1,-3).
$\therefore$ Required distance, $PL=\sqrt{(-4-2{)}^2+(1-4{)}^2+(-3+1{)}^2}=\sqrt{36+9+4}=7$ units