Question

If the length of shortest distance between the two lines $\frac{1}{2}(x-1)=\frac{1}{4}(y-3)=z+2$ and $3 x-y-2 z+4=0=2 x+y+z+1$ is $\sqrt{\frac{a}{b}}$ where $a$ and $b$ are coprime then find the value of $(a+b)$.

Answer 1

Any plane through the second line is given by
$(3 x-y-2 z+4)+1(2 x+y+z+1)=1 $
$(3+2 \lambda) x+(\lambda-1) y+(\lambda-2) z+(4+\lambda)=0$
If this plane is parallel to first line then its normal must be at right angles to first line
$(3+2 \lambda) 2+(\lambda-1) 4+(\lambda-2)=0 \Rightarrow \lambda=0$
$\therefore $ equation of plane through the second line and parallel to first line is $3 x-y-2 z+4=0$
Shortest distance $=\perp$ distance of a point $(1,3,-2)$ on the first line to the plane $3 x-y-2 z+4=0$
$ \text { S.D. }=\frac{|3-3+4+4|}{\sqrt{3^2+1^2+2^2}}=\frac{8}{\sqrt{14}}=\sqrt{\frac{64}{14}}=\sqrt{\frac{32}{7}}=\sqrt{\frac{a}{b}} $
$ a=32 ; b=7 $
$\therefore (a+b)=39 $