Question

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

Answer 1

Answer: 0039

Any plane through the second line is given by
$(3x-y-2z+4)+1(2x+y+z+1)=1$
$(3+2\lambda )x+(\lambda -1)y+(\lambda -2)z+(4+\lambda )=0$
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 $3x-y-2z+4=0$
Shortest distance $=\perp$ distance of a point $(1,3,-2)$ on the first line to the plane $3x-y-2z+4=0$
$\text{ S.D. }=\frac{|3-3+4+4|}{\sqrt{3^2+1^2+2^2}}=\frac{8}{\sqrt{14}}=\sqrt{\frac{64}{14}}=\sqrt{\frac{32}{7}}=\sqrt{\frac{a}{b}}$
$a=32;b=7$
$\therefore (a+b)=39$