If the square of the shortest distance between the lines $\frac{x-2}1=\frac{y-1}2=\frac{z+3}{-3}$ and $\frac{x+1}2=\frac{y+3}4=\frac{z+5}{-5}$ is $\frac mn$, where $m,n$ are coprime numbers, then $m+n$ is equal to :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\vec d_1	imes\vec d_2=(1,2,-3)	imes(2,4,-5)=(2,-1,0)$, $|\cdot|=\sqrt5$. The vector between points is $(-3,-4,-2)$, and its dot product with $(2,-1,0)$ is $-6+4=-2$. Distance $=\frac2{\sqrt5}$, so the square is $\frac45$ and $m+n=9$. Answer (2).

Q. If the square of the shortest distance between the lines $\frac{x-2}1=\frac{y-1}2=\frac{z+3}{-3}$ and $\frac{x+1}2=\frac{y+3}4=\frac{z+5}{-5}$ is $\frac mn$, where $m,n$ are coprime numbers, then $m+n$ is equal to :

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?