A $A(3,2,-1)$ and $B(1,4,3)$, then equation of the plane which bisects segment AB perpendicularly
A $A(3,2,-1)$ and $B(1,4,3)$, then equation of the plane which bisects segment
AB perpendicularly
$x+y+2 z+3=0$
$x-y+2 z-3=0$
$x+y-2 z-3=0$
$x-y-2 z+3=0$
Solution
Since the plane bisects seg $A B$, the plane meets the line $A B$ at the mid point i.e. $\left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right) \equiv(2,3,1)$
Now line $A B$ is $\perp$ er to the plane
Direction ratios of plane are $1-3,4-2,3+1$ i.e. $-2,2,4$ i.e. $-1,1,2$
Equation of plane passing through $(2,3,1)$ and having d.r.s. $(-1,1,2)$ are
$-(x-2)+(y-3)+2(z-1)=0 \Rightarrow-x+2+y-3+2 z-2=0$
$\therefore x-y-2 z+3=0$