Question

The line joining points $(2,-1,4)$ and $(4,3,2)$ intersects the plane $x+y+z-3=0$ at the point

• A. $(1,5,-3)$
• B. $(1,-3,5)$
• C. $(2,9,-8)$
• D. $(\frac{5}{2},0,7)$

Answer 1

The correct option is B $(1,-3,5)$

Given:
$x+y+z-3=0$ divides the line segment joining $A(2,-1,4)$ and $B(4,3,2)$
Let given plane divide $AB$ in the ratio $AP:PB$
$\Rightarrow \frac{AP}{PB}=\frac{-(a{x}_1+b{y}_1+c{z}_1+d)}{(a{x}_2+b{y}_2+c{z}_2+d)}$
$\Rightarrow \frac{AP}{PB}=\frac{-\left(1\right(2)+1(-1)+1(4)-3)}{\left(1\right(4)+1(3)+1(2)-3)}$
$\Rightarrow \frac{AP}{PB}=\frac{-2}{6}$
$\Rightarrow \frac{AP}{PB}=\frac{-1}{3}$
As the ratio is negative division is external.
$\therefore$Coordinates of $P$
$=(\frac{4-6}{1-3},\frac{3+3}{1-3},\frac{2-12}{1-3})$
$=(1,-3,5)$