Question

In what ratio does the plane $2x+y-z=3$ divide line segment joining the point $a=(2,1,3),b=(9,-2,5)$.

Answer 1

Plane:$2x+y-z=3$

Line passing through $a=(2,1,3)$ and $b=(9,-2,5)$

$L:\frac{x-2}{7}=\frac{y-1}{-3}=\frac{z-3}{2}=k$

Let $P(7r+2)+(-3r+1)-2r-3=3r=\frac{1}{9}$

$\therefore P(x,y,z)=(\frac{25}{9},\frac{2}{3},\frac{29}{9})$

Let $P$ divide $ab$ in $\lambda :1$

$\therefore (\frac{25}{9},\frac{2}{3},\frac{29}{9})=(\frac{2+9\lambda }{\lambda +1},\frac{1-2\lambda }{\lambda +1},\frac{3+5\lambda }{\lambda +1})\frac{2+9\lambda }{\lambda +1}=\frac{25}{9}\Rightarrow 25\lambda +25=18+81\lambda \Rightarrow \lambda =\frac{1}{8}$

$\therefore$ $P$ divide $ab$ in $1:8$.