Question

Find the coordinates of the point where the line joining the points $(2,-3,1)$ and $(3,-4,-5)$ cuts the p[lane $2x+y+z=7$.

Answer 1

The equation of a line passing through two points A($x_1,y_1,z_1$)

and B($x_2,y_2,z_2$) is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Given the lines passes through the points

A(2,-3,1)

$\therefore x_1=2,y_1=-3,z_1=1$

B(3,-4,-5)

$\therefore x_2=3,y_2=-4,z_2=-5$

So, the equation of line is

$\frac{x-2}{3-2}=\frac{y+3}{-4+3}=\frac{z-1}{-5-1}$

so,

$\begin{array}{c}\frac{x-2}{1}=\frac{y+3}{-1}=\frac{z-1}{-6}=k\\ x=k+2\\ y=-k+3\\ z=-6k+1\end{array}$

Let (x,y,z) be the coordinates of the point where the line crosses the plane $2x+y+z=7$

Putting values of x,y,z from the equation (1) in the plane,

$\begin{array}{c}2x+y+z=7\\ 2(k+2)+(-k+3)+(-6k+1)=7\\ 2k+4-k+3-6k+1=7\\ -5k+8=7\\ 5k=8-7\\ 5k=1\\ \therefore k=\frac{1}{5}\end{array}$

putting value of k in x,y,z

$\begin{array}{c}x=k+2=\frac{1}{5}+2=\frac{11}{5}\\ y=-k+3=-\frac{1}{5}+3=\frac{14}{5}\\ z=-6k+1=\frac{-6}{5}+1=\frac{-1}{5}\end{array}$

therefore, the coordinates of the given points are$\left(\frac{11}{5},\frac{14}{5},\frac{-1}{5}\right)$