Question

Find the coordinates of the point where the line through $(3,-4,-5)$ and $(2,-3,1)$ crosses the plane $2x+y+z=7$

Answer 1

Equation of line passes through the points $(3,-4,-5)$ and $(2,-3,1)$ is given by
$\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}$
$\Rightarrow$ $\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k$ (say)
$\Rightarrow$ $x=3-k,y=k,z=6k-5$
Therefore, any point on the line is of the form $(3-k,k-4,6k-5)$
This point lies on the plane $2x+y+z=7$
$\therefore$ $2(3-k)+(k-4)+(6k-5)=7$
$\Rightarrow$ $5k-3=7\Rightarrow k=2$
Hence, the coordinates of the required point are $(3-2,2-4,6\times 2-5)$ i.e,$(1,-2,7)$