Question

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

Answer 1

The equation of the line joining (3, -4, -5) and (2,-3, 1) is
$\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \left(say\right)\Rightarrow x=3-\lambda ,y=\lambda -4,z=6\lambda -5$
Therefore, any point on the line is of the form $(3-\lambda ,\lambda -4,6\lambda -5)$
This point lies on the plane, 2x + y + z = 7
Therefore, $2(3-\lambda )+(\lambda -4)+(6\lambda -5)=7\Rightarrow 5\lambda -3=7\Rightarrow \lambda =2$
Hence, the coordinates of the required point are (3 - 2, 2 - 4, 6 $\times$ 2-5) i.e., (1, -2, 7)