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

$$\begin{gathered} \text{The equation of the line through the points (3, -4, -5) and (2, -3, 1) is} \\ \frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5} \\ \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6} \\ \text{The coordinates of any point on this line are of the form} \\ \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \\ \Rightarrow x=-\lambda +3;y=\lambda -4;z=6\lambda -5 \\ \text{So, the coordinates of the point on the given line are}\left(-\lambda +3,\lambda -4,6\lambda -5\right)\text{.} \\ \text{Since this point lies on the plane}2x+y+z=7, \\ 2\left(-\lambda +3\right)+\lambda -4+6\lambda -5=7 \\ \Rightarrow -2\lambda +6+\lambda -4+6\lambda -5=7 \\ \Rightarrow 5\lambda =10 \\ \Rightarrow \lambda =2 \\ \mathrm{So,\; the\; coordinates\; of\; the\; point}\text{are} \\ \left(-\lambda +3,\lambda -4,6\lambda -5\right) \\ =\left(-2+3,2-4,6\left(2\right)-5\right) \\ =\left(1,-2,7\right) \end{gathered}$$