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

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