Question

Find the vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0. Hence find whether the plane thus obtained contains the line \[\frac{x + 2}{5} = \frac{y - 3}{4} = \frac{z}{5}\] or not.

Answer 1

\[\text { The equation of the plane passing through the line of intersection of the given planes is }\]

\[x + y + z - 1 + \lambda \left( 2x + 3y + 4z - 5 \right) = 0 \]

\[\left( 1 + 2\lambda \right)x + \left( 1 + 3\lambda \right)y + \left( 1 + 4\lambda \right)z - 1 - 5\lambda = 0 . . . \left( 1 \right)\]

\[\text { This plane is perpendicular to }x - y + z = 0 . So,\]

\[1 + 2\lambda - 1 \left( 1 + 3\lambda \right) + 1 + 4\lambda = 0 (\text { Because }a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]

\[ \Rightarrow 1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0\]

\[ \Rightarrow 3\lambda + 1 = 0\]

\[ \Rightarrow \lambda = \frac{- 1}{3}\]

\[\text { Substituting this in  (1), we get }\]

\[\left( 1 + 2 \left( \frac{- 1}{3} \right) \right)x + \left( 1 + 3 \left( \frac{- 1}{3} \right) \right)y + \left( 1 + 4 \left( \frac{- 1}{3} \right) \right)z - 1 - 5 \left( \frac{- 1}{3} \right) = 0\]

\[ \Rightarrow x - z + 2 = 0\]

We know that if 

\[\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}\] lies in the plane ax + by + cz + d = 0 then, 
ax₁ + by₁ + cz₁ + d = 0 and al + bm + cn = 0.
So, if the line 

\[\frac{x + 2}{5} = \frac{y - 3}{4} = \frac{z}{5}\] lies in the plane x − z + 2 = 0 then, 

\[1 \times \left( - 2 \right) + 0 \times 3 + \left( - 1 \right) \times 0 + 2 = - 2 + 2 = 0\]

Also, 

\[1 \times 5 + 0 \times 4 + \left( - 1 \right) \times 5 = 5 - 5 = 0\]

Hence, the given line lies in the plane  x − z + 2 = 0.