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.

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ x+y+z-1+\lambda \left(2x+3y+4z-5\right)=0 \\ \left(1+2\lambda \right)x+\left(1+3\lambda \right)y+\left(1+4\lambda \right)z-1-5\lambda =0...\left(1\right) \\ \text{This plane is perpendicular to}x-y+z=0.\text{So,} \\ 1+2\lambda -1\left(1+3\lambda \right)+1+4\lambda =0\text{(Because}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\text{)} \\ \Rightarrow 1+2\lambda -1-3\lambda +1+4\lambda =0 \\ \Rightarrow 3\lambda +1=0 \\ \Rightarrow \lambda =\frac{-1}{3} \\ \text{Substituting this in (1), we get} \\ \left(1+2\left(\frac{-1}{3}\right)\right)x+\left(1+3\left(\frac{-1}{3}\right)\right)y+\left(1+4\left(\frac{-1}{3}\right)\right)z-1-5\left(\frac{-1}{3}\right)=0 \\ \Rightarrow x-z+2=0 \end{gathered}$$