Question

Find the equation of the plane passing through the line of intersection of the planes 2x − y = 0 and 3z − y = 0 and perpendicular to the plane 4x + 5y − 3z = 8.

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ 2x-y+\lambda \left(3z-y\right)=0 \\ 2x+\left(-1-\lambda \right)y+3\lambda z=0...\left(1\right) \\ \text{This plane is perpendicular to}4x+5y-3z=8.\text{So,} \\ 2\left(4\right)+\left(-1-\lambda \right)5-9\lambda =0\text{(Because}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\text{)} \\ \Rightarrow 8-5-5\lambda -9\lambda =0 \\ \Rightarrow \lambda =\frac{3}{14} \\ \text{Substituting this in (1), we get} \\ 2x+\left(-1-\frac{3}{14}\right)y+3\left(\frac{3}{14}\right)z=0 \\ \Rightarrow 28x-17y+9z=0 \end{gathered}$$