Question

Find the equation of the plane passing through the point $(1,1,-1)$ and perpendicular to the planes $x+2y+3z-7=0$ and $2x-3y+4z=0$.

Answer 1

Vector equation of a line passing through a point and parallel to a vector is $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ where $\lambda \in R$

The equation of the plane containing the given point is

$A(x-1)+B(y-1)+C(z+1)=0$ --------$\left(1\right)$

Applying the condition of perpendicularity to the plane $\left(1\right)$ with the given planes $x+2y+3z-7=0$ and $2x-3y+4z=0$

$A+2B+3C=7$ and $2A-3B+4C=0$

Solving the equation,we get

$\frac{A}{\left|\begin{array}{cc}2& 3\\ -3& 4\end{array}\right|}=\frac{B}{\left|\begin{array}{cc}3& 1\\ 4& 2\end{array}\right|}=\frac{C}{\left|\begin{array}{cc}1& 2\\ 2& -3\end{array}\right|}$

$=\frac{A}{8+9}=\frac{B}{6-4}=\frac{C}{-3-4}$

$\Rightarrow \frac{A}{17}=\frac{B}{2}=\frac{C}{-7}=\lambda$(say)

$\therefore A=17\lambda ,b=2\lambda$ and $C=-7\lambda$

Substituting these values in equ$\left(1\right)$ we get,

$17\lambda (x-1)+2\lambda (y-1)-7\lambda (z+1)=0$

$\Rightarrow \lambda [17(x-1)+2(y-1)-7(z+1)]=0$

$\Rightarrow \lambda \ne 0,17x-17+2y-2-7z-7=0$

$\Rightarrow 17x+2y-7z=26$

This is the required solution to the plane.