Equation of the plane, through the points $(-1,2,-2)$ and $(-1,3,2)$ and perpendicular to $y z$ - plane, is

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Aicharya · Accepted Answer

Equation of plane passing though $(-1,2,-2)$ and $(-1,3,2)$ is $\frac{x+1}{(-1+1)}=\frac{y-3}{2-3}=\frac{z-2}{-2-2}$
Above plane is perpendicular to $y \mathrm{z}$ - plane $\begin{aligned} \therefore \quad & \frac{y-3}{-1}=\frac{z-2}{-4} \\ & \Rightarrow 4(y-3)=z-2 \\ & \Rightarrow 4 y-12-z+2=0 \\ & \Rightarrow 4 y-z=10 \end{aligned}$

Equation of the plane, through the points $(-1,2,-2)$ and $(-1,3,2)$ and perpendicular to $y z$ - plane, is

Solution