Finding equation of plane normal to a line.

Question

How to find the equation of the plane which passes through the point $(3, -3, 1)$ and normal to the line joining the points $(3, 2, -1)$ and $(2, -1, 5)$ ?

Answer 1

Notice:

Let the equation of the plane be $ax+by+cz+d=0$

The normal vector to the plane will be parallel to the given line passing through the points $(3, 2, -1)$ & $(2, -1, 5)$ Hence, the direction ratios of the normal vector to the plane are calculated as follows $$a=3-2=1, b=2-(-1)=3, c=-1-5=-6$$ hence, the equation of the plane is $$x+3y-6z+d=0$$ but the plane passes through the given point $(3, -3, 1)$ hence $$3+3(-3)-6(1)+d=0\iff d=12$$

substituting the value $d=12$, the equation of the plane is $$\color{red}{x+3y-6z+12=0}$$

Answer 2

the vector between (3,2,-1) and (2,-1,5) is $$v=(3-2)i+(2+1)j+(-1-5)k$$ $$v=i+3j-6k$$ the vector $v$ is normal to the plane so we can use the following equation to find the plane equation $$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$ so that $$A=v_x=1, B=v_y=3, and C=v_z=-6$$ $(x_0,y_0,z_0)=(3,-3,1)$