I need to get an equation of a plane

Question

1. Find equation of a plane which passes through the point M(2,-1,-1) and is normal with the planes 3x+2y-z+4=0 and x+y+z-3=0....? How can i get equation of required plane?

Answer 1

From the equations of the given two planes we get a vector normal to the first plane $\vec v_1=(3,2,-1)$, and a vector normal to the second plane $\vec v_2=(1,1,1)$. From these two vectors we get a vector $\vec n$ normal to both by doing the vector product:

$$\vec n = \vec v_1\times\vec v_2 = (3,-4,1)\,.$$

The desired plane is the plane normal to $\vec n$ and passing through the given point $\mathbf{M}$. The equation of this plane is then $\vec n\cdot(\vec r - \mathbf{\vec M})=0$ where $\vec r =(x,y,z)$ and $\mathbf{\vec M} = (2,-1,-1)$, the position vector of the given point.

$$3(x-2)-4(y+1)+(z+1)=0\,.$$

Simplifying:

$$\bbox[lightgreen,7px]{3x-4y+z=9}\,.$$

Answer 2

The normal vectors to each of the planes have coordinates $(3,2,-1)$ and $(1,1,1)$. A plane which is orthogonal to both planes has the cross-product of these vectors as a normal vector: $$\begin{bmatrix}2&2&-1&3&2\\1&1&1&1&1\end{bmatrix}\rightsquigarrow\;\vec n=(3,-4,1)$$ So, if the plane passes through the point $(2,-1,-1)$, we obtain the equation $$\color{red}{3x-4y+z=}3\cdot2-4\cdot(-1)+1\cdot(-1)=\color{red}9.$$