Finding cartesian equation of a plane that passes only two points and parallel to x-axis

Question

I need help with finding the cartesian equation for the plane for the following conditions:

$\pi_2$ passed B(4,2,-1) and C(2,2,3) and is also parallel to x-axis.

Usually if the question gives me three points, I will be able to obtain two vector from which I can find the normal vector of the plane (Conducting a cross product of these two vectors).

So where I can find the third point in this question?

The answer key says the right equation is $y=2$ but I am not sure how they obtained this?

Answer 1

General equation of a plane in cartesian form : $$ ax + by + cz = d$$ Given that the plane is parallel to $x$-axis $\Rightarrow a = 0 \Rightarrow$ , the equation becomes $ by + cz = d$, On putting the values $ (4 ,2 ,-1), (2,2,3)$ in the equation, and subtracting one equation from other, you would find $z = 0 \Rightarrow$ The equation is of form $by =d$, On putting the values in the equation you would find $ \dfrac{d}{b} = 2 \Rightarrow$ Equation of the plane is $y =2$

Answer 2

Three methods:

• The plane $\mathbf\pi_2$ is parallel to both the $x$-axis and the direction vector of the line $BC$, so its normal is perpendicular to both of them. A normal to the plane is therefore $(1,0,0)\times\left((4,2,-1)-(2,2,3)\right)=(1,0,0)\times(2,0,-4)=(0,4,0)$. Take $(0,1,0)$ for simplicity and use the point-normal form of plane equation.

• The plane contains the line through the two given points, which has direction vector $(2,0,-4)$. Two linearly-independent vectors perpendicular to this are $(0,1,0)$ and $(2,0,1)$, so every plane that contain the two points has an equation of the form $(1-\lambda)(y-2)+\lambda(2x-z-7)=0$, or $2\lambda x+(1-\lambda)y+\lambda z=5\lambda+2$, $\lambda\in\mathbb R$. If it’s parallel to the $x$-axis, then the plane’s normal is orthogonal to $(1,0,0)$. Taking the dot product yields the equation $2\lambda=0$, so the equation of the sought-after plane is $y=2$.

• Passing to homogeneous coordinates, you already have three points on the plane: the two finite points with homogeneous coordinates $4:2:-1:1$ and $2:2:3:1$, and the point at infinity $1:0:0:0$ corresponding to the direction of the $x$-axis. The plane that contains them is therefore a null vector of the matrix $$\begin{bmatrix}1&0&0&0\\4&2&-1&1\\2&2&3&1\end{bmatrix}.$$ Conventional Gaussian elimination yields $0:1:0:-2$, i.e., $y=2$.