Finding the Equation of a Plane

The coordinate planes

The $xy$ -plane is every point where the height above it is zero, so its equation is simply $z=0$ . Likewise the $yz$ -plane is $x=0$ and the $xz$ -plane is $y=0$ .

Worked example: a plane through an intersection

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

Solution: Any plane through that intersection has the form $(3x-y+2z-4) + \lambda(x+y+z-3) = 0$ for some constant $\lambda$ . Plugging in $(2,2,1)$ pins down $\lambda$ :

(6-2+2-4) + \lambda(2+2+1-3) = 0 \;\Rightarrow\; 2 + 2\lambda = 0 \;\Rightarrow\; \lambda = -1

Substituting back in:

(3x-y+2z-4) - (x+y+z-3) = 0 \;\Rightarrow\; 2x-2y+z-1 = 0

In vector form, this is:

\vec{r}\cdot(2\hat{i}-2\hat{j}+\hat{k}) = 1

Finding the Equation of a Plane

The coordinate planes

The $xy$ -plane is every point where the height above it is zero, so its equation is simply $z=0$ . Likewise the $yz$ -plane is $x=0$ and the $xz$ -plane is $y=0$ .

Worked example: a plane through an intersection

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

Solution: Any plane through that intersection has the form $(3x-y+2z-4) + \lambda(x+y+z-3) = 0$ for some constant $\lambda$ . Plugging in $(2,2,1)$ pins down $\lambda$ :

(6-2+2-4) + \lambda(2+2+1-3) = 0 \;\Rightarrow\; 2 + 2\lambda = 0 \;\Rightarrow\; \lambda = -1

Substituting back in:

(3x-y+2z-4) - (x+y+z-3) = 0 \;\Rightarrow\; 2x-2y+z-1 = 0

In vector form, this is:

\vec{r}\cdot(2\hat{i}-2\hat{j}+\hat{k}) = 1

Finding the Equation of a Plane

The coordinate planes

Worked example: a plane through an intersection

Continue learning

Finding the Equation of a Plane

The coordinate planes

Worked example: a plane through an intersection

Continue learning