Skill: Write scalar equations of planes

A plane can be written in the form

$$ax+by+cz=d,$$

where the vector $\mathbf n=(a,b,c)$ is a normal vector to the plane.

If you know a point $P(x_0,y_0,z_0)$ on the plane and a normal vector $\mathbf n=(a,b,c)$, use

$$\mathbf n\cdot\bigl((x,y,z)-P\bigr)=0.$$

This means that the vector from $P$ to any point $(x,y,z)$ on the plane is perpendicular to the normal vector.

Example

Write the scalar equation of the plane through

$$P=(1,-2,3)$$

with normal vector

$$\mathbf n=(2,1,-4).$$

Step 1: Substitute the point and normal vector into the point-normal form.

$$(2,1,-4)\cdot\bigl((x,y,z)-(1,-2,3)\bigr)=0$$

The difference of the points is

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

So,

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

Step 2: Calculate the dot product.

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

Step 3: Expand and simplify.

$$2x-2+y+2-4z+12=0$$ $$2x+y-4z+12=0.$$

Therefore, one scalar equation of the plane is

$$\boxed{2x+y-4z=-12}.$$

You can check that the point lies on the plane:

$$2(1)+(-2)-4(3)=-12.$$

Multiplying every coefficient by the same nonzero number gives an equivalent equation for the same plane.