Step by step
The question
Question: find the plane through (1, −2, 0) with normal vector n = ⟨0, 3, 1⟩.
(1, −2, 0) , ⟨0, 3, 1⟩
1 Draw it
A normal vector is an arrow at a square corner to the plane.
The plane goes through the point,
and the normal sticks straight out of it.
2 Write the plane
The rule: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.
The normal gives a, b, c.
The point gives x₀, y₀, z₀.
3(y + 2) + z = 0
n = ⟨a, b, c⟩, point (x0, y0, z0)
Careful!
The normal's numbers go in front.
The point's numbers go inside the brackets, with a minus.
Careful: n gives a, b, c; the point goes in the brackets
3 Open the brackets
What is in front of a bracket multiplies every part inside.
3y + 6 + z = 0
Add the plain numbers: they make 6.
3y + z + 6 = 0
Move the number to the right: take away 6 on both sides.
So the plane is 3y + z = −6.
3y + z = −6
4 Check
Let's check: put the point (1, −2, 0) in.
We get −6.
Right!
3(−2) + (0) = −6
Let's look back.
First: draw it.
Then: write the plane.
Then: open the brackets.
Last: check.
Answer
3y + z = −6