Step by step
The question
Question: find the plane through (1, 1, 1) with normal vector n = ⟨3, 2, 1⟩.
(1, 1, 1) , ⟨3, 2, 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(x − 1) + 2(y − 1) + (z − 1) = 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.
3x − 3 + 2y − 2 + z − 1 = 0
Add the plain numbers: they make −6.
3x + 2y + z − 6 = 0
Move the number to the right: add 6 on both sides.
So the plane is 3x + 2y + z = 6.
3x + 2y + z = 6
4 Check
Let's check: put the point (1, 1, 1) in.
We get 6.
Right!
3(1) + 2(1) + (1) = 6
Let's look back.
First: draw it.
Then: write the plane.
Then: open the brackets.
Last: check.
Answer
3x + 2y + z = 6