Step by step
The question
Question: find the plane through (−1, 2, 4) with normal vector n = ⟨2, 3, −1⟩.
(−1, 2, 4) , ⟨2, 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.
2(x + 1) + 3(y − 2) − (z − 4) = 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.
A minus in front changes the signs inside.
2x + 2 + 3y − 6 − z + 4 = 0
The numbers make 0.
So the plane is 2x + 3y − z = 0.
2x + 3y − z = 0
4 Check
Let's check: put the point (−1, 2, 4) in.
We get 0.
Right!
2(−1) + 3(2) − (4) = 0
Let's look back.
First: draw it.
Then: write the plane.
Then: open the brackets.
Last: check.
Answer
2x + 3y − z = 0