Step by step
The question
Question: find the plane through (1, 1, 0), (2, 0, 1), (0, 3, 2).
1 Two vectors in the plane
Draw the three points.
From P to Q: end minus start.
u = ⟨1, −1, 1⟩.
From P to R: v = ⟨−1, 2, 2⟩.
Both lie in the plane.
2 Find the normal vector
Write a grid.
On top: i, j, k, for the x, y and z directions.
Under them, u, then v.
Cover the i column.
Four numbers are left: −1, 1, 2, 2.
Write them as a small grid.
Its cross: down −1 times 2, minus up 1 times 2.
That is −2 minus 2: −4.
Now cover the j column.
Four numbers are left: 1, 1, −1, 2.
Write them as a small grid, with a minus in front: the middle part always gets one.
Its cross: down 1 times 2, minus up 1 times (−1).
2 minus (−1) is 3.
With the minus: −3.
Careful!
The y part, under j, always has a minus in front.
Forgetting it is the most common mistake.
Now cover the k column.
Four numbers are left: 1, −1, −1, 2.
Write them as a small grid.
Its cross: down 1 times 2, minus up −1 times (−1).
That is 2 minus 1: 1.
Put the three parts together: n = u × v = ⟨−4, −3, 1⟩.
3 Write the plane
The cross product is at a square corner to both, so it is a normal vector.
Use the point P.
The rule: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.
The normal gives a, b, c.
The point gives x₀, y₀, z₀.
4 Open the brackets
What is in front of a bracket multiplies every part inside.
A minus in front changes the signs inside.
Add the plain numbers: they make 7.
Move the number to the right: take away 7 on both sides.
So the plane is −4x − 3y + z = −7.
5 Check
Let's check: put the point (1, 1, 0) in.
We get −7.
Right!
(2, 0, 1): −7 again.
Right!
(0, 3, 2): −7 again.
Right!
Let's look back.
First: two vectors in the plane.
Then: find the normal vector.
Then: write the plane.
Then: open the brackets.
Last: check.