Step by step
The question
Question: find the plane through (0, 0, 0), (1, 2, 1), (2, −1, 3).
1 Two vectors in the plane
Draw the three points.
From P to Q: end minus start.
u = ⟨1, 2, 1⟩.
From P to R: v = ⟨2, −1, 3⟩.
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: 2, 1, −1, 3.
Write them as a small grid.
Its cross: down 2 times 3, minus up 1 times (−1).
That is 6 minus (−1): 7.
Now cover the j column.
Four numbers are left: 1, 1, 2, 3.
Write them as a small grid, with a minus in front: the middle part always gets one.
Its cross: down 1 times 3, minus up 1 times 2.
3 minus 2 is 1.
With the minus: −1.
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, 2, 2, −1.
Write them as a small grid.
Its cross: down 1 times (−1), minus up 2 times 2.
That is −1 minus 4: −5.
Put the three parts together: n = u × v = ⟨7, −1, −5⟩.
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.
The numbers make 0.
So the plane is 7x − y − 5z = 0.
5 Check
Let's check: put the point (0, 0, 0) in.
We get 0.
Right!
(1, 2, 1): 0 again.
Right!
(2, −1, 3): 0 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.