Step by step
The question
Question: find the plane through (1, 0, 2), (3, 1, 0), (0, 2, 1).
1 Two vectors in the plane
Draw the three points.
From P to Q: end minus start.
u = ⟨2, 1, −2⟩.
From P to R: v = ⟨−1, 2, −1⟩.
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, −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): 3.
Now cover the j column.
Four numbers are left: 2, −2, −1, −1.
Write them as a small grid, with a minus in front: the middle part always gets one.
Its cross: down 2 times (−1), minus up −2 times (−1).
−2 minus 2 is −4.
With the minus: 4.
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: 2, 1, −1, 2.
Write them as a small grid.
Its cross: down 2 times 2, minus up 1 times (−1).
That is 4 minus (−1): 5.
Put the three parts together: n = u × v = ⟨3, 4, 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.
Add the plain numbers: they make −13.
Move the number to the right: add 13 on both sides.
So the plane is 3x + 4y + 5z = 13.
5 Check
Let's check: put the point (1, 0, 2) in.
We get 13.
Right!
(3, 1, 0): 13 again.
Right!
(0, 2, 1): 13 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.