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