Step by step
The question
Question: find the plane through (2, 0, 0), (0, 4, 0), (0, 0, 1).
1 Two vectors in the plane
Draw the three points.
From P to Q: end minus start.
u = ⟨−2, 4, 0⟩.
From P to R: v = ⟨−2, 0, 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: 4, 0, 0, 1.
Write them as a small grid.
Its cross: down 4 times 1, minus up 0 times 0.
That is 4 minus 0: 4.
Now cover the j column.
Four numbers are left: −2, 0, −2, 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 0 times (−2).
−2 minus 0 is −2.
With the minus: 2.
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, 4, −2, 0.
Write them as a small grid.
Its cross: down −2 times 0, minus up 4 times (−2).
That is 0 minus (−8): 8.
Put the three parts together: n = u × v = ⟨4, 2, 8⟩.
Any arrow in the same direction works.
All parts can be divided by 2: n = ⟨2, 1, 4⟩.
Smaller numbers are easier.
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 −4.
Move the number to the right: add 4 on both sides.
So the plane is 2x + y + 4z = 4.
5 Check
Let's check: put the point (2, 0, 0) in.
We get 4.
Right!
(0, 4, 0): 4 again.
Right!
(0, 0, 1): 4 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.