The question
Question: find the plane through (1, 2, 3) with normal vector n = ⟨2, −1, 4⟩.
(1, 2, 3) , ⟨2, −1, 4⟩
1 Draw it
A normal vector is an arrow at a square corner to the plane.
The plane goes through the point,
and the normal sticks straight out of it.
2 Write the plane
The rule: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.
The normal gives a, b, c.
The point gives x₀, y₀, z₀.
2(x − 1) − (y − 2) + 4(z − 3) = 0
n = ⟨a, b, c⟩, point (x0, y0, z0)
Careful!
The normal's numbers go in front.
The point's numbers go inside the brackets, with a minus.
Careful: n gives a, b, c; the point goes in the brackets
3 Open the brackets
What is in front of a bracket multiplies every part inside.
A minus in front changes the signs inside.
2x − 2 − y + 2 + 4z − 12 = 0
Add the plain numbers: they make −12.
2x − y + 4z − 12 = 0
Move the number to the right: add 12 on both sides.
So the plane is 2x − y + 4z = 12.
2x − y + 4z = 12
4 Check
Let's check: put the point (1, 2, 3) in.
We get 12.
Right!
2(1) − (2) + 4(3) = 12
Let's look back.
First: draw it.
Then: write the plane.
Then: open the brackets.
Last: check.
▶ Watch it solved
The question
Question: find the plane through (2, 3, 1) with normal vector n = ⟨1, 1, 1⟩.
(2, 3, 1) , ⟨1, 1, 1⟩
1 Draw it
A normal vector is an arrow at a square corner to the plane.
The plane goes through the point,
and the normal sticks straight out of it.
2 Write the plane
The rule: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0.
The normal gives a, b, c.
The point gives x₀, y₀, z₀.
(x − 2) + (y − 3) + (z − 1) = 0
n = ⟨a, b, c⟩, point (x0, y0, z0)
Careful!
The normal's numbers go in front.
The point's numbers go inside the brackets, with a minus.
Careful: n gives a, b, c; the point goes in the brackets
3 Open the brackets
What is in front of a bracket multiplies every part inside.
x − 2 + y − 3 + z − 1 = 0
Add the plain numbers: they make −6.
x + y + z − 6 = 0
Move the number to the right: add 6 on both sides.
So the plane is x + y + z = 6.
x + y + z = 6
4 Check
Let's check: put the point (2, 3, 1) in.
We get 6.
Right!
(2) + (3) + (1) = 6
Let's look back.
First: draw it.
Then: write the plane.
Then: open the brackets.
Last: check.
▶ Watch it solved
The question
Question: find the plane through (1, 1, 0), (2, 0, 1), (0, 3, 2).
P(1, 1, 0), Q(2, 0, 1), R(0, 3, 2)
1 Two vectors in the plane
Draw the three points.
From P to Q: end minus start.
u = ⟨1, −1, 1⟩.
u = PQ = ⟨1, −1, 1⟩
From P to R: v = ⟨−1, 2, 2⟩.
Both lie in the plane.
v = PR = ⟨−1, 2, 2⟩
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.
n = u × v = |i, j, k; 1, −1, 1; −1, 2, 2|
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.
x part = |−1, 1; 2, 2| = −1 × 2 − 1 × 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.
y part = −|1, 1; −1, 2| = −(1 × 2 − 1 × (−1)) = −3
Careful!
The y part, under j, always has a minus in front.
Forgetting it is the most common mistake.
Careful: the middle part gets a minus
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.
z part = |1, −1; −1, 2| = 1 × 2 − −1 × (−1) = 1
Put the three parts together: n = u × v = ⟨−4, −3, 1⟩.
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(x − 1) − 3(y − 1) + z = 0
n = ⟨a, b, c⟩, point (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.
−4x + 4 − 3y + 3 + z = 0
Add the plain numbers: they make 7.
−4x − 3y + z + 7 = 0
Move the number to the right: take away 7 on both sides.
So the plane is −4x − 3y + z = −7.
−4x − 3y + z = −7
5 Check
Let's check: put the point (1, 1, 0) in.
We get −7.
Right!
−4(1) − 3(1) + (0) = −7
(2, 0, 1): −7 again.
Right!
−4(2) − 3(0) + (1) = −7
(0, 3, 2): −7 again.
Right!
−4(0) − 3(3) + (2) = −7
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.
▶ Watch it solved
