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Plane through (1, 2, 3), (2, 3, 1) and (0, 1, 2)

Step by step, with the reason for every step · Planes in space · Calculus 3

Step by step

The question

Question: find the plane through (1, 2, 3), (2, 3, 1), (0, 1, 2).

P(1, 2, 3), Q(2, 3, 1), R(0, 1, 2)

1 Two vectors in the plane

Draw the three points.

From P to Q: end minus start.

u = ⟨1, 1, −2⟩.

u = PQ = ⟨1, 1, −2⟩

From P to R: v = ⟨−1, −1, −1⟩.

Both lie in the plane.

v = PR = ⟨−1, −1, −1⟩

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, −2; −1, −1, −1|

Cover the i column.

Four numbers are left: 1, −2, −1, −1.

Write them as a small grid.

Its cross: down 1 times (−1), minus up −2 times (−1).

That is −1 minus 2: −3.

x part = |1, −2; −1, −1| = 1 × (−1) − −2 × (−1) = −3

Now cover the j column.

Four numbers are left: 1, −2, −1, −1.

Write them as a small grid, with a minus in front: the middle part always gets one.

Its cross: down 1 times (−1), minus up −2 times (−1).

−1 minus 2 is −3.

With the minus: 3.

y part = −|1, −2; −1, −1| = −(1 × (−1) − −2 × (−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, −1.

Write them as a small grid.

Its cross: down 1 times (−1), minus up 1 times (−1).

That is −1 minus (−1): 0.

z part = |1, 1; −1, −1| = 1 × (−1) − 1 × (−1) = 0

Put the three parts together: n = u × v = ⟨−3, 3, 0⟩.

n = u × v = ⟨−3, 3, 0⟩

Any arrow in the same direction works.

All parts can be divided by 3: n = ⟨−1, 1, 0⟩.

Smaller numbers are easier.

n = ⟨−1, 1, 0⟩

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.

−(x − 1) + (y − 2) = 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.

−x + 1 + y − 2 = 0

Add the plain numbers: they make −1.

−x + y − 1 = 0

Move the number to the right: add 1 on both sides.

So the plane is −x + y = 1.

−x + y = 1

5 Check

Let's check: put the point (1, 2, 3) in.

We get 1.

Right!

−(1) + (2) = 1

(2, 3, 1): 1 again.

Right!

−(2) + (3) = 1

(0, 1, 2): 1 again.

Right!

−(0) + (1) = 1

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.

Answer
−x + y = 1

Your turn

Try this one yourself, then check:

Find the plane through P(4, 1, 2), Q(2, 4, 2), R(0, 2, 2)
Show the answer
z = 2

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