Plane through (2, 1, -1) with normal <-1, -2, 3> Step by Step

Step by step

Question: find the plane through (2, 1, −1) with normal vector n = ⟨−1, −2, 3⟩.

(2, 1, −1) , ⟨−1, −2, 3⟩

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.

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) − 2(y − 1) + 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

What is in front of a bracket multiplies every part inside.

A minus in front changes the signs inside.

−x + 2 − 2y + 2 + 3z + 3 = 0

Add the plain numbers: they make 7.

−x − 2y + 3z + 7 = 0

Move the number to the right: take away 7 on both sides.

So the plane is −x − 2y + 3z = −7.

−x − 2y + 3z = −7

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

We get −7.

Right!

−(2) − 2(1) + 3(−1) = −7

Let's look back.

First: draw it.

Then: write the plane.

Then: open the brackets.

Last: check.

Answer

−x − 2y + 3z = −7