Plane through (3, 1, 2) perpendicular to <1, -1, 2> Step by Step

Step by step

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

(3, 1, 2) , ⟨1, −1, 2⟩

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 − 3) − (y − 1) + 2(z − 2) = 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 − 3 − y + 1 + 2z − 4 = 0

Add the plain numbers: they make −6.

x − y + 2z − 6 = 0

Move the number to the right: add 6 on both sides.

So the plane is x − y + 2z = 6.

x − y + 2z = 6

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

We get 6.

Right!

(3) − (1) + 2(2) = 6

Let's look back.

First: draw it.

Then: write the plane.

Then: open the brackets.

Last: check.

Answer

x − y + 2z = 6