Plane through (0, 3, -2) with normal <5, -2, 1> Step by Step

Step by step

Question: find the plane through (0, 3, −2) with normal vector n = ⟨5, −2, 1⟩.

(0, 3, −2) , ⟨5, −2, 1⟩

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.

5x − 2(y − 3) + (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.

5x − 2y + 6 + z + 2 = 0

Add the plain numbers: they make 8.

5x − 2y + z + 8 = 0

Move the number to the right: take away 8 on both sides.

So the plane is 5x − 2y + z = −8.

5x − 2y + z = −8

Let's check: put the point (0, 3, −2) in.

We get −8.

Right!

5(0) − 2(3) + (−2) = −8

Let's look back.

First: draw it.

Then: write the plane.

Then: open the brackets.

Last: check.

Answer

5x − 2y + z = −8