Plane through (1, 2, 3) with normal <2, -1, 4> Step by Step

Step by step

Question: find the plane through (1, 2, 3) with normal vector n = ⟨2, −1, 4⟩.

(1, 2, 3) , ⟨2, −1, 4⟩

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.

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

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

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.

Answer

2x − y + 4z = 12