# Thread: Finding the normal of a plane using the cross product.

1. ## Finding the normal of a plane using the cross product.

Hi all. I have attached the question and worked solution. I understand the steps for finding the normal. First, find two vectors on the plane, then using the vectors P1P2 x P1P3 to find the normal. After the normal is found the Cartesian equation can be obtained. What I need some clarification on is whether or not the inverse of the normal vector is still correct. In the worked solutions you can see the vector PQ x PR was used by the author, whereas when I did the question I used QP x QR. As a result, I obtained the inverse equation (35i, -170j, -180k). Is my answer still correct, or did I go wrong somewhere?

I should also add I know order matters when taking the cross product (reversing order reverses the result), though I'm not sure if that's my error or if my error is in some other step.

Any help would be appreciated.

2. ## Re: Finding the normal of a plane using the cross product.

The question did ask for an equation. Did you arrive at
7x - 14y -36z = -124 (?)

If so, that's correct. Same equation in another form. Just multiply through by -1.

Dividing through by 5 as your answer key suggests makes it look better, but doesn't change the result. I guess it depends on your professor's taste whether he marks you down.

3. ## Re: Finding the normal of a plane using the cross product.

Originally Posted by nalo6451
Hi all. I have attached the question and worked solution. I understand the steps for finding the normal. First, find two vectors on the plane, then using the vectors P1P2 x P1P3 to find the normal. After the normal is found the Cartesian equation can be obtained. What I need some clarification on is whether or not the inverse of the normal vector is still correct. ?? In the worked solutions you can see the vector PQ x PR was used by the author, whereas when I did the question I used QP x QR. As a result, I obtained the inverse equation (35i, -170j, -180k). Is my answer still correct, or did I go wrong somewhere?
I should also add I know order matters when taking the cross product (reversing order reverses the result), though I'm not sure if that's my error or if my error is in some other step.
By inverse do you mean the additive inverse i.e. the negative?
If $\vec{\bf{~N}}$ is the normal of a plane then so is $\vec{\bf{-N}}$

4. ## Re: Finding the normal of a plane using the cross product.

If $\vec{N}\cdot\vec{v}= 0$ then $(-\vec{N})\cdot\vec{v}= -(\vec{N}\cdot\vec{v})= -0= 0$.