Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following diagram: enter image description here

I want to find the normal vector for the polygon of points $abc$ and the plane highlighted in red with the points $bcde$.

To find the normal vector for the polygon of the the points $abc$, what I did was to find the vector $\vec{ac}=oc-oa$ and $\vec{ab}=ob-oa$, where $o$ is the origin, and cross them together. So it turns out to be $ac \times ab = \begin{bmatrix} 2\\ -2\\ 2 \end{bmatrix} \times \begin{bmatrix} 3\\ -3\\ -3 \end{bmatrix}= \begin{bmatrix} 12\\ 12 \\ 0 \end{bmatrix}$.

And for the plane highlighted in red wit hthe points $bcde$, I used the same method by finding any two vectors, $\vec{ed}$ and $\vec{ab}$ and cross them together. So, $\vec{ed} \times \vec{ab}=\begin{bmatrix} 0\\ 0\\ 2 \end{bmatrix} \times \begin{bmatrix} 3\\ -3\\ -3 \end{bmatrix}= \begin{bmatrix} 6\\ 6 \\ 0 \end{bmatrix}$.

But then now, by looking at the picture, how could the 2 planes have the same normal vector when they are so off in their own direction? What have I done wrong? Is what I have done finding the right normal vectors in the first place?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

The vector $\vec{ab}$ does not lie in the "red plane" defined by bcde. Try instead: $\vec{ed} \times \vec{dc}$

Also, be careful about whether your normals point inward or outward. The first vector you computed points somewhat upward (out of the polygon).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.