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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?

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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).

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