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**divinelogos** Problem:

Verify that the vectors <2,3,1>, <1,-1,0>, and <7,3,2> are coplanar.

Question:

Of course, the easiest way to do this is using the scalar triple product. At the time of my test, however, I was unaware of this. I attempted to find the angle between the vectors, and I got the result that the angle did not exist for any vectors I tried. So, if there is no angle between any of the vectors, does it follow that they are coplanar?

Call the three vectors $\displaystyle \displaystyle \vec{a},\ \vec{b},\ \vec{c}\ .$

Use the scalar product to find the angle between each pair of vectors.

The angle between $\displaystyle \displaystyle \vec{a}$ and $\displaystyle \displaystyle \vec{b}$ is: $\displaystyle \displaystyle \theta_{ab}=\cos^{-1}\left({{\vec{a} \cdot \vec{b}}\over{|\vec{a}|\ |\vec{b}|}}\right).$

Find $\displaystyle \displaystyle \theta_{bc}$ and $\displaystyle \displaystyle \theta_{ac}$ similarly.

If the vectors are co-planar, then one of the two following things will be true.

1. The sum of two of these angles is equal to the third.

2. The sum of all three angles will be 360° (or $\displaystyle \displaystyle 2\pi$ radians).

For the three given angles, #1 works.

You could do a similar process using the vector product. Then $\displaystyle \displaystyle \theta_{ab}=\sin^{-1}\left({{|\vec{a} \times \vec{b}|}\over{|\vec{a}|\ |\vec{b}|}}\right).$

In this case, criterion #2 would change to: The sum of the three angles being 180°, rather than 360°, would indicate that the vectors are co-planar.