# 3D | coplanar vectors

## Main Question or Discussion Point

I dont have the question with me so I ll just describe it to you. It was a test question.

I was given 3 vectors $$\vec u$$, $$\vec v$$, $$\vec w$$ and $$\vec w$$ had an unknown parameter p in it.

We were supposed to calculate a value for p such that the three vectors are coplanar.

What I did was find the cross product of $$\vec u$$and $$\vec v$$ and then take the scalar product of the product of $$\vec {uv}$$ and $$\vec w$$ and make it equal to 0 and solve for p, i.e. $$(\vec{u} \times \vec{v}).\vec w=0$$
Now apparently this was the correct answer but I do not understand how it proves that they are coplanar. Because scalar product just shows that they are perpendicular and thus the vector is parallel to the plane. But cant it just as well be below or above the plane ? Last edited:

Basically in a linear space all the vectors are starting at the same origin 0, (the components just indicate the lenght in every direction), so that the plane is containing 0 and the vector w is in your case parallel to the plane and also passing in 0, so it's completely in the plane. (0 has to be contained in every subspace, since it is in the definition of a vector space)

Working with affine space allow every vector to have a different origin.

robphy
Homework Helper
Gold Member
The scalar triple product (u x v) dot w can be interpreted as a determinant.
The determinant can be regarded as the signed volume of a parallelepiped ("a box with parallel sides") with sides given by the sequence of row vectors. If that volume is zero, then it's as if the box were flat. That is, the row vectors lie on a common plane.

I see. Now that makes sense. Thx a lot. :)
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Also, the cross product produces a vector perpendicular to both vectors. If the dot product of another vector with that cross product is zero, the vector is also orthogonal to the cross product, and hence coplanar with the first two vectors.

Just another way to think about it.