# Is the cross-product of two displacement vectors orthogonal to both of them?

I have the point P,Q and R given. I calculate the displacement vectors PQ and PR. If I then compute their cross product I get a vector orthogonal to the plane they're in. But the value of the cross product vector is not orthogonal to PQ and PR (it is not 0). Shouldn't it be 0?

Thank you

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Let $u = PQ$, $v = PR$, and $w = u \times v$. Then $w$ is orthogonal to $u$ and $v$ (and normal to the plane which contains the points P, Q, and R). So the dot product of $w$ with both $u$ and $v$ should be zero. If it isn't, you've made a mistake. – Bill Cook Dec 14 '11 at 20:41

The cross product of two vectors is always orthogonal to both vectors. In three dimensions, a vector does not have to be zero to be orthogonal to two other vectors; as you said, the result is orthogonal to the plane that $P$, $Q$, and $R$ are in, so it is also orthogonal to both $PQ$ and $PR$.

Here's a slightly more technical discussion.

In general, a vector is only necessarily zero if it is orthogonal to everything. The dimension of a subspace and the dimension of its 'orthogonal complement' (which just means everything that is orthogonal to everything in the subspace) must add to the dimension of the whole space. For example, in 3 dimensional space, two vectors define a 2 dimensional subspace (as long as they're linearly independent, but you don't have to worry about what that means if you don't know). So everything orthogonal to that whole 2 dimensional subspace, or its orthogonal complement, is a 1 dimensional subspace.

This also explains why the cross product can only be used in 3 dimensional space. Two vectors only determine a 1 dimensional orthogonal complement, and therefore a unique cross product, if the original space is three dimensional. It turns out there isn't a really useful generalization of this to higher dimensions.

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it is only in $R^2$ 0 when it's orthogonal? How can I know then in $R^3$? – Andrew Dec 14 '11 at 20:50
@Andrew does this expanded discussion answer your question? – smackcrane Dec 15 '11 at 20:58
yes it does, thank you – Andrew Dec 18 '11 at 9:32

The Cross Product $v \times w$ is always orthogonal to both $v$ and $w$. This is easy to see by a direct calculation:

Write $v = \langle v_1, v_2 , v_3\rangle$ and $w = \langle w_1, w_2, w_3\rangle$. Then

$$v \times w = \langle v_2 w_3 - v_3 w_2, v_3w_1 - v_1w_3, v_1w_2 - v_2 w_1 \rangle$$

which implies

\begin{align} v \cdot (v \times w) &= v_1 (v_2 w_3 - v_3 w_2) + v_2 (v_3 w_1 - v_1 w_3) + v_3 (v_1 w_2 - v_2 w_1) \\ &= v_1 v_2 w_3 - v_1 v_3 w_2 + v_2 v_3 w_1 - v_1 v_2 w_3 + v_1 v_3 w_2 - v_2 v_3 w_1 = 0. \end{align} The same calculation works to show $w \cdot (v \times w) = 0$.

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