Let $(V,\langle,\rangle)$ be the $\mathbb R^3$ with the standard bilinear-form and let $W \subset V$ be a two dimensional spanning set given by $v = (x_1,x_2,x_3)$ and $w = (y_1,y_2,y_3)$ and the cross product $v \times w := (x_2y_3-x_3y_2,x_3y_1-x_1y_3,x_1y_2-x_2y_1)$.

Why is the cross product $v \times w$ contained in the orthogonal complement $W^\perp$?

I know that the orthogonal complement contains all vectors that are orthogonal to both input vectors so I know it is valid, but I cannot prove it.


You can just compute that $v\times w$ is orthogonal to both $v$ and $w$, and hence it is orthogonal to the subspace $W$ generated by $v$ and $w$. This means it lies in the orthogonal complement of $W$.

There are a lot of ways to show $v\times w$ is orthogonal to both $v$ and $w$. If you have already proven the scalar triple product identity you could consider using that:

$(v\times w)\cdot w=(w\times w)\cdot v = 0$. etc.

If you are just very desperate for a proof using coordinates, you can just compute $(v\times w)\cdot w$ and $(v\times w)\cdot v$ directly from the coordinates.

  • $\begingroup$ Thank you for your response, so I just have to show that the components of the cross-product are orthogonal to the cross-product itself as it is here proofwiki.org/wiki/Cross_Product_is_Orthogonal_to_Factors? $\endgroup$ – sj134 Apr 14 '14 at 12:52
  • $\begingroup$ @sj134 I think it's a good idea to call $v$ and $w$ factors, but a bad idea to call them "components." The components are the $x_i$'s and the $y_i$'s as you wrote in your post. That link you provided could be very helpful to you. Do you see why being orthogonal to $v$ and $w$ implies it's orthogonal to the span of $v$ and $w$? $\endgroup$ – rschwieb Apr 14 '14 at 12:58
  • $\begingroup$ Yes because it means that it is then orthogonal to the whole subspace $\endgroup$ – sj134 Apr 14 '14 at 13:01
  • $\begingroup$ @sj134 Hm, yes, but I was asking if you know why that is true. Just let me know if I can help with anything else. $\endgroup$ – rschwieb Apr 14 '14 at 13:04
  • $\begingroup$ I have one other open question, you could help there, I dont get responses... $\endgroup$ – sj134 Apr 14 '14 at 13:07

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