2
$\begingroup$

I'm wondering if there's a relation between the length of the projection (vector) of a vector $u$ onto a plane $P$ to which another vector $v$ is the normal vector and $\|v \times u\|$.

I was thinking that both $u \times v$ and $\operatorname{proj}_P(u)$ lie on the plane $P$, but what about their length?

I know the length of $u \times v$ should be the area of the parallelogram, but apart from this I'm not having any other ideas.

I also know that usually by normal vector we mean a unit vector, so the area of the parallelogram and thus of the cross product between $u$ and $v$ should simply be its height...

The reason I'm asking this question is because I'm reading a paper where it's stated that the projection of two vectors, say $u$ and $w$, onto a plane $P$, to which another vector $v$ is the normal vector, have the same length, which is equivalent to $\|u \times v\| - \|w \times v\| = 0$.

$\endgroup$
2
$\begingroup$

Let $\theta$ be the angle between $v$ and $u$. We can see that $\| v \times u\| = |\sin(\theta)|\|v \| \|u\|$ and $\|\operatorname{proj}_P (u)\| = |\cos(\pi/ 2 - \theta)| \|u \| = |\sin(\theta)| \|u\|$.

Diagram

As for the reason you were asking, if you could provide more context about what $u$ and $w$ are then it would be easier to help with that.

$\endgroup$
  • $\begingroup$ Actually, I had already arrived at this result in the meantime...but, maybe you didn't notice, but this result tells us that the length of the projection of $u$ onto $P$ is equal to the length of the cross product between $u$ and $v$ when $||v|| = 1$. $\endgroup$ – nbro Mar 24 '17 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.