# Relation between $\|\operatorname{proj}_P(u)\|$ and $\|u \times v\|$ where $v$ is the normal to $P$

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 $\|u \times v\| - \|w \times v\| = 0$.

## 1 Answer

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

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.

• 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 $||v|| = 1$. – nbro Mar 24 '17 at 22:55