Please explain (visually would be appreciated) how $u{\times}v$ is orthogonal to both $u$ and $v$.

Question

$u{\times}v$ is orthogonal to both $u$ and $v$.

I tried to visualise the above, but couldn't do it. If $u{\times}v$ is orthogonal to both $u$ and $v$, then provided that the angle between $u$ and $v$ $\theta \neq 0$, then the triangle formed between $u$, $v$ and $u{\times}v$ would have a total internal angle value greater than $180^o$.

Isn't the vector $u \times v$ perpendicular to the plane that contains $u$ and $v?$ — green frog, Jul 03 '17 at 13:27
See https://en.wikipedia.org/wiki/Cross_product#/media/File:Cross_product_vector.svg — Widawensen, Jul 03 '17 at 13:30
There is no "triangle formed between $\vec u,\vec v$ and $\vec u\times \vec v$". These vectors do not lie in a single plane. — lulu, Jul 03 '17 at 13:33
This is related https://math.stackexchange.com/questions/420006/why-does-cross-product-give-a-vector-which-is-perpendicular-to-a-plane — futurebird, Jul 03 '17 at 13:34

score 2 · Answer 1 · answered Jul 03 '17 at 13:37

2

Try this: if $u$ is a vector on the ground, pointing North, and $v$ is a vector on the ground, pointing in some other direction, say southwest, from the same point, then $u\times v$ would be a vector pointing straight up to the zenith of the sky. Thus, it is perpendicular to both $u$ and $v$.

Does that visualization help?

answered Jul 03 '17 at 13:37

G Tony Jacobs

31,218

Thanks, that the vector was in a different plane and what I was missing. – Tobi Alafin Jul 03 '17 at 14:03

score 2 · Accepted Answer · answered Jul 03 '17 at 13:41

2

Hold you finger's like this:

Suppose you want to do $B \times I$. The angle between $B$ and $I$ does not have to be 90. It is that the vector that the cross product produces, is $90$ to the plane that B and I span. Notice that $F$ and $B$ are $90$ degrees apart, and same for $F$ and $I$.

Now consider this:

These are three dimensional planes, that you two vectors that you are crossing span. When you cross the two vectors, it produces a vector in another plane that is perpendicular ($90$ degrees) to the other two. The other two need not be perpendicular too. It's just that the result that you get is.

answered Jul 03 '17 at 13:41

K Split X

6,565

I don't think this answers the question. It was not "how" so much as "why" – futurebird Jul 03 '17 at 13:55
I mistakenly downvoted (I was trying to accept your answer) please edit so I can accept and upvote. – Tobi Alafin Jul 03 '17 at 14:02

Please explain (visually would be appreciated) how $u{\times}v$ is orthogonal to both $u$ and $v$.

2 Answers2