Calculus 3 introduction to cross product

[lonewolf219]

Just learning about vectors in 3 dimensions. Would it be correct to think of the cross product vector like a resultant vector? Is it similar to the displacement? If so, why is it always 90 degrees from both original vectors?

If anyone has some deeper insight into any of these vector topics I would appreciate it. Also, I'm wondering what the dot product is actually measuring. And what the relationship is beteen the cosine of dot product and the sine of cross product...

THANK YOU!

 

[Char. Limit]

Just learning about vectors in 3 dimensions. Would it be correct to think of the cross product vector like a resultant vector? Is it similar to the displacement? If so, why is it always 90 degrees from both original vectors?

If anyone has some deeper insight into any of these vector topics I would appreciate it. Also, I'm wondering what the dot product is actually measuring. And what the relationship is beteen the cosine of dot product and the sine of cross product...

THANK YOU!

In order...

No, the cross product is not a resultant vector, at least not as I understand the term. As I understand it, the resultant vector is a sum of two other vectors. Or more.

I'm not sure if it'd be similar to the displacement, myself. That depends on what you mean by displacement.

The cross product is always orthogonal to the two vectors being cross producted because that's how it's defined. As for what they're measuring...

The magnitude of the cross product can be thought of as measuring the area of the parallelogram defined by its two component vectors. Picture below:

As you can see, the cross product vector itself will always be orthogonal to that parallelogram.

 

[HallsofIvy]

Another way to define the cross product is to define
1) $\vec{i}\times\vec{j}= \vec{k}$
2) $\vec{j}\times\vec{k}= \vec{i}$
3) $\vec{k}\times\vec{i}= \vec{k}$

and defining it to be associative, distributive, and anti-commutative so
that $(a\vec{i}+ b\vec{j}+ c\vec{k})\times(u\vec{i}+ v\vec{j}+ w\vec{k})= a\vec{i}\times(u\vec{i}+ v\vec{j}+ w\vec{k})+ b\vec{j}\times(u\vec{i}+ v\vec{j}+ w\vec{k})+ c\vec{k}(u\vec{i}+ v\vec{j}+ w\vec{k})$

 

[A. Bahat]

Another way to think about it is that, while the (absolute value of the) dot product tells you "how parallel" two vectors are, the (magnitude of the) cross product tells you "how perpendicular" two vectors are.

 

[lonewolf219]

Thanks for the replies, guys! Hmm, the angle of how parallel and the angle of how perpendicular. That's interesting!

 

[Fredrik]

Another way to define the cross product is to define
1) $\vec{i}\times\vec{j}= \vec{k}$
2) $\vec{j}\times\vec{k}= \vec{i}$
3) $\vec{k}\times\vec{i}= \vec{k}$

and defining it to be [strike]associative[/strike], distributive, and anti-commutative so
that $(a\vec{i}+ b\vec{j}+ c\vec{k})\times(u\vec{i}+ v\vec{j}+ w\vec{k})= a\vec{i}\times(u\vec{i}+ v\vec{j}+ w\vec{k})+ b\vec{j}\times(u\vec{i}+ v\vec{j}+ w\vec{k})+ c\vec{k}(u\vec{i}+ v\vec{j}+ w\vec{k})$

There's one word too many in that sentence. The cross product isn't associative. It satisfies the Jacobi identity instead: a×(b×c)+b×(c×a)+c×(a×b)=0.