- #1

- 39

- 0

When multiplying two vectors, why do we multiply them with the cosine of the angle between them? why not the sine of the angle?

Thanks.

- Thread starter racer
- Start date

- #1

- 39

- 0

When multiplying two vectors, why do we multiply them with the cosine of the angle between them? why not the sine of the angle?

Thanks.

- #2

- 65

- 1

When using cosine you actually multiply the length of one vector by the projection of the other vector on the first one. So when using cosine two perpendicular vectors would produce 0. And this is what the mathematicians wanted, a scalar that would describe the projection of one vector on the other. Notice that it doesn't matter which vector to project on the other, it is equivalent.

However, there is another kind of multiplication, with a vector product. This uses the sine function and would produce a maximum value for perpendicular vectors, and with that definition the order in which you multiply the vectors count.

- #3

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

So, for any two vectors v and w, (v + w)(v + w) = 0,

which means that v.v + w.w + v.w + w.v = 0,

so 0 + 0 + v.w + w.v = 0.

Since we want the scalar product not to depend on the order of the vectors (v.w = w.v), that would mean v.w = 0 for all vectors - in other words, every product is zero, which is pretty useless!

The only way out is to allow v.w = -w.v. Do you think that can work?

- #4

- 836

- 13

- #5

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Hold it! The cross-product is aHey! Isn't that the cross-product!

Do you think you can make a

- #6

- 102

- 0

- #7

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Racer, IRacer, I really didn't understand your question.

We don't have to accept everything at face value, and sometimes it helps to check something even if it's right.

Racer, just keep on doubting, and it can only strengthen your understanding! (gosh, that sounds a bit religious, doesn't it?)

- #8

- 836

- 13

Hold it! The cross-product is avector.

Do you think you can make ascalarproduct (in other words, a number) which has v.w = -w.v?

Well, wouldn't it work equally well if we were talking about vectors and cross products?

- #9

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Yes, I wasWell, wouldn't it work equally well if we were talking about vectors and cross products?

But race originally asked about a scalar product, and it's interesting to work out whether

Erm … well … is it?

- #10

- 836

- 13

- #11

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

That's missing the point - you might as well say "any

But, as you say, we can, and it's called the vector product.

So … ?

- #12

- 836

- 13

- #13

- 836

- 13

If we're talking about the cross product, it maps from vectors to vectors, so v.w and w.v map to vectors, not necessarily the same vectors, depending on how we define this mapping by vector product.

- #14

- 488

- 2

http://en.wikipedia.org/wiki/Dot_product#Proof_of_the_geometric_interpretation

- #15

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

Racer was right: wearescalars.

If we're talking about the cross product, it maps from vectors to vectors, so v.w and w.v map to vectors, not necessarily the same vectors, depending on how we define this mapping by vector product.

But racer was only right

In three or more directions, it doesn't work, basically because there's no systematic way of deciding whether the sine is positive or negative.

(In two dimensions, that isn't a problem, because you can always choose the clockwise direction from v to w to define sine(vw).)

- #16

- 836

- 13

- #17

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

I think you're being misled by the fact that we've been writing v.w, which looks like ordinary multiplication (for which of course we can't have v.w = -w.v).

Suppose we call it v∆w.

Then there's no reason we shouldn't have v∆w a scalar (an ordinary number), with w∆v its negative: v∆w = -w∆v.

In fact, in ordinary two-dimensional Euclidean space, we can define v∆w (or v.w) to be the magnitude of the ordinary right-handed cross product vxw - that works fine, and does have ∆w = -w∆v (and it also uses the sine, as originally asked).

- #18

- 836

- 13

- #19

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

You're very welcome, qspeechc (and obviously interested)!

- #20

- 102

- 0

------------------------------------------------------------------------------Racer, Ididunderstand your question, and I thought it was a perfectly sensible one!

We don't have to accept everything at face value, and sometimes it helps to check something even if it's right.

Racer, just keep on doubting, and it can only strengthen your understanding! (gosh, that sounds a bit religious, doesn't it?)

We should doubt definitions, too? Dot product was defined with cosine with an application in mind - that of finding component of one vector along another vector. That's all. New functions can be defined if we find other interesting applications for them.

Why is cosine defined adjacent side/hypotenuse?

- #21

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

[size=-2](goodness, that was 13 posts ago!)[/size]We should doubt definitions, too?

Absolutely!

Definitions can be improved, or can have alternatives, or be given a better name!

Anyway, Racer was actually right:

He obviously checked it out in two dimensions, found that it worked, and came to us for a second opinion!Racer was right: wecanuse the sine of the angle to form a scalar product of two vectors (though it needs v.w = -w.v).

But racer was only rightin two dimensions.

Should make a good PhD student!

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 832

- Replies
- 3

- Views
- 811

- Replies
- 5

- Views
- 7K

- Replies
- 4

- Views
- 955

- Last Post

- Replies
- 13

- Views
- 928

- Last Post

- Replies
- 9

- Views
- 788

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 493

- Last Post

- Replies
- 3

- Views
- 604