# Question about Scalar Product of Vectors

1. Feb 22, 2008

### racer

Hello there

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. Feb 22, 2008

### Yoni

Hi,
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. Feb 22, 2008

### tiny-tim

Yes, if you use $$\Large\sin\theta$$, then v.v = 0 for any vector v (because sin0 = 0).

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. Feb 22, 2008

### qspeechc

Hey! Isn't that the cross-product! That's amazing! Why don't they teach us this stuff about vectors, instead of giving it in the just-so way that they do?

5. Feb 22, 2008

### tiny-tim

vector or scalar?

Hold it! The cross-product is a vector.
Do you think you can make a scalar product (in other words, a number) which has v.w = -w.v?

6. Feb 26, 2008

### manjuvenamma

Racer, I really didn't understand your question. There are so many well-written books on vector algebra and so much literature and help on the same subject is available on the internet. Dot product and cross product of two vectors are just two concepts of vector algebra. If you know this already and questioning why dot product is defined that way, the answer is application. All mathematical rules and definitions are made using some application in mind. Dot product is used for finding the component of one vector along the other vector, so cosine is right. We have other concepts like triple product etc which again are defined with application in mind.

7. Feb 26, 2008

### tiny-tim

Keep on doubting!

Racer, I did understand 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?)

8. Feb 26, 2008

### qspeechc

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

9. Feb 26, 2008

### tiny-tim

Yes, I was agreeing with you!!

But race originally asked about a scalar product, and it's interesting to work out whether that's possible with v.w = -w.v.

Erm … well … is it?

10. Feb 26, 2008

### qspeechc

No, it's not possible with scalars, unless either v or w are zero, since any number cannot equal its negative.(v.w = z, some other number)

11. Feb 26, 2008

### tiny-tim

not logical, captain …

That's missing the point - you might as well say "any vector cannot equal its negative", and thereby prove that we can't have v.w = -w.v for vectors!

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

So … ?

12. Feb 26, 2008

### qspeechc

No, but for scalars v.w = w.v, not necessarily so for vectors, since we must define vector multiplication in some manner, dropping certain properties of multiplication of scalars.

13. Feb 26, 2008

### qspeechc

Sorry, if we're discussing the scalar product, v.w, and w.v, the vectors are mapped to scalars, so v.w and w.v are scalars.
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. Feb 26, 2008

### John Creighto

15. Feb 26, 2008

### tiny-tim

sine scalar product works in 2 dimensions!

Racer was right: we can use the sine of the angle to form a scalar product of two vectors (though it needs v.w = -w.v).

But racer was only right in two dimensions.

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. Feb 27, 2008

### qspeechc

Sorry if I'm coming across as thick, but wont v.w and w.v be scalars, so for scalars, how can v.w = - w.v? Zero is trivial. I don't think I quite get your answer.

17. Feb 27, 2008

### tiny-tim

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. Feb 27, 2008

### qspeechc

Aha! Me thinks I gets it! Thanks a lot for going through all that trouble to explain it to stupid ol' me:tongue:

19. Feb 27, 2008

### tiny-tim

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

20. Mar 4, 2008

### manjuvenamma

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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?