Hi,

I know HOW to use a scalar product. However I don't know why do we need it and need a image of the result.

thanks,

Vinh

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- Dec 25th 2008, 05:22 AMvinh48scalar product
Hi,

I know HOW to use a scalar product. However I don't know why do we need it and need a image of the result.

thanks,

Vinh - Dec 25th 2008, 08:36 AMSoroban
Hello, vinh48!

Quote:

I know HOW to use a scalar product.

However I don't know why do we need it and need a image of the result.

As far as I know, the scalar product has no immediate meaning or image,

. . but it is used in a variety of formulas.

Given two vectors: .$\displaystyle \vec u\text{ and }\vec v$

. . $\displaystyle [1]\;\;\vec u \perp \vec v \:\text{ if and only if }\:\vec u\cdot\vec v$

. . $\displaystyle [2]\;\;\text{The angle between }\vec u\text{ and }\vec v\text{ is given by: }\cos\theta \:=\:\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}$

. . $\displaystyle [3]\;\;\text{The projection of }\vec u\text{ onto }\vec v\text{ is: }\;\text{proj}_{\vec v}\vec u \;=\;\frac{\vec u \cdot\vec v}{|\vec v|^2}\,\vec v$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I suggest that the scalar product is "similar" to a determinant.

A determinant has a value which, by itself, is meaningless.

However, it can be used in a variety of ways (e.g. Cramer's rule)

. . for a number of mathematical tasks.

- Dec 25th 2008, 08:43 AMvinh48
Thanks Soroban for your quick reply.

Yes, you just answered for the immediate meaning or image.

It is use if we want to have a projection of a vector on another one.

Can be use to calculate the shadow.

Vinh - Dec 25th 2008, 12:22 PMmr fantastic