# Homework Help: Dot product

1. Sep 13, 2006

### mbrmbrg

Nice, simple, straighforward problem:
$j \cdot {-j}$

Difficulty: I solved it two ways and got two answers.

$\vec{a} \cdot \vec{b}=ab\cos\phi$

So in this case, I should get $$j \cdot {-j}=(1)(-1)(cos(180^o))=(1)(-1)(-1)=1[/itex] However, using the formula $\vec{a} \cdot \vec{b}=a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}$, I got $j \cdot {-j}=a_{y}b_{y}=(1)(-1)=-1$ Can anyone tell me why? Last edited: Sep 13, 2006 2. Sep 13, 2006 ### FrogPad Recall that the dot product is: [tex] \vec A \cdot \vec B = |\vec A| |\vec B| \cos \theta_{AB}$$

let $\vec A = (0,-1,0)$
$$A=|\vec A| = \sqrt{(0)^2 +(-1)^2 +(0)^2}$$

What is $(-1)^2$ :)

3. Sep 13, 2006

### mbrmbrg

Neither my book nor my professor mentions the absolute value thing. I guess that's completely implied, though, because the negative sign has to do with direction, not magnitude.

Thank you so much!

4. Sep 13, 2006

### Astronuc

Staff Emeritus
$$|\vec A|$$ means magnitude of $$\vec A$$, which is a scalar and is given by the square root of the sum of the squares of the magnitudes of the component vectors. $$|\vec{a}|\,=\,\sqrt{a^2_x+a^2_y+a^2_z}$$

One could also use the fact that -j * j = - (j * j) = - (1) = -1, where I use * to mean the dot product.

Presumable j is a unit vector (0, 1, 0).

Back to the OP:

$$j \cdot {-j}=(1)(1)(cos(180^o))=(1)(1)(-1)=-1$$

Last edited: Sep 13, 2006
5. Sep 13, 2006

### professorlucky

I think basically your mistake lies in your calculation of the modulus(magnitude) of -j which is 1 not -1.For use in the formula a*b=abcos#
where * means dot product
a/b mean the magnitudes(moduli) of a & b

6. Sep 13, 2006

### mbrmbrg

7. Sep 13, 2006

The dot product operation has an associative property, that says:

$$(k \vec A) \cdot \vec B = k (\vec A \cdot \vec B)$$
Where $k$ is a scalar, and $\vec A, \,\, \vec B$ are vectors.

Last edited: Sep 13, 2006
8. Sep 13, 2006

### mbrmbrg

Riiiiiight....
I've got to start consistently thinking of a negative sign as -1 multiplied by what follows, don't I.
Thanks!

9. Sep 14, 2006