# Questions about Dot Product (Vectors)

• Mar 9th 2011, 10:14 AM
StuckOnVectors
This is my first post here so I'm sorry if this isn't the right section (didn't see a vectors subforum). I am also not sure how to put arrows on top of letters, so assume that every letter in the following question is a vector. I will also use | | for magnitude.

1. |a| = 3 , |b| = 2 , the angle in between these vectors is 60
Determine the numerical value of (3a + 2b) dot (4a - 3b). - In order to do this I tried distributing.

I know that (3a dot 4a) = 12|a|² and (2b dot -3b) = -6|b|²
I however don't know what to do in the case of (3a dot -3b) and (2b dot 4a)

The formula given is ( |a||b|cosθ = a dot b ) and the final answer is 81 as listed in the back of the book.

2. A regular hexagon has sides of 3cm, as shown below. Determine a dot b.

http://i180.photobucket.com/albums/x...onquestion.png

This is my drawing - I slid b over resulting in an angle of 120 between a and b.
I used the above formula to yield: |3||3|cos120 = -4.5

The answer in the back of the book is however 4.5 (positive, not negative). This has also occurred in other questions where I am getting negative answers where they should be positive - is there something I'm not doing right?

All help would be greatly appreciated.
• Mar 9th 2011, 10:23 AM
Plato
Quote:

Originally Posted by StuckOnVectors
1. |a| = 3 , |b| = 2 , the angle in between these vectors is 60
Determine the numerical value of (3a + 2b) dot (4a - 3b). - In order to do this I tried distributing.

I know that (3a dot 4a) = 12|a|² and (2b dot -3b) = -6|b|²
I however don't know what to do in the case of (3a dot -3b) and (2b dot 4a)

The formula given is ( |a||b|cosθ = a dot b ) and the final answer is 81 as listed in the back of the book.

$\displaystyle \frac{a\cdot b}{\|a\|\|b\|}=\cos(60^o)=\frac{1}{2}$

$\displaystyle \alpha a\cdot \beta b=(\alpha\beta)a\cdot b$

$\displaystyle a\cdot a=\|a\|^2$
• Mar 9th 2011, 10:32 AM
TheChaz
So, roughly speaking (!), we have $\displaystyle 12a^2 - ab - 6b^2$
after distributing.
"a" = 3, "b" = 2, and "ab" = 3 by following Plato's computation.
• Mar 9th 2011, 10:57 AM
StuckOnVectors
I'm still just not understanding.. How did you come up with -ab = -3? I know that you have to use the formula, I just don't get how.
• Mar 9th 2011, 12:46 PM
earboth
Quote:

Originally Posted by StuckOnVectors

2. A regular hexagon has sides of 3cm, as shown below. Determine a dot b.

This is my drawing - I slid b over resulting in an angle of 120 between a and b.
I used the above formula to yield: |3||3|cos120 = -4.5

The answer in the back of the book is however 4.5 (positive, not negative). This has also occurred in other questions where I am getting negative answers where they should be positive - is there something I'm not doing right?

All help would be greatly appreciated.

The vectors including an angle had to be placed tail to tail and not - as you did - head to tail. Or in other words: The arrows representing vectors had to start at the vertex of the angle.
• Mar 9th 2011, 05:08 PM
StuckOnVectors
earboth thank you for the help, I see where I went wrong with the hexagon

thechaz the problem was that I didn't know how to get to $\displaystyle 12a^2 - a \cdot b - 6b^2$, but i figured it out

anyone reading this: $\displaystyle 3a \cdot -3b$ is the same thing as $\displaystyle -9a \cdot b$ - not knowing this was my problem