# Perpendicular Vectors

1. Dec 29, 2009

### thomas49th

1. The problem statement, all variables and given/known data
If a and b are perpendicular, simplify $$(a - 2b) \cdot (3a + 5b)$$

3. The attempt at a solution

Not really sure what they're asking, but to be perp. the angle between them is 90. So using the scalar

so a.b = 0??

The answer in the book it (3a² - 10b²) and I can see that they just multiplied the as and bs together, but why?

Thanks
Thomas

2. Dec 29, 2009

### rock.freak667

You are to expand it as follows $3a \cdot (a - 2b)+5b \cdot(a-2b)$ and then use the fact that a.b=0

3. Dec 29, 2009

### tiny-tim

Hi Thomas!

Yes, a.b = 0 …

so just expand (a-2b).(3a+5b) in the usual way, and use a.b = 0

4. Dec 30, 2009

### thomas49th

oh it's really easy ;) I was thinking that the each of the brackets were vectors when of course a and b are the vectors. I'm a silly billy.

Use the definition of the scalar product to show that if two vectors are perpendicular, their scalar product is zero.

well a.b = |a||b|Cosx

and when x = 90° => cos x = 0 and that means

a.b = |a||b|.0
a.b = 0

but that's a bit easy isn't it?

Thanks

5. Dec 30, 2009

### rock.freak667

yes that would be correct

6. Dec 30, 2009

### Staff: Mentor

Each of the quantities inside parentheses is a vector, made up of scalar multiples of a and b, which are themselves vectors.