Solving Perpendicular Vectors: (a - 2b) \cdot (3a + 5b) Explanation and Solution

[thomas49th]

Homework Statement

If a and b are perpendicular, simplify (a - 2b) \cdot (3a + 5b)

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

 

[rock.freak667]

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

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

 

[tiny-tim]

so a.b = 0??

Hi Thomas!

Yes, a.b = 0 …

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

 

[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.

How about this one

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

 

[rock.freak667]

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.

How about this one

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

yes that would be correct

 

[Mark44]

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.

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

How about this one

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