The vector resolute.Why doesn't it cancel out?

[sameeralord]

Hello everyone ,

I'm reviewing vectors and I thought the best way to go about it is understand the exact definition without blindingly using formulas. Anyway here is my question

The vector resoulute of a on b

* - unit vector

(a . b*) b*

My question is why doesn't this cancel out to a like this.

a. b x b
----- ---
|b| |b|

|b|² = b.b

So wouldn't this be

a.b x b
----
b.b
which is equal to a

I know this doesn't happen but I think there is something wrong in my understanding. I can't understand what happens when two vectors are multiplied. Why does it give a scalar. How do you explain it. Can you multiply 3 vectors. Can you divide vectors?

It seems a lot of questions but I think by answering my first question you would probably answer these questions. Thanks a lot in advance

 

[tiny-tim]

Hello everyone ,

I'm reviewing vectors and I thought the best way to go about it is understand the exact definition without blindingly using formulas. Anyway here is my question

The vector resoulute of a on b

* - unit vector

(a . b*) b*

My question is why doesn't this cancel out to a like this.

a. b x b
----- ---
|b| |b|

|b|² = b.b

So wouldn't this be

a.b x b
----
b.b
which is equal to a

Hi sameeralord!

You're asking why doesn't (a.b)b/b² = a.

Because (a.b)b isn't a(b.b) …

the first one is a scalar multiple of b, and the second is a scalar multiple of a.

I know this doesn't happen but I think there is something wrong in my understanding. I can't understand what happens when two vectors are multiplied.
Why does it give a scalar. How do you explain it.
Can you multiply 3 vectors.
Can you divide vectors?

You can either dot-product two vectors, or cross-product them:

a.b and a x b.

the first is a scalar, the second is a vector.

Why? Because that's how we define them, and we can define anything we like.

You can't divide vectors unless one is a scalar times the other (in other words, they're parallel).

You can't multiply more than two vectors, except that you can keep cross-producting vectors in pairs, as many times as you like:

((axb)xc)xd etc,

and you can also do the scalar "triple product" (axb).c

 

[sameeralord]

Thanks heaps tiny tam Your answer was very clear. So basically you can't apply all the algebra techniques to vectors. I mean if there is (b.b) you take it as a scalar then do your calculations. So |b2| = (b.b) is also a scalar then. Thanks a lot again