Vector algebra question(s)

Hi MHF

This is my first post, I'm faaaairly certain this is in the right subforum, sorry if it isn't.

Anyway, I'm having difficulty with vector algebra, more specifically, proving one expression equals another

I have this question:

"Prove (**a** - **b**) x (**a** +** b**) = 2(**a** x **b**)"

Where **a** and **b** are both vectors and x= cross product

The problem is, I am getting confused over what rules I can use from traditional algebra (ie taking a and b as just variables) and the vector specific rules, (ie commutative laws, distributive laws, etc)

I have another one too:

" if (**c**-0.5**a**).**a** = (**c**-0.5**b**).**b** = 0 then, prove **c**-0.5(**a**+**b**) is perpendicular to **a - b**"

where . = dot product

(Sorry if that one is confusing, typing out 1/2 was a bit messy compared to just typing 0.5)

I know that vectors are perpendicular if they have a dot product of 0, but (again) I'm just quite confused about what rules I can use here.

Thanks very much

Re: Vector algebra question(s)

Quote:

Originally Posted by

**willsor** "Prove (**a** - **b**) x (**a** +** b**) = 2(**a** x **b**)"

Where **a** and **b** are both vectors and x= cross product

First, the cross product distributes over addition. So

$\displaystyle ( \bold{a} - \bold{b} ) \times ( \bold{a} + \bold{b} )$

$\displaystyle =~\bold{a} \times \bold{a} + \bold{a} \times \bold{b} - \bold{b} \times \bold{a} - \bold{b} \times \bold{b}$

Recall that $\displaystyle \left | \bold{x} \times \bold{y} \right | = xy~sin(\theta)$

What is the angle between vector **a** and itself?

-Dan

Re: Vector algebra question(s)

You can't simply treat vectors as variables (first off, variables are scalar). Also, cross products are anti-commutative, unlike regular numbers.

For the first question, you could try drawing two random vectors a and b, as well as a+b, a-b. The magnitude of the cross product of two vectors is the area of the parallelogram spanned by those vectors.

Edit: topsquark's solution is faster...I forgot cross products distribute over addition.

Re: Vector algebra question(s)

Quote:

Originally Posted by

**topsquark** First, the cross product distributes over addition. So

$\displaystyle ( \bold{a} - \bold{b} ) \times ( \bold{a} + \bold{b} )$

$\displaystyle =~\bold{a} \times \bold{a} + \bold{a} \times \bold{b} - \bold{b} \times \bold{a} - \bold{b} \times \bold{b}$

Recall that $\displaystyle \left | \bold{x} \times \bold{y} \right | = xy~sin(\theta)$

What is the angle between vector **a** and itself?

-Dan

so that works out to:

**a** x **a** + **a** x **b** - **b** x **a** - **b** x **b**

The first and last terms both work out to be 0?

Then the third term - **b** x **a** can be made to equal **a** x **b** under the "anti-commutative law", so you're left with 2 (**a** x **b**)

Is that correct?

Re: Vector algebra question(s)

Quote:

Originally Posted by

**willsor** " if (**c**-0.5**a**).**a** = [/SIZE][/SIZE] (**c**-0.5**b**).**b** = 0 then, prove **c**-0.5(**a**+**b**) is perpendicular to **a - b**"

where . = dot product

This one comes a bit harder. First let's take the top equation apart into two pieces.

$\displaystyle \left ( \bold{c} - \frac{1}{2}\bold{a} \right ) \cdot \bold{a} = \bold{c} \cdot \bold{a} - \frac{1}{2}a^2 = 0$

Again

$\displaystyle \left ( \bold{c} - \frac{1}{2}\bold{b} \right ) \cdot \bold{b} = \bold{c} \cdot \bold{b} - \frac{1}{2}b^2 = 0$

Keep these under your hat for a moment.

You must show that $\displaystyle \left ( \bold{c} - \frac{1}{2} ( \bold{a} + \bold{b} ) \right ) \cdot ( \bold{a} - \bold{b} ) = 0$

To go from here: Expand out this last line and you will find $\displaystyle \bold{c} \cdot \bold{a}$ and $\displaystyle \bold{c} \cdot \bold{b}$. Plug in the equations we showed above. Then simplify.

-Dan

Re: Vector algebra question(s)

Quote:

Originally Posted by

**willsor** so that works out to:

**a** x **a** + **a** x **b** - **b** x **a** - **b** x **b**

The first and last terms both work out to be 0?

Then the third term - **b** x **a** can be made to equal **a** x **b** under the "anti-commutative law", so you're left with 2 (**a** x **b**)

Is that correct?

That's it exactly. :)

-Dan

Re: Vector algebra question(s)

Quote:

Originally Posted by

**willsor** so that works out to:

**a** x **a** + **a** x **b** - **b** x **a** - **b** x **b**

The first and last terms both work out to be 0?

Then the third term - **b** x **a** can be made to equal **a** x **b** under the "anti-commutative law", so you're left with 2 (**a** x **b**) Is that correct?

That is certainly the way I would do.

Re: Vector algebra question(s)

Yay, I got the second one to come out too :)

Thank you so much for your help :)