# Thread: Orthogonal Vectors (Calc. 3)

1. ## Orthogonal Vectors (Calc. 3)

The question is: Suppose the vectors (a+b) and (a-b) are orthogonal. What, if anything, can you conclude about a and b?

So far all I know is that since they are orthogonal, the dot product of (a+b).(a-b)=0. I'm not so sure what to do after that. I've been trying to substitute a=<a1,a2,a3> and b=<b1,b2,b3> to do the dot product and somehow get a.b=0 but no such luck. I don't even think that approach is right. Is there another way to approach this problem?

Thanks!!

2. ## Re: Orthogonal Vectors (Calc. 3)

You can also write (a+b).(a-b) as:

(a+b)·(a-b) = a·a - a·b + b·a - b·b .

Do you know if the dot product is commutative? That will help you to simplify the above expression.

3. ## Re: Orthogonal Vectors (Calc. 3)

Originally Posted by mishdo
The question is: Suppose the vectors (a+b) and (a-b) are orthogonal. What, if anything, can you conclude about a and b?
Can you show that
$\displaystyle (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\|\vec{a}\|-\|\vec{b}\|~?$

4. ## Re: Orthogonal Vectors (Calc. 3)

Originally Posted by Plato
Can you show that
$\displaystyle (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\|\vec{a}\|-\|\vec{b}\|~?$
I had a feeling it had to do with magnitude. I tried that out, but I'm a little stuck. So far I have:

$\displaystyle a1^2+a2^2+a3^2-b1^2-b2^2-b3^2=\|\vec{a}\|-\|\vec{b}\|$

I can see the magnitude formula in there, but can't figure out how to pull it out

5. ## Re: Orthogonal Vectors (Calc. 3)

Originally Posted by SammyS
You can also write (a+b).(a-b) as:

(a+b)·(a-b) = a·a - a·b + b·a - b·b .

Do you know if the dot product is commutative? That will help you to simplify the above expression.
Thanks, I tried it but I sadly didn't get anywhere

6. ## Re: Orthogonal Vectors (Calc. 3)

Originally Posted by mishdo
Thanks, I tried it but I sadly didn't get anywhere
Looking an SammyS's post # 2, the dot product there is commutative. That is,

$\displaystyle \mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a }.$

That leads to a simplification. Can you see it? That should lead you to be able to obtain Plato's result.