Prove that (A × B)^2 + (A · B)^2 = A^2B^2
A and B are vectors. (Don't know how to notate them)

- wolfe8

Prove that (A × B)^2 + (A · B)^2 = A^2B^2
A and B are vectors. (Don't know how to notate them)

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- katieb

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- anonymous

Oh OK !
Wait ...

- wolfe8

So is it then somehow, (A × B)^2 is = 0?

- anonymous

yes I can be :) just when you put A=0 or B=o it will be :)

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- anonymous

sorry it can be not I can be XD

- anonymous

@E.ali , just please show him steps,and don't give the answer at once :)

- anonymous

OK !
@wolfe8
We have :
(A.B)^2=(A.B).(A.B)

- wolfe8

Well I can technically show it with examples step-by-step, but the question is only worth 3 points. So I'm thinking I must be missing some key identity or something. And I'd really like to know which one.

- anonymous

so you want examples?

- wolfe8

Not really. Can you tell me if I just need to substitute in some properties in? Is that what I'm missing?

- anonymous

I think you don't need "substitute" at all!

- wolfe8

I mean 'apply' then

- anonymous

Do u want start ?????!!!!!

- anonymous

maybe I can't understand you :D I can't understand that what exactly do you want!

- anonymous

the proof?

- wolfe8

Ah sorry. Well yes I need the proof.

- wolfe8

I'll be in the Mathematics chat

- anonymous

No problem I think:)

- wolfe8

|dw:1378715095452:dw|

- blockcolder

What does the square of a vector mean?

- wolfe8

@blockcolder Exactly. I'm stuck at that.

- anonymous

I will google it for you :) wait...

- anonymous

@ganeshie8 , @Callisto , can you help us about square of a vector ?

- anonymous

@ganeshie8

- ganeshie8

first of all A.B is dot product, which is a scalar. so question is bit wrong.

- ganeshie8

question must be like this :-
\(\large (\vec{A} \times \vec{B})^2 + (\vec{A} \vec{B})^2 = \vec{A}^2 \vec{B}^2\)

- wolfe8

Well here is a snapshot of the question:

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- ganeshie8

ohk nvm im wrong square is making it scalar it seems

- ganeshie8

square of a vector is defined to be the dot product,
\(\vec{A}^2 = \vec{A} . \vec{A}\)
all terms in the given equation become scalars, so its ok :)

- ganeshie8

here is a lil proof :-
\(\large (\vec{A} \times \vec{B})^2 + (\vec{A} . \vec{B})^2 = \vec{A}^2 \vec{B}^2 \)
take left side,
\(\large (\vec{A} \times \vec{B})^2 + (\vec{A} . \vec{B})^2 \)
\(\large (AB \sin \theta)^2 + (AB \cos \theta)^2 \)
simplify

- wolfe8

One question: If I expand (ABsinθ)^2 will it be (AB)^2sin θ or A^2B^2 sinθ?

- wolfe8

Oh I left the square for the sin

- ganeshie8

since there are no arrows on top of A, they're just scalars. AB = BA
so. (AB)^2 = A^2B^2

- ganeshie8

\(\vec{A} \times \vec{B} = ||\vec{A}|| \ ||\vec {B}|| \sin \theta \vec{u}\)
= \( AB\sin \theta \vec{u}\)

- ganeshie8

here, A and B are scalars

- ganeshie8

you can simplify them as u wud simplify any trig expression

- wolfe8

Alright I think I got the left side. I get (AB)^2
Now for the right side, I made it so it is (A.A) . (B.B) A and B vectors. And then?

- wolfe8

Ohhhh. I think I got it! Thanks everyone!

- ganeshie8

\(\large (\vec{A} \times \vec{B})^2 + (\vec{A} . \vec{B})^2 = \vec{A}^2 \vec{B}^2 \)
take left side
\(\large (\vec{A} \times \vec{B})^2 + (\vec{A} . \vec{B})^2 \)
\(\large (\vec{A} \times \vec{B}) . (\vec{A} \times \vec{B}) + (||\vec{A}||\ ||\vec{B}|| \cos \theta)^2 \)
\(\large (||\vec{A}||\ ||\vec{B}|| \sin \theta \vec{u}) . (||\vec{A}||\ ||\vec{B}|| \sin \theta \vec{u}) + (||\vec{A}||\ ||\vec{B}|| \cos \theta)^2 \)
\(\large (AB \sin \theta \vec{u}) . (AB \sin \theta \vec{u}) + (AB \cos \theta)^2 \)
\(\large (A^2B^2 \sin^2 \theta) \vec{u} . \vec{u} . + A^2B^2 \cos^2 \theta \)
\(\large (A^2B^2 \sin^2 \theta) . + A^2B^2 \cos^2 \theta \)
\(\large (A^2B^2 \)

- ganeshie8

but right hand side is a vector in ur question, so it is better you show it as vector

- ganeshie8

\(\large A^2B^2 \)
\(\large (\vec{A} . \vec{A} ) (\vec{B} . \vec{B}) \)
\(\large \vec{A}^2\vec{B}^2 \)

- wolfe8

Yes but (picture) so I end up with scalars in the end. I think I got it from here. Thanks! I am going to close this question now :D

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- wolfe8

Well, I showed the wrong property...it's this one

##### 1 Attachment

- ganeshie8

okay good :)
square of a vector is defined as, dot product of a vector and itself.
\(\large \vec{a} . \vec{a} = ||\vec{a}|| \ ||\vec{a}|| \cos 0 = ||\vec a ||^2 = a^2\)

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