wolfe8
  • wolfe8
Prove that (A × B)^2 + (A · B)^2 = A^2B^2 A and B are vectors. (Don't know how to notate them)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Oh OK ! Wait ...
wolfe8
  • wolfe8
So is it then somehow, (A × B)^2 is = 0?
anonymous
  • anonymous
yes I can be :) just when you put A=0 or B=o it will be :)

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anonymous
  • anonymous
sorry it can be not I can be XD
anonymous
  • anonymous
@E.ali , just please show him steps,and don't give the answer at once :)
anonymous
  • anonymous
OK ! @wolfe8 We have : (A.B)^2=(A.B).(A.B)
wolfe8
  • 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
  • anonymous
so you want examples?
wolfe8
  • wolfe8
Not really. Can you tell me if I just need to substitute in some properties in? Is that what I'm missing?
anonymous
  • anonymous
I think you don't need "substitute" at all!
wolfe8
  • wolfe8
I mean 'apply' then
anonymous
  • anonymous
Do u want start ?????!!!!!
anonymous
  • anonymous
maybe I can't understand you :D I can't understand that what exactly do you want!
anonymous
  • anonymous
the proof?
wolfe8
  • wolfe8
Ah sorry. Well yes I need the proof.
wolfe8
  • wolfe8
I'll be in the Mathematics chat
anonymous
  • anonymous
No problem I think:)
wolfe8
  • wolfe8
|dw:1378715095452:dw|
blockcolder
  • blockcolder
What does the square of a vector mean?
wolfe8
  • wolfe8
@blockcolder Exactly. I'm stuck at that.
anonymous
  • anonymous
I will google it for you :) wait...
anonymous
  • anonymous
@ganeshie8 , @Callisto , can you help us about square of a vector ?
anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
first of all A.B is dot product, which is a scalar. so question is bit wrong.
ganeshie8
  • 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
  • wolfe8
Well here is a snapshot of the question:
1 Attachment
ganeshie8
  • ganeshie8
ohk nvm im wrong square is making it scalar it seems
ganeshie8
  • 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
  • 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
  • wolfe8
One question: If I expand (ABsinθ)^2 will it be (AB)^2sin θ or A^2B^2 sinθ?
wolfe8
  • wolfe8
Oh I left the square for the sin
ganeshie8
  • ganeshie8
since there are no arrows on top of A, they're just scalars. AB = BA so. (AB)^2 = A^2B^2
ganeshie8
  • ganeshie8
\(\vec{A} \times \vec{B} = ||\vec{A}|| \ ||\vec {B}|| \sin \theta \vec{u}\) = \( AB\sin \theta \vec{u}\)
ganeshie8
  • ganeshie8
here, A and B are scalars
ganeshie8
  • ganeshie8
you can simplify them as u wud simplify any trig expression
wolfe8
  • 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
  • wolfe8
Ohhhh. I think I got it! Thanks everyone!
ganeshie8
  • 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
  • ganeshie8
but right hand side is a vector in ur question, so it is better you show it as vector
ganeshie8
  • ganeshie8
\(\large A^2B^2 \) \(\large (\vec{A} . \vec{A} ) (\vec{B} . \vec{B}) \) \(\large \vec{A}^2\vec{B}^2 \)
wolfe8
  • 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
1 Attachment
wolfe8
  • wolfe8
Well, I showed the wrong property...it's this one
1 Attachment
ganeshie8
  • 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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