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anonymous
 5 years ago
a x (b x a) . c ?
how to proceed?
anonymous
 5 years ago
a x (b x a) . c ? how to proceed?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that how your problem starts? and is a*(b*a)*c the expression?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. the expression is a x (b x a) . c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and yes it starts like that. i am not sure what to do first: the dot product or the cross product.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In algebra, the dot and the cross express multiplication. In don't get it. What I can see there is a simple product. So it will result in (a*a)bc.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am talking about dot and cross product x = cross product . = dot product

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to take the cross product first. You don't know what (b x a) is yet, so you can't take the dot product (unless you know about tensors...do you?).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Once you have that result, you have to take the cross product with a (but keep the order of the vectors). You can't dot product until the end because the dot product punches out a scalar...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So (b x a) first. Say it equals Z. Then a x Z = W, say. Then W . c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see here a,b,c are indeed vectors otherwise there is no meaning of . and X. here obviously u have to do X product first. if you dont do that then dot product will give you a scaler and you can't make X product of a vector with scaler...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay thanks. but can you please elaborate on tensor?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you ever seen the dot and cross product defined in this form:\[a.b \iff (a.b)_i=a_ib_i\]\[a \times b \iff (a \times b)_i =\epsilon_{ijk}a_jb_k\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They're the tensor definitions for dot and cross product. If you haven't seen them, I wouldn't worry about it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this for school or university?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0university. freshman.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. Just punch the vectors out the way you're used to for now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the thing is we are just given the vectors as a and b we have to prove or disprove a result unless i know how to go about it i would not be able to

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is there a full question, with words?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i tried using the assumed vectors but it gets messy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a x (b x a) = (a.a)b  (a.b)a so [a x (b x a)].c = [(a.a)b  (a.b)a].c = (a.a)(b.c)  (a.b)(a.c)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. the question is prove or disprove the following: a x (a x(a x b)).c = a^2 a.b x c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, okay, that's a full question...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea. just give me a clue as to how to solve it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In general we have a x (b x c) = (a.c)b  (a.b)c so try it slowly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i am doin it that way..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a x (b(a.a)a(a.b)) . c what now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can a.a be written as a^2 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the last part, i assume it is a.(bxc)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a x (ba^2  a(a.b)) . c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wat to do abt a(a.b) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To solve this, use a x (b x c) = (a.c)b  (a.b)c and the scalar triple product a.(bxc) = (axb).c [a x (a x (a x b))].c = [(a.(axb)) a  (a.a)(axb)].c =[a.(axb)](a.c)  (a.a)(axb).c =[(axa).b](a.c)  a^2[a.(bxc)] = a^2[a.(bxc)] because axa =0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm glad *you* typed it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hm.. ok thanks a lot !

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just one more question. i want to practice these questions. can you suggest some website with answers?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, don't know of any. I'm sure you can find something, though. Schaum's Outlines are usually pretty good for practicing skills.
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