Question

a x (b x a) . c ?
how to proceed?

Answer 1

is that how your problem starts? and is a*(b*a)*c the expression?

Answer 2

No. the expression is a x (b x a) . c

Answer 3

WTH????

Answer 4

and yes it starts like that. i am not sure what to do first: the dot product or the cross product.

Answer 5

In 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.

Answer 6

i am talking about dot and cross product
x = cross product
. = dot product

Answer 7

You 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?).

Answer 8

Once 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...

Answer 9

So (b x a) first. Say it equals Z. Then a x Z = W, say. Then W . c

Answer 10

see 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...

Answer 11

okay thanks. but can you please elaborate on tensor?

Answer 12

Have you ever seen the dot and cross product defined in this form:\[a.b \iff (a.b)_i=a_ib_i\]\[a \times b \iff (a \times b)_i =\epsilon_{ijk}a_jb_k\]?

Answer 13

no.

Answer 14

They're the tensor definitions for dot and cross product. If you haven't seen them, I wouldn't worry about it.

Answer 15

Is this for school or university?

Answer 16

university. freshman.

Answer 17

Okay. Just punch the vectors out the way you're used to for now.

Answer 18

but 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

Answer 19

Is there a full question, with words?

Answer 20

i tried using the assumed vectors but it gets messy

Answer 21

a 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)

Answer 22

okay. the question is prove or disprove the following:
a x (a x(a x b)).c = -|a|^2 a.b x c

Answer 23

Ah, okay, that's a full question...

Answer 24

yea. just give me a clue as to how to solve it.

Answer 25

In general we have
a x (b x c) = (a.c)b - (a.b)c

so try it slowly

Answer 26

yea i am doin it that way..

Answer 27

a x (b(a.a)-a(a.b)) . c what now?

Answer 28

can a.a be written as |a|^2 ?

Answer 29

Yes

Answer 30

the last part, i assume it is a.(bxc)

Answer 31

a x (b|a|^2 - a(a.b)) . c

Answer 32

wat to do abt a(a.b) ?

Answer 33

To 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

Answer 34

I'm glad *you* typed it out

Answer 35

hm.. ok thanks a lot !

Answer 36

just one more question. i want to practice these questions. can you suggest some website with answers?

Answer 37

Sorry, don't know of any. I'm sure you can find something, though. Schaum's Outlines are usually pretty good for practicing skills.