Question

Prove: (more vectors!)

Answer 1

\[u/\left| u \right|^2-v/\left| v \right|^2=u/\left| u \right|\left| v \right|-v/\left| u \left| v \right| \right|\]

Answer 2

Evaluating the left side using the algebraic definition yields zero, but it's not so clear on the right

Answer 3

What does it mean if there is a backslash? O___O

Answer 4

/ means divide, or fraction

Answer 5

So divide the absolute value of "u" squared - "v" over the absolute value of "v" squared....and so on?

Answer 6

close; the extra ab bars makes it a bit confusing:

\[\frac{u}{|u|^2}-\frac{v}{|v|^2}=\frac{u}{|u|\ |v|}-\frac{v}{|u|\
|v|}\]

Answer 7

:D That's cool.

Answer 8

yea, so... would multiplying the magnitudes on the denominator on the right yield \[\sqrt (u _{1}v_{1}+u_{2}v_2+...+u_nv_n)\] ?

Answer 9

the right is a sum with common denoms; i see no reason to multiply

Answer 10

but the algebraic expansion of the magnitudes of the vectors would be what I just wrote, right?

Answer 11

maybe, my brains starting to shut down i think :)

\[\frac{u|v|^2-v|u|^2}{|u|^2|v|^2}=\frac{u-v}{|u|\ |v|}\]

Answer 12

oops I completely forgot to state part of the question

Answer 13

\[\frac{(|v|-|u|)}{|u||v|}=\frac{u-v}{|u|\ |v|}\]

Answer 14

It should be that the magnitude of the expression on the left should equal the magnitude of the expression on the right

Answer 15

i factored out a uv from the top; and canceled it with a uv vrom the bottom; and now they look at least more doable

Answer 16

its that subtraction up top that might get ya

Answer 17

good luck with it, i gotta go

Answer 18

this one turned out pretty cool. I'll post my ideas on here for those who are interested