Question

Show that if u + v and u - v are perpendicular, then the vectors u and v must have the same length.

Answer 1

now how you prove it =0 me clueless:

Answer 2

Could probably do this with geometry. Do you have to use vector properties in the proof?

Answer 3

This problem is under vector. so im not really sure if geometry is used

Answer 4

im guessing vector properties, like a dot b, magnitude a, etc etc idk

Answer 5

It can be done with geometry of isosceles triangles in a couple steps. If you want to show off your vector skills start with what it means for vectors to be perpendicular: their dot product = zero.

Answer 6

I cant do it with out numbers =/ cant even start by staring at u and v.. head spins

Answer 7

If u+v is perpendicular to u-v then (u+v)*(u-v)=0.

Answer 8

I'm pretty sure you can treat them as regular algebraic variables since you're given that u and v themselves are perpendicular. Solve it by reducing it to the form u=v.

Answer 9

nvm lost my self lol

Answer 10

let me know if you came up with something : =)

Answer 11

brb; worked all after noon on math VECTORS!- heading to get food to fill my tummy.

Answer 12

It's not difficult to show algebraically:
We define the Norm of a vector v to be
\[N(\vec v)=|\vec{v}|^2\]And, we know, for two vectors u, v:\[\left(\vec u +\vec v\right)\cdot\left(\vec u -\vec v\right)=0\]Writing this in normal form:
\[\vec u=\langle x_u,y_u\rangle, \vec v=\langle x_v,y_v\rangle\]And, we write the above expliticly:\[\left(\langle x_u,y_u\rangle+\langle x_v,y_v\rangle\right)\cdot\left(\langle x_u,y_u\rangle-\langle x_v,y_v\rangle\right)=\langle x_u+x_v,y_u+y_v\rangle\cdot \langle x_u-x_v,y_u-y_v\rangle\]Which is\[(x_u+x_v)(x_u-x_v)+(y_u+y_v)(y_u-y_v)=x_u^2-x_v^2+y_u^2-y_v^2=0\]Rearranging gives us:\[x_u^2+y_u^2=x_v^2+y_v^2 \implies N(\vec v)=N(\vec u) \implies |\vec v|=|\vec u|\]Thus we are done.

Answer 13

Nooo... it ran partly off the page... took me a ridiculous amount of time to write in LaTeX... dang, whatever, it's just an angled bracket.

Answer 14

Thanks, i thought about doing that but i didn't think i would get it right and its a mess -- but i toke a look and its awesome!

Answer 15

remember that magnitude squared of u is
\[ u \cdot u= |u|^2 \]
We are given
\[ (u+v)\cdot (u-v)= 0\]
expanding the dot product:
(I would just write |u|^2 - |v|^2, but here are the details:)
\[ (u+v)\cdot (u-v)= u \cdot u + u \cdot -v +v \cdot u+ v \cdot -v =0\]
\[ |u|^2 -(u \cdot v) +(u \cdot v)- |v|^2=0\]
\[ |u|^2 - |v|^2=0\]
\[ |u|^2= |v|^2 \]
or
\[ |u|= |v| \]
magnitude of u equals the magnitude of v