1. ## Vector Proof

Show that for any two nonzero vectors u and v, the vectors x = |v|u + |u|v and y = |v|u - |u|v are perpendicular.

2. Originally Posted by tdat1979
Show that for any two nonzero vectors u and v, the vectors x = |v|u + |u|v and y = |v|u - |u|v are perpendicular.
$\displaystyle x.y = (|v|u + |u|v) . (|v|u - |u|v)$

= $\displaystyle |v|^2|u|^2+|u||v|v.u-|v||u|u.v-|u|^2|v|^2$

= $\displaystyle 0$

Hence, x and y are perpendicular.

3. Originally Posted by tdat1979
Show that for any two nonzero vectors u and v, the vectors x = |v|u + |u|v and y = |v|u - |u|v are perpendicular.
I hope you're familiar with the matrix notation of vectors...

Let $\displaystyle u = [a, b]$ and $\displaystyle v = [c, d]$.

Then

$\displaystyle |u| = \sqrt{a^2 + b^2}$ and $\displaystyle |v| = \sqrt{c^2 + d^2}$.

Therefore

$\displaystyle x = |v|u + |u|v$

$\displaystyle = \sqrt{c^2 + d^2}[a, b] + \sqrt{a^2 + b^2}[c, d]$

$\displaystyle = [\sqrt{c^2 + d^2}a + \sqrt{a^2 + b^2}c, \sqrt{c^2 + d^2}b + \sqrt{a^2 + b^2}d]$.

$\displaystyle y = |v|u - |u|v$

$\displaystyle = \sqrt{c^2 + d^2}[a, b] - \sqrt{a^2 + b^2}[c, d]$

$\displaystyle = [\sqrt{c^2 + d^2}a - \sqrt{a^2 + b^2}c, \sqrt{c^2 + d^2}b - \sqrt{a^2 + b^2}d]$

To show $\displaystyle x$ and $\displaystyle y$ are perpendicular, we need to show that their dot product is 0.

So $\displaystyle x \cdot y = [\sqrt{c^2 + d^2}a + \sqrt{a^2 + b^2}c, \sqrt{c^2 + d^2}b + \sqrt{a^2 + b^2}d]$
$\displaystyle \cdot[\sqrt{c^2 + d^2}a - \sqrt{a^2 + b^2}c, \sqrt{c^2 + d^2}b - \sqrt{a^2 + b^2}d]$

$\displaystyle = (\sqrt{c^2 + d^2}a + \sqrt{a^2 + b^2}c)(\sqrt{c^2 + d^2}a - \sqrt{a^2 + b^2}c)$
$\displaystyle + (\sqrt{c^2 + d^2}b + \sqrt{a^2 + b^2}d)(\sqrt{c^2 + d^2}b - \sqrt{a^2 + b^2}d)$

$\displaystyle = (c^2 + d^2)a^2 - (a^2 + b^2)c^2 + (c^2 + d^2)b^2 - (a^2 + b^2)d^2$

$\displaystyle a^2c^2 + a^2d^2 - a^2c^2 - b^2c^2 + b^2c^2 + b^2d^2 - a^2d^2 - b^2d^2$

$\displaystyle = 0$

So $\displaystyle x$ and $\displaystyle y$ are perpendicular (PHEW!)