Vector Proof

**tdat1979** · September 7th 2009, 06:10 PM

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.

**alexmahone** · September 7th 2009, 06:22 PM

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.

$x.y = (|v|u + |u|v) . (|v|u - |u|v)$

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

= $0$

Hence, x and y are perpendicular.

**Prove It** · September 7th 2009, 06:32 PM

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

Then

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

Therefore

$x = |v|u + |u|v$

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

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

$y = |v|u - |u|v$

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

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

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

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

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

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

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

$= 0$

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

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