(adsbygoogle = window.adsbygoogle || []).push({}); Find a basis for U perp given U = {(3,-1,2),(2,0,-3)}

to find U perp i have to find a vector such that it perpendicular to BOTH of those vectors up there

3x - 2y + 2z = 0

2x - 3z = 0

i get z = t

y = 13t/2

x = 2t

so the only basis for U perp is

{2t,13t/2,t}?? ALl values of t are simply multiples of each other anyway...

AM i right? Only one vector?

Also

If [tex] R^n = span\{X_{1},X_{2},...,X_{N}\ [/tex]} and [tex] X \bullet X_{i} = Y \bullet X_{i} [/tex]

for all i, then show that X = Y

WEll all the Xi form an orthogonal basis right? So from the given condition couldnt we say

for n not i

[tex] X \bullet \left(\frac{X_{i} \bullet X_{1}}{||X_{1}||^2} X_{1} + ... + \frac{X_{i} \bullet X_{n}}{||X_{n}||^2} X_{n}\right) [/tex]

and [tex] Y \bullet \left(\frac{X_{i} \bullet X_{1}}{||X_{1}||^2} X_{1} + ... + \frac{X_{i} \bullet X_{n}}{||X_{n}||^2} X_{n}\right) [/tex]

since the two are equal.. and Xi dot X and Xi dot are all equal... then X must be same as Y

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# Homework Help: Find a basis for U perp given U = {(3,-1,2),(2,0,-3)}

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