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Mathematics 16 Online
OpenStudy (anonymous):

Find ||u|| if u=[1,1,0]

OpenStudy (anonymous):

Also, what does ||u|| mean?

OpenStudy (anonymous):

(1+1)^1/2=sqrt(2)

OpenStudy (anonymous):

are you trying to find the magnitude?

OpenStudy (anonymous):

Sorry, \[u=\left(\begin{matrix}1\\1 \\ 0\end{matrix}\right)\]Not sure if that makes a difference... Answer says sqrt(3)...

OpenStudy (anonymous):

Erm I'm using it as part of the Gram-Schmidt process, but can't figure out how it ends up being sqrt3...

OpenStudy (anonymous):

hey man are u trying to find the square root or unit vectors?

OpenStudy (anonymous):

i meant the magnitude*

OpenStudy (anonymous):

Dunno - using it as part of \[v_1 = u_1/||u_1||\]Where\[u_1 = \left(\begin{matrix}1\\1 \\ 0\end{matrix}\right)\]The answer for this step shows \[\left(\begin{matrix}1\\1 \\ 0\end{matrix}\right)/\sqrt{1+2+0}\]

OpenStudy (anonymous):

what does the double bar stand for have any idea?

OpenStudy (anonymous):

does it mean magnitude or something else?

OpenStudy (anonymous):

\[=\left(\begin{matrix}{1/{\sqrt{3}}}\\{1/{\sqrt{3}}} \\ 0\end{matrix}\right)\]

OpenStudy (anonymous):

I think it is magnitude or norm

OpenStudy (anonymous):

pellet never mind - I got it! Didn't see we are given the inner product relation... Cheers anyway bro!

OpenStudy (anonymous):

something is fishy about ur problem

OpenStudy (anonymous):

(hahaha "s***" automatically changes to "pallet")

OpenStudy (anonymous):

oh ok this is for linear algebra right?

OpenStudy (anonymous):

Yeah

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