Linear Combination of vectors

I have a previous post here. There is a part b to that question and it asks:
Let $x=(1,1,1)^T$. Write x as a linear combination of $u_1, u_2, u_3$ using Parseval's formula to compute $||x||$.

I know how to compute $||x||$, it's simply the magnitude. However I am totally unsure how do to x as a linear combo, I've read through my book and tried looking online with no luck. Any ideas?

-

Hint: Since you have orthonormal basis $\{u_1,u_2,u_3\}$, you can express any vector $x$ as a linear combination of the basis vectors as $$x=\alpha_1 u_1+\alpha_2 u_2+\alpha_3 u_3$$ where you have to determine the coefficients (scalars) $\alpha_i$'s.
Since, $u_i$'s are orthonormal, $<u_i,u_j>=u_i^t.u_j=\delta_{ij}$ (=1, only if $i=j$, else $0$). So,
$$x^t.u_i=<x,u_i>=(\sum\alpha_j u_j^t).u_i=\alpha_i$$.
you completely lost me after the "Since, $u_i$'s are". Can you please rephrase it a little differently, i'm not getting it –  Charlie Yabben Oct 12 '12 at 0:48
Sorry... I assumed you know about inner product, otherwise how could the Perseval's identity come? The $<x,y>$ symbol stands for "inner product of x and y" which, in $\mathbb{R}^3$ can be taken as $x^T.y$. Is it clear now? –  Tapu Oct 12 '12 at 0:56
Inner product and dot product are different terminology of the same thing $<x,y>$. Perseval's formula says $\|x\|^2=\sum\alpha_i^2$ –  Tapu Oct 12 '12 at 1:05
okay so the first step I is determine the coefficients $\alpha_i$'s which can be found by doing $<x,u_i>$. Am I right so far? –  Charlie Yabben Oct 12 '12 at 1:09