# Gram schmidt, incorrect answer

I projected y onto x giving me a vector y1. I then subtracted y1 from y which gave me a vector orthogonal to x, I called this vector y2. This vector was $y_2 = (-1, -7, -1)$. I then normalized x and y2 and entered them as my answers. But it says my answer for y2 is incorrect.

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Is your arithmetic correct? I calculate $y_1=(-2, -3/2, -1/2)$, $y_2=(3,-4,0)$. (Note your $y_2$ is not orthogonal to $x$; you erred either when calculating $y_1$ or $y_2$.) – David Mitra Apr 11 '12 at 18:23
The description of what you did is the correct procedure, though. – David Mitra Apr 11 '12 at 18:26
Left out a minus sign along the way...cheers mate. Note to self: write out answers fully and stop scribbling crap. – Jim_CS Apr 11 '12 at 18:34

You have the right procedure. But \eqalign{ y_1 ={\rm proj}_x(y)&={x\cdot y\over x\cdot x} x \cr &={ 4\cdot 1+3(-11/2)+1(-1/2)\over 4\cdot4+3\cdot 3+1\cdot1 }x\cr &={-13\over26 }x\cr &={-1\over2}\Biggl(\matrix{4\cr3\cr1}\Biggr)\cr &= \Biggl(\matrix{-2\cr-3/2\cr-1/2}\Biggr). } So, $$y_2= \Biggl(\matrix{1\cr-11/2\cr-1/2}\Biggr) - \Biggl(\matrix{-2\cr-3/2\cr-1/2}\Biggr)= \Biggl(\matrix{3\cr-4\cr0}\Biggr).$$
Here's the salient point I want to make: It's a good idea to check your work when you can. As a quick spot check, we have $$y_2\cdot x=3\cdot4+(-4)(3)+0\cdot1=0,$$ as it should (the $y_2$ of your calculation isn't orthogonal to $x$).