Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

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.

share|improve this question
1  
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
1  
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

1 Answer 1

up vote 1 down vote accepted

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 $x$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.