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When finding the norm of the vector:

Find $\|2w-2y\|$ such that $w=(1/2,3,1)$ and $y=(0,-1,3/2)$.

answer:

$$\begin{align*} &2(1/2,3,1)= (1,6,2)\\ &2(0,-1,3/2) =(0,-2,3)\\ &(1,6,2)- (0,-2,3) = (1,8,-1)\\ &\sqrt{ 1^2 +8^2+(-1)^2}=\sqrt{66}= 8.124 \end{align*}$$

Is this the correct way of doing it?

Thanks

  • 0
    It is indeed; my only criticism is that $\sqrt{66}$ isn’t **equal** to $8.124$, but only **approximately** equal.2012-05-14
  • 0
    Okay, i'll bear that in mind, thanks!2012-05-14

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That's correct, you can also do the subtraction first:

$\|2w-2y\|= \|2(w-y)\|= 2(1/2,4,-1/2)=(1,8,-1)$