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When finding the norm of the vector:

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


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

Is this the correct way of doing it?


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It is indeed; my only criticism is that $\sqrt{66}$ isn’t equal to $8.124$, but only approximately equal. – Brian M. Scott May 14 '12 at 8:13
Okay, i'll bear that in mind, thanks! – Xabi May 14 '12 at 8:13
up vote 1 down vote accepted

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

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

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