How do you normalize <2,0,-1>?

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How do you normalize <2,0,-1>?

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Answer 1 · 2017-09-09T20:45:45

$ˆ u = < \frac{2 \sqrt{5}}{5}, 0, \frac{- \sqrt{5}}{5} >$ $hatu=<(2sqrt(5))/5,0,(-sqrt(5))/5>$

Explanation:

To normalize a vector is to find unit vector (vector with magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

$ˆ u = \frac{\to v}{∣ ∣ \to v ∣ ∣}$ $hatu=vecv/(|vecv|)$

Given $\to v = < 2, 0, - 1 >$ $vecv=<2,0,-1>$ , we can calculate the magnitude of the vector:

$∣ ∣ \to v ∣ ∣ = \sqrt{{(v_{x})}_{x}^{2} + {(v_{y})}_{y}^{2} + {(v_{z})}_{z}^{2}}$ $abs(vecv)=sqrt((v_x)^2+(v_y)^2+(v_z)^2)$

$\Rightarrow = \sqrt{{(2)}^{2} + {(0)}^{2} + {(- 1)}^{2}}$ $=>=sqrt((2)^2+(0)^2+(-1)^2)$

$\Rightarrow = \sqrt{4 + 0 + 1}$ $=>=sqrt(4+0+1)$

$\Rightarrow = \sqrt{5}$ $=>=sqrt(5)$

We now have:

$ˆ u = \frac{< 2, 0, - 1 >}{\sqrt{5}}$ $hatu=(<2,0,-1>)/sqrt(5)$

$\Rightarrow ˆ u = < \frac{2}{\sqrt{5}}, 0, - \frac{1}{\sqrt{5}} >$ $=>hatu=<2/sqrt(5),0,-1/sqrt(5)>$

We can also rationalize the denominator on the $ˆ x$ $hatx$ $(ˆ i)$ $(hati)$ and $ˆ z$ $hatz$ $(ˆ k)$ $(hatk)$ component:

$\Rightarrow ˆ u = < \frac{2 \sqrt{5}}{5}, 0, \frac{- \sqrt{5}}{5} >$ $=>hatu=<(2sqrt(5))/5,0,(-sqrt(5))/5>$

Hope that helps!

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How do you normalize <2,0,-1>?

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