How do you normalize $< 3, - 6, 2 >$ $<3, -6, 2>$ ?

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How do you normalize $< 3, - 6, 2 >$ $<3, -6, 2>$ ?

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Answer 1 · 2017-01-23T22:47:53

$u = < \frac{3}{7}, - \frac{6}{7}, \frac{2}{7} >$ $u=<3/7,-6/7,2/7>$

Explanation:

In normalizing the vector we are finding a unit vector (magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

$u = \frac{v}{| v |}$ $u=v/(|v|)$

Given $v = < 3, - 6, 2 >$ $v=<3,-6,2>$ , we can calculate the magnitude of the vector:

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

$| v | = \sqrt{{(3)}^{2} + {(- 6)}^{2} + {(2)}^{2}}$ $|v|=sqrt((3)^2+(-6)^2+(2)^2)$

$| v | = \sqrt{9 + 36 + 4}$ $|v|=sqrt(9+36+4)$

$| v | = \sqrt{49}$ $|v|=sqrt(49)$

$| v | = 7$ $|v|=7$

We now have:

$u = \frac{< 3, - 6, 2 >}{7}$ $u=(<3,-6,2>)/7$

$\Rightarrow u = < \frac{3}{7}, - \frac{6}{7}, \frac{2}{7} >$ $=>u=<3/7,-6/7,2/7>$

Hope that helps!

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How do you normalize $< 3, - 6, 2 >$ $<3, -6, 2>$ ?

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