How do you normalize $(2i - 3 j +k)$ ?

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How do you normalize $(2i - 3 j +k)$ ?

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Answer 1 · 2016-02-09T11:30:32

$bar x=1/sqrt 14$
$bar y=-3/sqrt 14$
$bar z=1/sqrt 14$

Explanation:

$a_x=2$
$a_y=-3$
$a_z=1$
$|\a|=a_x*a_x+a_y*a_y+a_z*a_z$
$|\a|=sqrt(2*2+(-3)*(-3)+1*1)$
$|\a|=sqrt (4+9+1)$
$|\a|=sqrt 14$
$bar x=2/|\a|$
$bar y=-3/|\a|$
$bar z=1/|\a|$
$bar x=1/sqrt 14$
$bar y=-3/sqrt 14$
$bar z=1/sqrt 14$

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How do you normalize $(2i - 3 j +k)$ ?

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