How do you normalize $(2i -3j + 4k)$ ?

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Answer 1 · 2015-12-26T21:21:43

If you have a vector $veca$ with coordinates (x,y,z) then the normalized vector $vecn$ is

$vecn = (veca)/(absa) = (<< x","y","z >>)/absa$

where $absa=sqrt(x^2+y^2+z^2)$

Hence for $x=2,y=-3,z=4$ we have that

$absa=sqrt(2^2+3^2+4^2)=sqrt(4+9+16)=sqrt29$

Thus the normalized vector is

$vecn = color(blue)(<< 2/sqrt29,-3/sqrt29,4/sqrt29 >>)$

How do you normalize (2i -3j + 4k) (2i -3j + 4k)?