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

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Answer 1 · 2017-02-26T15:27:41

Divide by its length to get:

$(3sqrt(14))/14i-sqrt(14)/14j-sqrt(14)/7k$

Divide it by its length:

$abs(abs(3i-j-2k)) = sqrt(3^2+(-1)^2+(-2)^2) = sqrt(9+1+4) = sqrt(14)$

So the unit length vector in the same direction as $3i-j-2k$ is:

$1/sqrt(14) (3i-j-2k) = sqrt(14)/14(3i-j-2k)$

$color(white)(1/sqrt(14) (3i-j-2k)) = (3sqrt(14))/14i-sqrt(14)/14j-sqrt(14)/7k$

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