How do you normalize $(-4i + 0j + 3k)$ ?

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Answer 1 · 2016-01-23T22:07:38

To normalise a vector we divide each component by the length of the original vector, so the resultant vector is:

$(-4/5i+0/5j+3/5k)$

Normalising a vector means dividing each of its components by its length, to yield a unit vector (a vector 1 unit long) in the same direction.

The length of a 3D vector (a $i$ +b $j$ +c $k$ ) is given by:

$l=sqrt(a^2+b^2+c^2)$

In this case:

$l=sqrt((-4)^2+0^2+3^2) = sqrt(25) = 5$ units

To normalise the vector, then, we divide each component by 5, so the resultant vector is:

$(-4/5i+0/5j+3/5k)$

How do you normalize (-4i + 0j + 3k) (-4i + 0j + 3k)?