# How do you normalize ( - 4 i - 5 j + 2 k )?

Jan 17, 2016

Normalizing a vector means changing it from its current magnitude (length) to a 'unit vector' with a length of 1. This yields vecu )=(-0.66i-0.82j+0.33k), or, if you prefer: $\vec{u} = \left(- \frac{4}{\sqrt{45}} i - \frac{5}{\sqrt{45}} j + \frac{2}{\sqrt{45}} k\right)$

#### Explanation:

Normalizing a vector - making it into a unit vector - involves dividing each of its components (coefficients) by the length of the vector.

Our first step, then, is to find the length of the vector. This video is very clear in discussing how to find the length of a vector in 3 dimensions:

The length of our vector, then, is:

$l = \sqrt{\left(- {4}^{2}\right) + \left(- {5}^{2}\right) + {2}^{2}} = \sqrt{45} = 6.7$

Two notes here:

1. the length, $l$, is itself a scalar, not a vector

2. you could leave the
answer as $\sqrt{45}$ if you like, rather than calculating and rounding. It's a matter of taste that's between you and your marker. ;-)

Now that we know the length, we need to divide each coefficient by it. The new, normalised vector, $\vec{u}$, is:

$\vec{u} = \left(- \frac{4}{6.7} i - \frac{5}{6.7} j + \frac{2}{6.7} k\right) = \left(- 0.66 i - 0.82 j + 0.33 k\right)$

Or, if you prefer:

$\vec{u} = \left(- \frac{4}{\sqrt{45}} i - \frac{5}{\sqrt{45}} j + \frac{2}{\sqrt{45}} k\right)$