# The vector U=2i +3j -2k, how do you find a unit vector parallel to U?

Jun 19, 2016

The unit vector is $v = \left(\frac{2}{\sqrt{17}} , \frac{3}{\sqrt{17}} , - \frac{2}{\sqrt{17}}\right)$

#### Explanation:

If you have a vector $\left(a , b , c\right)$ and multiply all the components by a same scalar $\alpha$, you obtain the vector $\left(\alpha a , \alpha b , \alpha c\right)$. This vector is parallel to the original one, and it is smaller than the original if $\alpha < 1$, and greater than the original if $\alpha > 1$.

Now, let's talk about norms. The norm of a vector is the quantity

$\left\lVert a , b , c \right\rVert = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}}$.

So, we want to multiply the vector $\left(2 , 3 , - 2\right)$ by some scalar $\alpha$ such that the new vector $\left(2 \setminus \alpha , 3 \setminus \alpha , - 2 \setminus \alpha\right)$ has norm equal to one.

To do so, choose $\alpha$ as the inverse of the norm of the vector: we have

$\left\lVert 2 , 3 , - 2 \right\rVert = \sqrt{4 + 9 + 4} = \sqrt{17}$

and choosing $\alpha = \frac{1}{\sqrt{17}}$ we have

$\left\lVert 2 \setminus \alpha , 3 \setminus \alpha , - 2 \setminus \alpha \right\rVert = \left\lVert \frac{2}{\sqrt{17}} , \frac{3}{\sqrt{17}} , - \frac{2}{\sqrt{17}} \right\rVert$

$= {\left(\frac{2}{\sqrt{17}}\right)}^{2} + {\left(\frac{3}{\sqrt{17}}\right)}^{2} + {\left(- \frac{2}{\sqrt{17}}\right)}^{2}$

$= \frac{4}{17} + \frac{9}{17} + \frac{4}{17} = 1$