# How do you find a unit vector u in the same direction as the vector ⟨1,−2,−3⟩?

Jun 20, 2016

$\vec{v} = \frac{\left\{1 , - 2 , - 3\right\}}{\sqrt{14}}$

#### Explanation:

A unit vector is a vector $\vec{u}$ such that

$\left\lVert \vec{u} \right\rVert = 1$

Given a vector $\vec{V} = \left\{1 , - 2 , - 3\right\}$ the way to find its associated unit vector $\vec{v}$ is normalizing it. Then

$\vec{v} = \frac{\vec{V}}{\left\lVert \vec{V} \right\rVert} = \frac{\left\{1 , - 2 , - 3\right\}}{\sqrt{{1}^{2} + {\left(- 2\right)}^{2} + {\left(- 3\right)}^{2}}} = \frac{\left\{1 , - 2 , - 3\right\}}{\sqrt{14}}$.

We know that $\left\langle\vec{V} , \vec{V}\right\rangle = {\left\lVert \vec{V} \right\rVert}^{2}$ then

$\left\langle\vec{v} , \vec{v}\right\rangle = \left\langle\frac{\vec{V}}{\left\lVert \vec{V} \right\rVert} , \frac{\vec{V}}{\left\lVert \vec{V} \right\rVert}\right\rangle = \frac{\left\langle\vec{V} , \vec{V}\right\rangle}{\left\lVert \vec{V} \right\rVert} ^ 2 = {\left\lVert \vec{V} \right\rVert}^{2} / {\left\lVert \vec{V} \right\rVert}^{2} = 1$