# How do you compute the dot product to find the magnitude of u=<2, -4>?

Nov 1, 2016

Inner Product Definition
If $\vec{u} = \left\langle\left({u}_{1} , {u}_{2} , {u}_{3}\right)\right\rangle$, and $\vec{v} = \left\langle\left({v}_{1} , {v}_{2} , {v}_{3}\right)\right\rangle$, then the inner product (or dot product), a scaler quantity, is given by:
$\vec{u} \cdot \vec{v} = {u}_{1} {v}_{1} + {u}_{2} {v}_{2} + {u}_{3} {v}_{3}$

Inner Product = 0 $\Leftrightarrow$ vectors are perpendicular
Also, $| \vec{u} | = \sqrt{\vec{u} \cdot \vec{u}}$

With $\vec{u} = \left\langle2 , - 4\right\rangle \implies | \vec{u} | = \sqrt{\left(2\right) \left(2\right) + \left(- 4\right) \left(- 4\right)}$
$\therefore | \vec{u} | = \sqrt{4 + 16}$
$\therefore | \vec{u} | = \sqrt{20}$
$\therefore | \vec{u} | = \sqrt{4 \times 5}$
$\therefore | \vec{u} | = 2 \sqrt{5}$