Notes: `u = ` Use the dot product to find the magnitude of u.

Monday, 5 June 2017

The magnitude of a vector $\displaystyle{u}$ is the square root of its product by itself, because

$\displaystyle{u}\cdot{u}={\left|{\left|{u}\right|}\right|}\cdot{\left|{\left|{u}\right|}\right|}\cdot{\cos{{\left(\theta\right)}}},$

$\displaystyle{\quad\text{and}\quad}\theta={0},$ $\displaystyle{\cos{{\left(\theta\right)}}}={1}.$ So

$\displaystyle{\left|{\left|{u}\right|}\right|}=\sqrt{{{u}\cdot{u}}}=\sqrt{{\lt{4},-{6}\gt\cdot\lt{4},-{6}\gt}}=$

$\displaystyle=\sqrt{{{4}\cdot{4}+{\left(-{6}\right)}\cdot{\left(-{6}\right)}}}=\sqrt{{{16}+{36}}}=\sqrt{{{52}}}={2}\sqrt{{{13}}}\approx{7.2}.$

The magnitude of a vector $\displaystyle{u}$ is the square root of its product by itself, because

$\displaystyle{u}\cdot{u}={\left|{\left|{u}\right|}\right|}\cdot{\left|{\left|{u}\right|}\right|}\cdot{\cos{{\left(\theta\right)}}},$

$\displaystyle{\quad\text{and}\quad}\theta={0},$ $\displaystyle{\cos{{\left(\theta\right)}}}={1}.$ So

$\displaystyle{\left|{\left|{u}\right|}\right|}=\sqrt{{{u}\cdot{u}}}=\sqrt{{\lt{4},-{6}\gt\cdot\lt{4},-{6}\gt}}=$

$\displaystyle=\sqrt{{{4}\cdot{4}+{\left(-{6}\right)}\cdot{\left(-{6}\right)}}}=\sqrt{{{16}+{36}}}=\sqrt{{{52}}}={2}\sqrt{{{13}}}\approx{7.2}.$

Notes