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

Wednesday, 18 March 2015

$\displaystyle{u}=$ Use the dot product to find the magnitude of u.

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

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

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

$\displaystyle{\left|{\left|{u}\right|}\right|}=\sqrt{{{u}\cdot{u}}}=\sqrt{{\lt-{8},{15}\gt\cdot\lt-{8},{15}\gt}}=$ $\displaystyle=\sqrt{{{\left(-{8}\right)}\cdot{\left(-{8}\right)}+{15}\cdot{15}}}=\sqrt{{{64}+{225}}}=\sqrt{{{289}}}={17}.$

This is the answer.

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

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

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

$\displaystyle{\left|{\left|{u}\right|}\right|}=\sqrt{{{u}\cdot{u}}}=\sqrt{{\lt-{8},{15}\gt\cdot\lt-{8},{15}\gt}}=$
$\displaystyle=\sqrt{{{\left(-{8}\right)}\cdot{\left(-{8}\right)}+{15}\cdot{15}}}=\sqrt{{{64}+{225}}}=\sqrt{{{289}}}={17}.$

This is the answer.