# Question #a7f09

Sep 11, 2017

$| \vec{\text{A}} | = \sqrt{106}$

#### Explanation:

The vector $\text{A}$ has components ${a}_{x} = - 5$ $\text{m}$ and ${a}_{y} = 9$ $\text{m}$.

This can be written as:

$R i g h t a r r o w \vec{\text{A}} = - 5 i + 9 j$

The magnitude of a vector is of the form:

$R i g h t a r r o w | \vec{\text{A}} | = \sqrt{{x}^{2} + {y}^{2}}$

Substituting the relevant values:

$R i g h t a r r o w | \vec{\text{A}} | = \sqrt{{\left(- 5\right)}^{2} + {\left(9\right)}^{2}}$

$R i g h t a r r o w | \vec{\text{A}} | = \sqrt{25 + 81}$

$\therefore | \vec{\text{A}} | = \sqrt{106}$

Therefore, the magnitude of vector $\text{A}$ is $\sqrt{106}$.