Answer
$|\overline{V}|=13$
$\theta=337.38^{\circ}$
Work Step by Step
$\because \overline{V}=〈a_1,a_2〉=〈-12,5〉$
The magnitude of the vector: $|\overline{V}|=\sqrt{a_1^2+a_2^2}$
$\therefore |\overline{V}|=\sqrt{(-12)^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$
The direction of the vector: $\theta =tan^{-1}\dfrac{a_2}{a_1}$
$\theta =tan^{-1}\dfrac{5}{(-12)}=-22.62^{\circ}=360-22.62=337.38^{\circ}$
