2D vector Magnitude:

$\overline{)\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}{\mathbf{=}}\sqrt{{{\mathbf{v}}_{\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{v}}_{\mathbf{y}}}^{\mathbf{2}}}}$

2D vector direction:

$\overline{){\mathbf{tan}}{\mathit{\theta}}{\mathbf{=}}\frac{{\mathit{v}}_{\mathit{y}}}{{\mathit{v}}_{\mathit{x}}}}$

Vector decomposition:

$\overline{)\begin{array}{rcl}{\mathbf{v}}_{\mathbf{x}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\theta}\\ {\mathbf{v}}_{\mathbf{y}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{\theta}\end{array}}$

V is a vector 24.8 units in magnitude and points an angle of 23.4° above the negative axis.**(a)** Sketch this vector.**(b)** Calculate V_{x} and V_{y}.**(c) **Use V_{x} and V_{y} to obtain (again) the magnitude of V. [ Note: Part (c) is a good way to check if you've resolved your vector correctly.]**(d)** Use V_{x} and V_{y} to obtain (again) the direction of V.

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