Three forces $\vec{F_{1}}, \vec{F_{2}} and \vec{F_{3}}$ are such that $\vec{F_{1}} = \vec{F_{2}} + \vec{F_{3}}$ and their magnitudes are 5, 4 and 3 respectively. Find the angle between $\vec{F_{1}}$ and $\vec{F_{3}}$ in degrees.

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Detailed Solution

$\vec{F_{1}} = \vec{F_{2}} + \vec{F_{3}} \Rightarrow | \vec{F_{1}} | = | \vec{F_{2}} + \vec{F_{3}} | .$

Let $θ$ be angle between $\vec{F_{2}}$ and $\vec{F_{3}}$ so

$F_{1} = \sqrt{F_{2}^{2} + F_{3}^{2} + 2 F_{2} F_{3} \cos θ} F_{1}^{2} = F_{2}^{2} + F_{3}^{2} + 2 F_{2} F_{3} \cos θ 5^{2} = 4^{2} + 3^{2} + 2 (4) (3) \cos θ$

$\cos θ = 0 \Rightarrow θ = 90 °$ means $\vec{F_{2}}$ and $\vec{F_{3}}$ are perpendicular. We can represent the given vectors as

Clearly if $β$ be angle between $\vec{F_{1}}$ and $\vec{F_{3}}$ then, $\tan β = \frac{| \vec{F_{2}} |}{| \vec{F_{3}} |} = \frac{4}{3} = \frac{4}{3}$ or $\cos β = \frac{| \vec{F_{3}} |}{| \vec{F_{1}} |} = \frac{3}{5}, β = \cos^{- 1} \frac{3}{5} .$