Magnitudes of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are $4$ units and $3$ units respectively. If these vectors are $(i)$ in the same direction $(\theta = 0^{\circ})$ and $(ii)$ in the opposite direction $(\theta = 180^{\circ})$,find the magnitude of the resultant vector in each case.
The magnitude of the resultant vector $\overrightarrow{R}$ of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
$(i)$ When the vectors are in the same direction,$\theta = 0^{\circ}$. Thus,$R = \sqrt{4^2 + 3^2 + 2(4)(3) \cos 0^{\circ}} = \sqrt{16 + 9 + 24(1)} = \sqrt{49} = 7$ units.
$(ii)$ When the vectors are in the opposite direction,$\theta = 180^{\circ}$. Thus,$R = \sqrt{4^2 + 3^2 + 2(4)(3) \cos 180^{\circ}} = \sqrt{16 + 9 + 24(-1)} = \sqrt{25 - 24} = \sqrt{1} = 1$ unit.
$(i)$ When the vectors are in the same direction,$\theta = 0^{\circ}$. Thus,$R = \sqrt{4^2 + 3^2 + 2(4)(3) \cos 0^{\circ}} = \sqrt{16 + 9 + 24(1)} = \sqrt{49} = 7$ units.
$(ii)$ When the vectors are in the opposite direction,$\theta = 180^{\circ}$. Thus,$R = \sqrt{4^2 + 3^2 + 2(4)(3) \cos 180^{\circ}} = \sqrt{16 + 9 + 24(-1)} = \sqrt{25 - 24} = \sqrt{1} = 1$ unit.
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