Question

if the sum of two unit vectors is also a unit vector , then the magnitude of their differences and angle between the two given unit vectors is

Answer 1

$$\begin{gathered} \mathrm{Suppose}\mathrm{two}\mathrm{unit}\mathrm{vectors}\mathrm{are}\overrightarrow{\mathrm{A}}\mathrm{and}\overrightarrow{\mathrm{B}}\mathrm{and}\mathrm{angle}\mathrm{between}\mathrm{them}\mathrm{is}\mathrm{\theta }. \\ \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{X}} \\ \mathrm{The}\mathrm{resultant}\mathrm{of}\overrightarrow{\mathrm{A}}\mathrm{is}\overrightarrow{\mathrm{B}}\mathrm{is}\overrightarrow{\mathrm{X}} \\ \left|\overrightarrow{\mathrm{X}}\right|=\sqrt{|\overrightarrow{\mathrm{A}}{|}^2+|\overrightarrow{\mathrm{B}}{|}^2+2\left|\overrightarrow{\mathrm{A}}\right|\left|\overrightarrow{\mathrm{B}}\right|\mathrm{cos\theta }} \\ {1}^2=1^2+1^2+2\mathrm{cos\theta } \\ \mathrm{or}\mathrm{cos\theta }=-\frac{1}{2} \\ \mathrm{or}\mathbf{}\mathbf{\theta }\mathbf{=}{\mathbf{120}}^{\mathbf{0}} \\ \overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{Y}} \\ \left|\overrightarrow{\mathrm{Y}}\right|=\sqrt{|\overrightarrow{\mathrm{A}}{|}^2+|\overrightarrow{\mathrm{B}}{|}^2+2\left|\overrightarrow{\mathrm{A}}\right|\left|\overrightarrow{\mathrm{B}}\right|\cos (\mathrm{\pi }-\mathrm{\theta })} \\ =\sqrt{1^2+1^2+2\times \frac{1}{2}} \\ =\sqrt{\mathbf{3}} \end{gathered}$$