If $\overline{\mathrm{e}}_1, \overline{\mathrm{e}}_2$ and $\overline{\mathrm{e}}_1+\overline{\mathrm{e}}_2$…

If $\overline{\mathrm{e}}_1, \overline{\mathrm{e}}_2$ and $\overline{\mathrm{e}}_1+\overline{\mathrm{e}}_2$ are unit vectors, then the angle between $\overline{\mathrm{e}}_1$ and $\bar{e}_2$ is
  1. $150^{\circ}$
  2. $120^{\circ}$
  3. $90^{\circ}$
  4. $135^{\circ}$

Solution

$\begin{aligned} & \left|\overline{\mathrm{e}}_1+\overline{\mathrm{e}}_2\right|^2=\left|\overline{\mathrm{e}}_1^2\right|+\left|\overline{\mathrm{e}}_2^2\right|+2 \overline{\mathrm{e}}_1 \overline{\mathrm{e}}_2 \cos \theta \\ & \text { Here }\left|\overline{\mathrm{e}}_1\right|=1,\left|\overline{\mathrm{e}}_2\right|=1 \text { and }\left|\overline{\mathrm{e}}_1+\overline{\mathrm{e}}_2\right|=1 \\ & \therefore(1)=(2)^2+(1)^2+2(1)(1) \cos \theta \\ & \therefore \frac{-1}{2}=\cos \theta \Rightarrow \theta=120^{\circ}\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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