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If $\vec A,\vec B$ and $\vec C$ are vectors having a unit magnitude. If $\vec A + \vec B + \vec C = \vec 0$ then $\vec A.\vec B + \vec B.\vec C + \vec C.\vec A$ will be
$1$
$ - 1.5$
$ -0.5$
$0$
Solution
$\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}+\overrightarrow{\mathrm{C}} \cdot \overrightarrow{\mathrm{A}}$
$=\mathrm{AB} \cos 120^{\circ}+\mathrm{BC} \cos 120^{\circ}+\mathrm{AC} \cos 120^{\circ}$
as $A=B=C=1$
$\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}+\overrightarrow{\mathrm{C}} \cdot \overrightarrow{\mathrm{A}}=3 \cos 120^{\circ}=-3 / 2$
Similar Questions
If $\left| {\vec A } \right|\, = \,2$ and $\left| {\vec B } \right|\, = \,4$ then match the relation in Column $-I$ with the angle $\theta $ between $\vec A$ and $\vec B$ in Column $-II$.
| Column $-I$ | Column $-II$ |
| $(a)$ $\vec A \,.\,\,\vec B \, = \,\,0$ | $(i)$ $\theta = \,{0^o}$ |
| $(b)$ $\vec A \,.\,\,\vec B \, = \,\,+8$ | $(ii)$ $\theta = \,{90^o}$ |
| $(c)$ $\vec A \,.\,\,\vec B \, = \,\,4$ | $(iii)$ $\theta = \,{180^o}$ |
| $(d)$ $\vec A \,.\,\,\vec B \, = \,\,-8$ | $(iv)$ $\theta = \,{60^o}$ |
