Q65 / 180 · KCET 2010 · Mathematics

If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are unit vectors, such that $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}$, then $3 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}$ is

  • A
    $-1$
  • B
    1
  • Correct answer
    C
    $-3$
  • D
    3

Solution

$\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are unit vectors, then $|\overrightarrow{\mathbf{a}}|=|\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{c}}|=1$ Given, $$ \begin{aligned} \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}} &=0 \\ \overrightarrow{\mathbf{a}} &=-(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}) \end{aligned} $$ Squaring on both sides $$ \begin{aligned} \overrightarrow{\mathbf{a}^{2}} &=(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})^{2} \\ \overrightarrow{\mathbf{a}^{2}} &=\overrightarrow{\mathbf{b}^{2}}+\overrightarrow{\mathbf{c}^{2}}+2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} \\ |\overrightarrow{\mathbf{a}}|^{2} &=|\overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{c}}|^{2}+2(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}) \quad\left(\because \overrightarrow{\mathbf{a}^{2}}=|\overrightarrow{\mathbf{a}}|^{2}\right) \end{aligned} $$ $$ \begin{aligned} &1=1+1+2(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}})\\ &\Rightarrow \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}=-1 / 2\\ &\text { Similarly, } \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}=-1 / 2\\ &\text { Hence, } 3 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}\\ &=3(-1 / 2)+2(-1 / 2)+(-1 / 2)\\ &=(3+2+1)(-1 / 2)\\ &=6(-1 / 2)=-3 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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