Q46 / 120 · VITEEE 2016 · Mathematics

If $|\vec{a}|=3,|\vec{b}|=2,|\vec{c}|=1$ then the value of $|\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}|$ is (given that $\vec{a}+\vec{b}+\vec{c}=0$ )

  • A
    $-7$
  • Correct answer
    B
    7
  • C
    14
  • D
    $-14$

Solution

$\begin{aligned} & \vec{a}+\vec{b}+\vec{c}=0 \\ & \Rightarrow(\vec{a}+\vec{b}+\vec{c})^{2}=0 \\ & \Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\ & \Rightarrow 9+4+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\ & \Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=7 \end{aligned}$

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

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