Q217 / 270 · NDA 2016 (Phase 2) · Mathematics

Paragraph: Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\overrightarrow{\mathrm{a}}|=10,|\overrightarrow{\mathrm{~b}}|=6$ and $|\overrightarrow{\mathrm{c}}|=14$.
Question: What is $\overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} . \overrightarrow{\mathrm{c}} .+\overrightarrow{\mathrm{c}}$. $\overrightarrow{\mathrm{a}}$. equal to?

  • A
    $-332$
  • Correct answer
    B
    $-166$
  • C
    $0$
  • D
    $166$

Solution

We have $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ So $|\vec{a}+\vec{b}+\vec{c}|=0$ $\begin{aligned} & \Rightarrow|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\ & \Rightarrow 0=(10)^2+(6)^2+(14)^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\ & \Rightarrow 0=100+36+196+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\ & \Rightarrow-\frac{332}{2}=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} \\ & \Rightarrow-166=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} \end{aligned}$

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

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