Q218 / 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 the angle between $\vec{a}$ and $\vec{b}$ ?

  • A
    $30^{\circ}$
  • B
    $45^{\circ}$
  • Correct answer
    C
    $60^{\circ}$
  • D
    $75^{\circ}$

Solution

$\begin{aligned} & \text { Since } \vec{a}+\vec{b}+\vec{c}=0 \\ & \Rightarrow \vec{a}+\vec{b}=-\vec{c} \\ & \Rightarrow|\vec{a}+\vec{b}|=|-\vec{c}|=|\vec{c}|\end{aligned}$ $\begin{aligned} & \Rightarrow|\vec{a}+\vec{b}|^2=|\vec{c}|^2 \\ & \Rightarrow|\vec{a}|^2+|\vec{b}|^2+2(\vec{a} \cdot \vec{b})=|\vec{c}|^2 \\ & \Rightarrow(10)^2+(6)^2+2(\vec{a} \cdot \vec{b})=(14)^2 \\ & \Rightarrow 2(\vec{a} \cdot \vec{b})=60 \\ & \Rightarrow \vec{a} \cdot \vec{b}=30 \\ & \Rightarrow \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{30}{10 \times 6}=\frac{1}{2} \\ & \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow \theta=60^{\circ}\end{aligned}$

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

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