Q36 / 90 · JEE Main 2012 (19 May Online) · Mathematics

If $a+b+c=0,|\vec{a}|=3,|\vec{b}|=5$ and $|\vec{c}|=7$, then the angle between $\vec{a}$ and $\vec{b}$ is

  • Correct answer
    A
    $\frac{\pi}{3}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{6}$
  • D
    $\frac{\pi}{2}$

Solution

Let $a+b+c=0 \Rightarrow(a+b)=-c$ $\Rightarrow(a+b)^2=c^2$ $$ \begin{aligned} & \Rightarrow a^2+b^2+2 a \cdot b=c^2 \\ & \Rightarrow 9+25+2.3 .5 \cos \theta=49 \\ & (\because|\vec{a}|=3,|\vec{b}|=5 \text { and }|\vec{c}|=7) \\ & \therefore \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \end{aligned} $$

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

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