If $\vec{a}, \vec{b}, \vec{c}$ are 3 vectors such that $|\vec{a}|=5,|\vec{b}|=8,|\vec{c}|=11$ and…

If $\vec{a}, \vec{b}, \vec{c}$ are 3 vectors such that $|\vec{a}|=5,|\vec{b}|=8,|\vec{c}|=11$ and $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$ then the angle between the vectors $\vec{a}$ and $\vec{b}$ is
  1. $\cos ^{-1} \frac{2}{5}$
  2. $\cos ^{-1} \frac{10}{11}$
  3. $\cos ^{-1} \frac{41}{55}$
  4. $\frac{\pi}{3}$

Solution

$\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \Rightarrow \vec{a}+\vec{b}=-\vec{c}$
Squaring both sides $\begin{aligned} & |\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=|\vec{c}|^2 \\ & \Rightarrow 25+64+2|\vec{a} \| \vec{b}| \cos \theta=121 \\ & \Rightarrow(5)(8) \cos \theta=16 \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{5}\right) \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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