If $\vec{a}, \vec{b}, \vec{c}$ are 3 vectors such that $|\vec{a}|=5,|\vec{b}|=8,|\vec{c}|=11$ and…
- $\cos ^{-1} \frac{2}{5}$
- $\cos ^{-1} \frac{10}{11}$
- $\cos ^{-1} \frac{41}{55}$
- $\frac{\pi}{3}$
Solution
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)