Let $\vec{a}, \vec{b}$ be two unit vector. If $\vec{c}=\vec{a}+2 \vec{b}$ and $\vec{d}=5 \vec{a}-4 \vec{b}$…

Let $\vec{a}, \vec{b}$ be two unit vector. If $\vec{c}=\vec{a}+2 \vec{b}$ and $\vec{d}=5 \vec{a}-4 \vec{b}$ are perpendicular to each other, then the angle between $\vec{a}$ and $\vec{b}$ is
  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{8}$

Solution

Since, $|\vec{a}|=|\vec{b}|-1$ $\therefore \vec{c} \cdot \vec{d}=0 \Rightarrow(\vec{a}+2 \vec{b}) \cdot(5 \vec{a}-4 \vec{b})=0$ $\begin{aligned} & \Rightarrow 5|\vec{a}|^2-4 \vec{a} \cdot \vec{b}+10 \vec{a} \cdot \vec{b}-8|\vec{b}|^2=0 \\ & \Rightarrow 5+6 \vec{a} \cdot \vec{b}-8=0 \Rightarrow 6 \vec{a} \cdot \vec{b}=3\end{aligned}$ $\Rightarrow|\vec{a}||\vec{b}| \cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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