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Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a}$ is perpendicular to $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{b}}$ is perpendicular to $\overrightarrow{\mathrm{c}}$. If $|\overrightarrow{\mathrm{a}}|=2,|\overrightarrow{\mathrm{b}}|=3,|\overrightarrow{\mathrm{c}}|=5$ and $|\vec{a}+\vec{b}+\vec{c}|=4 \sqrt{3}$, then the angle between $\vec{a}$ and $\vec{c}$ is
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The correct answer is:
$\frac{\pi}{3}$
Given that $\vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0,|\vec{a}|=2,|\vec{b}|=3$
$\begin{aligned} & |\vec{c}|=5 \text { and }|\vec{a}+\vec{b}+\vec{c}|=4 \sqrt{3} \\ & \text { Now, }|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a} \\ & \Rightarrow 48=4+9+25+0+0+2|\vec{c} \| \vec{a}| \cos \theta \\ & \Rightarrow 48-38=2.5 .2 \cos \theta \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\end{aligned}$
$\begin{aligned} & |\vec{c}|=5 \text { and }|\vec{a}+\vec{b}+\vec{c}|=4 \sqrt{3} \\ & \text { Now, }|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a} \\ & \Rightarrow 48=4+9+25+0+0+2|\vec{c} \| \vec{a}| \cos \theta \\ & \Rightarrow 48-38=2.5 .2 \cos \theta \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\end{aligned}$
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