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If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that $|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=1$, $\overline{\mathrm{c}}=\lambda(\overline{\mathrm{a}} \times \overline{\mathrm{b}})$ and $|\overline{\mathrm{a}}|=\frac{1}{\sqrt{3}},|\overline{\mathrm{b}}|=\frac{1}{\sqrt{2}},|\overline{\mathrm{c}}|=\frac{1}{\sqrt{6}}$, then the angle between $\bar{a}$ and $\bar{b}$ is
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2095 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{2}$
Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$.
$$
\begin{aligned}
& \text { Since } \overline{\mathrm{c}}=\lambda(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \\
& \Rightarrow \overline{\mathrm{c}} \perp \overline{\mathrm{a}}, \overline{\mathrm{c}} \perp \overline{\mathrm{b}} \\
& \Rightarrow \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0
\end{aligned}
$$
Now,
$$
\begin{aligned}
& |\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=1 \\
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=1 \\
& \Rightarrow|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=1 \\
& \Rightarrow \frac{1}{3}+\frac{1}{2}+\frac{1}{6}+2\{|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta\}=1 \\
& \Rightarrow \cos \theta=0 \\
& \Rightarrow \theta=\frac{\pi}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Since } \overline{\mathrm{c}}=\lambda(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \\
& \Rightarrow \overline{\mathrm{c}} \perp \overline{\mathrm{a}}, \overline{\mathrm{c}} \perp \overline{\mathrm{b}} \\
& \Rightarrow \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0
\end{aligned}
$$
Now,
$$
\begin{aligned}
& |\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=1 \\
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=1 \\
& \Rightarrow|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=1 \\
& \Rightarrow \frac{1}{3}+\frac{1}{2}+\frac{1}{6}+2\{|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta\}=1 \\
& \Rightarrow \cos \theta=0 \\
& \Rightarrow \theta=\frac{\pi}{2}
\end{aligned}
$$
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