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Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be three vectors such that the magnitude of $\mathbf{b}$ is twice that of $\mathbf{a}$ and magnitude of $\mathbf{c}$ is three times that of $\mathbf{a}$. If the angle between each pair of vectors is $\frac{\pi}{3}$ and $|a+b+c|=5$, then $|c|+|a|+|b|=$
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Verified Answer
The correct answer is:
6
Given, $|\mathbf{b}|=2|\mathbf{a}|$ and $|\mathbf{c}|=3|\mathbf{a}|$
Angle between each pair of vector is $\pi / 3$.
$\begin{aligned}
& |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2|\mathbf{a}||\mathbf{b}| \cos \pi / 3 \\
& +2|\mathbf{b}||\mathbf{c}| \cos \frac{\pi}{3}+2|\mathbf{a}||\mathbf{c}| \cos \frac{\pi}{3} \\
& 25=|\mathbf{a}|^2+4|\mathbf{a}|^2+9|\mathbf{a}|^2+4|\mathbf{a}|^2 \times \frac{1}{2}+2 \times 6|\mathbf{a}|^2 \times \frac{1}{2}+6|\mathbf{a}|^2 \times \frac{1}{2} \\
& 25=25|\mathbf{a}|^2 \Rightarrow|\mathbf{a}|=1 \\
& \therefore \quad|\mathbf{b}|=2 \Rightarrow|\mathbf{c}|=3 \Rightarrow|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|=1+2+3=6 \\
&
\end{aligned}$
Angle between each pair of vector is $\pi / 3$.
$\begin{aligned}
& |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2|\mathbf{a}||\mathbf{b}| \cos \pi / 3 \\
& +2|\mathbf{b}||\mathbf{c}| \cos \frac{\pi}{3}+2|\mathbf{a}||\mathbf{c}| \cos \frac{\pi}{3} \\
& 25=|\mathbf{a}|^2+4|\mathbf{a}|^2+9|\mathbf{a}|^2+4|\mathbf{a}|^2 \times \frac{1}{2}+2 \times 6|\mathbf{a}|^2 \times \frac{1}{2}+6|\mathbf{a}|^2 \times \frac{1}{2} \\
& 25=25|\mathbf{a}|^2 \Rightarrow|\mathbf{a}|=1 \\
& \therefore \quad|\mathbf{b}|=2 \Rightarrow|\mathbf{c}|=3 \Rightarrow|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|=1+2+3=6 \\
&
\end{aligned}$
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