If |overline{a}|=2, |overline{b}|=3, |overlin | vedclass

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(A) Given that $|\overline{a}|=2, |\overline{b}|=3, |\overline{c}|=5$ and the angle between each pair of vectors is $60^{\circ}$.We know that $\overli

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$\sqrt{69}$

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(A) Given that $|\overline{a}|=2, |\overline{b}|=3, |\overline{c}|=5$ and the angle between each pair of vectors is $60^{\circ}$.We know that $\overline{a} \cdot \overline{b} = |\overline{a}||\overline{b}| \cos 60^{\circ} = (2)(3)(\frac{1}{2}) = 3$.Similarly,$\overline{b} \cdot \overline{c} = |\overline{b}||\overline{c}| \cos 60^{\circ} = (3)(5)(\frac{1}{2}) = \frac{15}{2}$.And $\overline{c} \cdot \overline{a} = |\overline{c}||\overline{a}| \cos 60^{\circ} = (5)(2)(\frac{1}{2}) = 5$.Now,$|\overline{a}+\overline{b}+\overline{c}|^2 = |\overline{a}|^2+|\overline{b}|^2+|\overline{c}|^2 + 2(\overline{a} \cdot \overline{b} + \overline{b} \cdot \overline{c} + \overline{c} \cdot \overline{a})$.Substituting the values: $|\overline{a}+\overline{b}+\overline{c}|^2 = 2^2 + 3^2 + 5^2 + 2(3 + \frac{15}{2} + 5)$.$|\overline{a}+\overline{b}+\overline{c}|^2 = 4 + 9 + 25 + 2(\frac{6+15+10}{2}) = 38 + 31 = 69$.Therefore,$|\overline{a}+\overline{b}+\overline{c}| = \sqrt{69}$.

If $|\overline{a}|=2, |\overline{b}|=3, |\overline{c}|=5$ and each of the angles between the vectors $\overline{a}$ and $\overline{b}$,$\overline{b}$ and $\overline{c}$,and $\overline{c}$ and $\overline{a}$ is $60^{\circ}$,then the value of $|\overline{a}+\overline{b}+\overline{c}|$ is

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