If the vectors a→1,b→,c→ are such that each is inclined at an angle π3 with the other and |a→|=1,|b→|=2,|c→|=3, then |a→+b→+c→|=

a→⋅b→=|a→||b→|cos⁡π3=1⋅2⋅12=1b→⋅c→=|b→||c→|cos⁡π3=2⋅3⋅12=3c→⋅a→=|c→||a→|cos⁡π3=3⋅1⋅12=32|a→+b→+c→|2=(a→+b→+c→)⋅(a→+b→+c→)=|a→|2+|b→|2+|c→|2+2(a→⋅b→+b→⋅c→+c→⋅a→)=12+22+32+21+3+32=25∴|a→+b→+c→|=5.

If the vectors a→1,b→,c→ are such that each is inclined at an angle π3 with the other and |a→|=1,|b→|=2,|c→|=3, then |a→+b→+c→|=

a→⋅b→=|a→||b→|cos⁡π3=1⋅2⋅12=1b→⋅c→=|b→||c→|cos⁡π3=2⋅3⋅12=3c→⋅a→=|c→||a→|cos⁡π3=3⋅1⋅12=32|a→+b→+c→|2=(a→+b→+c→)⋅(a→+b→+c→)=|a→|2+|b→|2+|c→|2+2(a→⋅b→+b→⋅c→+c→⋅a→)=12+22+32+21+3+32=25∴|a→+b→+c→|=5.

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If the vectors ${\vec{a}}_{1}, \vec{b}, \vec{c}$ are such that each is inclined at an angle $\frac{π}{3}$ with the other and
$| \vec{a} | = 1, | \vec{b} | = 2, | \vec{c} | = 3$ , then $| \vec{a} + \vec{b} + \vec{c} | =$

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$3 \sqrt{2}$

answer is C.

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Detailed Solution

$\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos \frac{π}{3} = 1 \cdot 2 \cdot \frac{1}{2} = 1$

$\begin{array}{l} \vec{b} \cdot \vec{c} = | \vec{b} | | \vec{c} | \cos \frac{π}{3} = 2 \cdot 3 \cdot \frac{1}{2} = 3 \\ \vec{c} \cdot \vec{a} = | \vec{c} | | \vec{a} | \cos \frac{π}{3} = 3 \cdot 1 \cdot \frac{1}{2} = \frac{3}{2} \\ | \vec{a} + \vec{b} + \vec{c} |^{2} = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \\ = | \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \\ = 1^{2} + 2^{2} + 3^{2} + 2 (1 + 3 + \frac{3}{2}) = 25 \\ ∴ | \vec{a} + \vec{b} + \vec{c} | = 5 . \end{array}$