Each of the angle between vectors a→,b→and c→, is equal to 60∘.if |a→|=4,|b→|=2 and |c→|=6then the modulus of a→+b→+c→, is

We know that |a→+b→+c→|2=|a→|2+|b→|2+|c→|2+2(a→⋅b→+b→⋅c→+c→⋅a→) .....(1)we find that a→⋅b→=|a→||b→|cos⁡π3=4×2×12=4b→⋅c→=|b→||c→|cos⁡π3=2×6×12=6c→⋅a→=|c→||a→|cos⁡π3=6×4×12=12Substituting these values and |a→|=4,|b→|=2 and |c→|=6 in (I) we obtian.|a→+b→+c→|2=42+22+62+2(4+6+12)=100⇒ |a→+b→+c→|=10

Each of the angle between vectors a→,b→and c→, is equal to 60∘.if |a→|=4,|b→|=2 and |c→|=6then the modulus of a→+b→+c→, is

We know that |a→+b→+c→|2=|a→|2+|b→|2+|c→|2+2(a→⋅b→+b→⋅c→+c→⋅a→) .....(1)we find that a→⋅b→=|a→||b→|cos⁡π3=4×2×12=4b→⋅c→=|b→||c→|cos⁡π3=2×6×12=6c→⋅a→=|c→||a→|cos⁡π3=6×4×12=12Substituting these values and |a→|=4,|b→|=2 and |c→|=6 in (I) we obtian.|a→+b→+c→|2=42+22+62+2(4+6+12)=100⇒ |a→+b→+c→|=10

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Each of the angle between vectors $\vec{a}, \vec{b}$ and $\vec{c}$ , is equal to $60^{\circ} .$ if $| \vec{a} | = 4, | \vec{b} | = 2$ and $| \vec{c} | = 6$ then the modulus of $\vec{a} + \vec{b} + \vec{c}$ , is

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Detailed Solution

We know that

$| \vec{a} + \vec{b} + \vec{c} |^{2}$

$= | \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) . . . . . (1)$

we find that

$\begin{array}{l} \vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos \frac{π}{3} = 4 \times 2 \times \frac{1}{2} = 4 \\ \vec{b} \cdot \vec{c} = | \vec{b} | | \vec{c} | \cos \frac{π}{3} = 2 \times 6 \times \frac{1}{2} = 6 \\ \vec{c} \cdot \vec{a} = | \vec{c} | | \vec{a} | \cos \frac{π}{3} = 6 \times 4 \times \frac{1}{2} = 12 \end{array}$

Substituting these values and $| \vec{a} | = 4, | \vec{b} | = 2 and | \vec{c} | = 6$ in (I) we obtian.

$\begin{array}{l} | \vec{a} + \vec{b} + \vec{c} |^{2} = 4^{2} + 2^{2} + 6^{2} + 2 (4 + 6 + 12) = 100 \\ \Rightarrow | \vec{a} + \vec{b} + \vec{c} | = 10 \end{array}$