a¯,b¯ and c¯ re three vectors such that a¯+b¯+c¯=0¯ |a¯|=1,|b¯|=2 |c¯|=3. Then a¯⋅b¯+b¯⋅c¯+c¯⋅a¯ is

|a→+b→+c→|2=(a→+b→+c→)⋅(a→+b→+c→)⇒0=|a→|2+|b→|2+|c→|2+2(a→⋅b→+b→⋅c→+c→⋅a→)⇒0=1+4+9+2(a→⋅b→+b→⋅c→+c→⋅a→)⇒a→⋅b→+b→⋅c→+a→⋅c→=−7

a¯,b¯ and c¯ re three vectors such that a¯+b¯+c¯=0¯ |a¯|=1,|b¯|=2 |c¯|=3. Then a¯⋅b¯+b¯⋅c¯+c¯⋅a¯ is

|a→+b→+c→|2=(a→+b→+c→)⋅(a→+b→+c→)⇒0=|a→|2+|b→|2+|c→|2+2(a→⋅b→+b→⋅c→+c→⋅a→)⇒0=1+4+9+2(a→⋅b→+b→⋅c→+c→⋅a→)⇒a→⋅b→+b→⋅c→+a→⋅c→=−7

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$\bar{a}, \bar{b} and \bar{c}$ re three vectors such that $\bar{a} + \bar{b} + \bar{c} = \bar{0}$ $| \bar{a} | = 1, | \bar{b} | = 2$ $| \bar{c} | = 3 . Then \bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}$ is

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-2

-7

-14

-6

answer is C.

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Detailed Solution

$| \vec{a} + \vec{b} + \vec{c} |^{2} = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})$
$\begin{array}{l} \Rightarrow 0 = | \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2} + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \\ \Rightarrow 0 = 1 + 4 + 9 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \\ \Rightarrow \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{a} \cdot \vec{c} = - 7 \end{array}$