Let a→, b→ and c→ be three vectors such that a→+b→+c→=0→. If |a→|=3, | b→|=4 and |c→|=5, then find the value of (a→⋅b→+b→⋅c→+c→⋅a→)

Since a→+b→+c→=0→⇒a→+b→=−c→ ⇒a→. c→+b→.c→=−c→.c→ ⇒a→. c→+b→.c→=−c2 ….(1) Similarly a→+ c→=−b→ ⇒a→. b→+c→.b→=−b2 …(2)Again b→+c→=-a→⇒b→. a→+c→. a→=−a2 …(3)Adding (1) (2) and (3), we get 2(a→ . b→+b→ . c→+c →. a→)=−(a2+b2+c2)⇒(a→ . b→+b→ . c→+c→ . a→)=−12(32+42+52)=−25⇒(a→ . b→+b→ . c→+c→ . a→)=−25

Let a→, b→ and c→ be three vectors such that a→+b→+c→=0→. If |a→|=3, | b→|=4 and |c→|=5, then find the value of (a→⋅b→+b→⋅c→+c→⋅a→)

Since a→+b→+c→=0→⇒a→+b→=−c→ ⇒a→. c→+b→.c→=−c→.c→ ⇒a→. c→+b→.c→=−c2 ….(1) Similarly a→+ c→=−b→ ⇒a→. b→+c→.b→=−b2 …(2)Again b→+c→=-a→⇒b→. a→+c→. a→=−a2 …(3)Adding (1) (2) and (3), we get 2(a→ . b→+b→ . c→+c →. a→)=−(a2+b2+c2)⇒(a→ . b→+b→ . c→+c→ . a→)=−12(32+42+52)=−25⇒(a→ . b→+b→ . c→+c→ . a→)=−25

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Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ . If $| \vec{a} | = 3, | \vec{b} | = 4 a n d | \vec{c} | = 5$ , then find the value of $(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$

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-5

-25

answer is D.

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Detailed Solution

Since $\vec{a} + \vec{b} + \vec{c} = \vec{0}$
$\Rightarrow \vec{a} + \vec{b} = - \vec{c}$
$\Rightarrow \vec{a} . \vec{c} + \vec{b} . \vec{c} = - \vec{c} . \vec{c}$
$\Rightarrow \vec{a} . \vec{c} + \vec{b} . \vec{c} = - c^{2}$ ….(1)
Similarly $\vec{a} + \vec{c} = - \vec{b}$
$\Rightarrow \vec{a} . \vec{b} + \vec{c} . \vec{b} = - b^{2}$ …(2)
Again $\vec{b} + \vec{c} = - \vec{a}$
$\Rightarrow \vec{b} . \vec{a} + \vec{c} . \vec{a} = - a^{2}$ …(3)
Adding (1) (2) and (3), we get
$2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = - (a^{2} + b^{2} + c^{2})$
$\Rightarrow (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = - \frac{1}{2} (3^{2} + 4^{2} + 5^{2}) = - 25$
$\Rightarrow (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = - 25$