Let a→,b→,c→ be any three vectors. Then a→+b← b→+c→ c→+a→ is always equal to

a+b b+c c+a =a b d 110011101=a b c 1(1-0)-1(-1) =2a b c

Let a→,b→,c→ be any three vectors. Then a→+b← b→+c→ c→+a→ is always equal to

a+b b+c c+a =a b d 110011101=a b c 1(1-0)-1(-1) =2a b c

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Let $\vec{a}, \vec{b}, \vec{c}$ be any three vectors. Then $[\vec{a} + \overset{\leftarrow}{b} \vec{b} + \vec{c} \vec{c} + \vec{a}]$ is always equal to

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$[\vec{a} \vec{b} \vec{c}]$

$3 [\vec{a} \vec{b} \vec{c}]$

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$2 [\vec{a} \vec{b} \vec{c}]$

answer is A.

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Detailed Solution

$[a + b b + c c + a] = [a b d] |\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{matrix}| = [a b c] (1 (1 - 0) - 1 (- 1)) = 2 [a b c]$

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Let a→,b→,c→ be any three vectors. Then a→+b← b→+c→ c→+a→ is always equal to

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Let $\vec{a}, \vec{b}, \vec{c}$ be any three vectors. Then $[\vec{a} + \overset{\leftarrow}{b} \vec{b} + \vec{c} \vec{c} + \vec{a}]$ is always equal to