The volume of Parallelopiped with edges a¯+b¯,b¯+c¯,c¯+a¯ is V1 and with edges a, b , c is V2 then V2V1=

V1=a→+b→b→+c→c+a→=1 1 00 1 11 0 1a→b→c→ ⇒V1=1(1-0)-1(0-1)+0a→b→c→ ⇒V1=2a→b→c→ ⇒V1=2V2⇒V2V1=12

The volume of Parallelopiped with edges a¯+b¯,b¯+c¯,c¯+a¯ is V1 and with edges a, b , c is V2 then V2V1=

V1=a→+b→b→+c→c+a→=1 1 00 1 11 0 1a→b→c→ ⇒V1=1(1-0)-1(0-1)+0a→b→c→ ⇒V1=2a→b→c→ ⇒V1=2V2⇒V2V1=12

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The volume of Parallelopiped with edges $\bar{a} + \bar{b}, \bar{b} + \bar{c}, \bar{c} + \bar{a}$ is V₁ and with edges a, b , c is V₂ then $\frac{V_{2}}{V_{1}} =$

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$\frac{1}{2}$

- $\frac{1}{2}$

answer is A.

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Detailed Solution

$\begin{aligned} V_{1} = [\begin{matrix} \vec{a} + \vec{b} & \vec{b} + \vec{c} & \vec{c + a} \end{matrix}] = |\begin{array}{lll} 1 1 0 \\ 0 1 1 \\ 1 0 1 \end{array}| [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] \\ \Rightarrow V_{1} = (1 (1 - 0) - 1 (0 - 1) + 0) [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] \\ \Rightarrow V_{1} = 2 [\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] \\ \Rightarrow V_{1} = 2 V_{2} \\ \Rightarrow \frac{V_{2}}{V_{1}} = \frac{1}{2} \end{aligned}$