JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 4)

The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors $$\overrightarrow a \,,\,\overrightarrow b ,\overrightarrow c $$ such that $$\widehat a\,.\,\widehat b = \widehat b\,.\,\widehat c = \widehat c\,.\,\widehat a = {1 \over 2}.$$ Then, the volume of the parallelopiped is :
$${1 \over {\sqrt 2 }}$$
$${1 \over {2\sqrt 2 }}$$
$${{\sqrt 3 } \over 2}$$
$${1 \over {\sqrt 3 }}$$

Explanation

The important thing to remember in this is the formula

$${\left[ {\overrightarrow x \,.\,\overrightarrow y \,.\,\overrightarrow z } \right]^2} = \left| {\matrix{ {\overrightarrow x \,.\,\overrightarrow x } & {\overrightarrow x \,.\,\overrightarrow y } & {\overrightarrow x \,.\,\overrightarrow z } \cr {\overrightarrow y \,.\,\overrightarrow x } & {\overrightarrow y \,.\,\overrightarrow y } & {\overrightarrow y \,.\,\overrightarrow z } \cr {\overrightarrow z \,.\,\overrightarrow x } & {\overrightarrow z \,.\,\overrightarrow y } & {\overrightarrow z \,.\,\overrightarrow z } \cr } } \right|$$

Volume of the parallelopiped $$v = \left[ {\matrix{ {\widehat a} & {\widehat b} & {\widehat c} \cr } } \right]$$

$$\therefore$$ $${v^2} = {\left[ {\matrix{ {\widehat a} & {\widehat b} & {\widehat c} \cr } } \right]^2} = \left| {\matrix{ {\widehat a\,.\,\widehat a} & {\widehat a\,.\,\widehat b} & {\widehat a\,.\,\widehat c} \cr {\widehat b\,.\,\widehat a} & {\widehat b\,.\,\widehat b} & {\widehat b\,.\,\widehat c} \cr {\widehat c\,.\,\widehat a} & {\widehat c\,.\,\widehat b} & {\widehat c\,.\,\widehat c} \cr } } \right|$$

$$ = \left| {\matrix{ 1 & {{1 \over 2}} & {{1 \over 2}} \cr {{1 \over 2}} & 1 & {{1 \over 2}} \cr {{1 \over 2}} & {{1 \over 2}} & 1 \cr } } \right| = {1 \over 2}$$ (on evaluation)

$$ \Rightarrow v = \left[ {\matrix{ {\widehat a} & {\widehat b} & {\widehat c} \cr } } \right] = {1 \over {\sqrt 2 }}$$

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