Question

If the volume of a parallelepiped with $ \overrightarrow{a}\times \overrightarrow{b},\text{ }\overrightarrow{b}\times \overrightarrow{c},\text{ }\overrightarrow{c}\times \overrightarrow{a} $ as co-termmus edges is $ 9\text{ }cu $ units, then the volume of the parallelepiped with $ (a\times b)\times (b\times c),(b\times c)\times (c\times a), $ $ (c\times a)\times (a\times b) $ as co-terminus edges is

• A. 9 cu unit
• B. 729 cu unit
• C. 81 cu unit
• D. 27 cu unit
• E. 243 cu unit

Answer 1

Answer: (C)

Let $ \overrightarrow{A}=\overrightarrow{a}\times \overrightarrow{b},\overrightarrow{B}=\overrightarrow{b}\times \overrightarrow{c},\overrightarrow{C}=\overrightarrow{c}\times \overrightarrow{a} $
Given, $ [\overrightarrow{A}\overrightarrow{B}\overrightarrow{C}]=9\,cu\,unit $ Now, $ [\overrightarrow{A}\times \overrightarrow{B}\,\,\,\overrightarrow{B}\times \overrightarrow{C}\,\,\,\overrightarrow{C}\times \overrightarrow{A}] $
$=(\overrightarrow{A}\times \overrightarrow{B}).[(\,\overrightarrow{B}\times \overrightarrow{C})\times (\,\,\overrightarrow{C}\times \overrightarrow{A})] $
$=(\overrightarrow{A}\times \overrightarrow{B}).[[(\,\overrightarrow{B}\times \overrightarrow{C}).\overrightarrow{A}]\overrightarrow{C}-[(\overrightarrow{B}\times \overrightarrow{C}).\,\,\overrightarrow{C}]\overrightarrow{A}] $
$=(\overrightarrow{A}\times \overrightarrow{B}).[[(\,\overrightarrow{B}\overrightarrow{C}\overrightarrow{A}]\overrightarrow{C}-0] $
$={{[\overrightarrow{A}\overrightarrow{B}\overrightarrow{C}]}^{2}}={{9}^{2}} $
$=81\text{ }cu\text{ }unit $