Question

The volume of a tetrahedron formed by the coterminus edges $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ is $3$, then the volume of the parallelepiped formed by the coterminus edges $\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{c}+\overrightarrow{a}$ is

• A. $6$
• B. $18$
• C. $36$
• D. $9$

Answer 1

The correct option is C $36$

$3=\frac{1}{6}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$

or $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=18$

Volume of the required parallelepiped
$=[\overrightarrow{a}+\overrightarrow{b}\overrightarrow{b}+\overrightarrow{c}\overrightarrow{c}+\overrightarrow{a}]$

$=\left|\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 1\end{array}\right|$

$=2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=36$