Question

Let $OABC$ be a regular tetrahedron with side length unity, then its volume (in cubic units) is

• A. $3\sqrt{2}$
• B. $6\sqrt{2}$
• C. $\frac{1}{3\sqrt{2}}$
• D. $\frac{1}{6\sqrt{2}}$

Answer 1

Answer: (D)

Let the position vector of $O,A,B,C$ are $\overrightarrow{0},\vec{a},\vec{b},\vec{c}$ respectively $\overrightarrow{OA}=\vec{a},\overrightarrow{OB}=\vec{b},\overrightarrow{OC}=\vec{c}$
$|\overrightarrow{OA}|=|\overrightarrow{OB}|=|\overrightarrow{OC}|=1$ [Given
Volume $=\frac{1}{6}\left[\begin{array}{lll}\overrightarrow{OA}& \overrightarrow{OB}& \overrightarrow{OC}\end{array}\right]=\frac{1}{6}\left[\begin{array}{lll}\vec{a}& \vec{b}& \vec{c}\end{array}\right]\dots \dots$ (i)
$\therefore {\left[\begin{array}{lll}\vec{a}& \vec{b}& \vec{c}\end{array}\right]}^2=\left|\begin{array}{ccc}\vec{a}\cdot \vec{a}& \vec{a}\cdot \vec{b}& \vec{a}\cdot \vec{c}\\ \vec{b}\cdot \vec{a}& \vec{b}\cdot \vec{b}& \vec{b}\cdot \vec{c}\\ \vec{c}\cdot \vec{a}& \vec{c}\cdot \vec{b}& \vec{c}\cdot \vec{c}\end{array}\right|$
$\Rightarrow {\left[\begin{array}{lll}\vec{a}& \vec{b}& \vec{c}\end{array}\right]}^2=\left|\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& 1& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& 1\end{array}\right|$
$\Rightarrow {\left[\begin{array}{lll}\vec{a}& \vec{b}& \vec{c}\end{array}\right]}^2=1\left(\frac{3}{4}\right)-\frac{1}{2}\left(\frac{1}{4}\right)+\frac{1}{2}\left(-\frac{1}{4}\right)=\frac{1}{2}$
$\Rightarrow \left[\begin{array}{lll}\vec{a}& \vec{b}& \vec{c}\end{array}\right]=\frac{1}{\sqrt{2}}$
From (i),
Volume of the tetrahedron $=\frac{1}{6\sqrt{2}}$ cubic units