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