Question

The volume (in cubic unit) of the tetrahedron with edges $\mathrm{i}+\mathrm{j}+\mathrm{k},\mathrm{i}-\mathrm{j}+\mathrm{k}$ and $\mathrm{i}+2\mathrm{j}-\mathrm{k}$ is?

• A. $4$
• B. $\frac{2}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B

$\frac{2}{3}$

Explanation of the correct option:

Given: The edges of of the tetrahedron are $\mathrm{i}+\mathrm{j}+\mathrm{k},\mathrm{i}-\mathrm{j}+\mathrm{k}$ and $\mathrm{i}+2\mathrm{j}-\mathrm{k}$.

SInce, Volume of tetrahedron$=\frac{1}{6}\left[\begin{array}{ccc}{u}_1& u_2& u_3\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right]$

Volume of tetrahedron$=\frac{1}{6}\left[\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 1& 2& -1\end{array}\right]$

$$\begin{gathered} =\frac{1}{6}\left[1(1-2)-1(-1-1)+1(2+1)\right] \\ =\frac{1}{6}\left[-1+2+3\right] \\ =\frac{4}{6} \\ =\frac{2}{3} \end{gathered}$$

Therefore, volume of the given tetrahedron is $\frac{2}{3}$ cubic unit.

Hence option B is the correct option.