Mathematics - Vector Question with Solution | TestHub

The volume of a tetrahedron is given by:

$$V=\frac{1}{6}|(\overrightarrow{\mathrm{OA}}\times \overrightarrow{\mathrm{OB}})\cdot \overrightarrow{\mathrm{OC}}|$$

Given the vectors, the volume is:

$$V=\frac{1}{6}\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|=\frac{1}{6}cubicunits$$

The area of the base is calculated as:

$$A=\frac{1}{2}|(\hat{i}+\hat{j}-\hat{i})\times (\hat{j}+\hat{k}-\hat{i})|=\frac{1}{2}|\hat{i}+\hat{k}|=\frac{1}{\sqrt{2}}squareunits$$

Therefore, the height of the tetrahedron is:

$$h=\frac{3\times V}{A}=\frac{3\times \frac{1}{6}}{\frac{1}{\sqrt{2}}}=\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}units$$