Mathematics - Vector Question with Solution | TestHub

To find the volume of the tetrahedron, we use the formula $V=\frac{1}{6}|[\bar{a}\bar{b}\bar{c}]|$.

Given that $\bar{a},\bar{b},\bar{c}$ are unit vectors, $|\bar{a}|=|\bar{b}|=|\bar{c}|=1$.

The angle between any two vectors is $30^{\circ }$.

So, $\bar{a}\cdot \bar{b}=|\bar{a}||\bar{b}|\cos 30^{\circ }=1\cdot 1\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$.

Similarly, $\bar{b}\cdot \bar{c}=\frac{\sqrt{3}}{2}$ and $\bar{c}\cdot \bar{a}=\frac{\sqrt{3}}{2}$.

The square of the scalar triple product is given by the determinant of the Gram matrix:

$$[\bar{a}\bar{b}\bar{c}{]}^2=\left|\begin{array}{ccc}\bar{a}\cdot \bar{a}& \bar{a}\cdot \bar{b}& \bar{a}\cdot \bar{c}\\ \bar{b}\cdot \bar{a}& \bar{b}\cdot \bar{b}& \bar{b}\cdot \bar{c}\\ \bar{c}\cdot \bar{a}& \bar{c}\cdot \bar{b}& \bar{c}\cdot \bar{c}\end{array}\right|$$

Substitute the dot product values:

$$[\bar{a}\bar{b}\bar{c}{]}^2=\left|\begin{array}{ccc}1& \frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& 1& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}& 1\end{array}\right|$$

Calculate the determinant:

$$[\bar{a}\bar{b}\bar{c}{]}^2=1\left(1-\frac{3}{4}\right)-\frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}}{2}-\frac{3}{4}\right)+\frac{\sqrt{3}}{2}\left(\frac{3}{4}-\frac{\sqrt{3}}{2}\right)$$

$$[\bar{a}\bar{b}\bar{c}{]}^2=1\left(\frac{1}{4}\right)-\frac{\sqrt{3}}{2}\left(\frac{2\sqrt{3}-3}{4}\right)+\frac{\sqrt{3}}{2}\left(\frac{3-2\sqrt{3}}{4}\right)$$

$$[\bar{a}\bar{b}\bar{c}{]}^2=\frac{1}{4}-\frac{6-3\sqrt{3}}{8}+\frac{3\sqrt{3}-6}{8}$$

$$[\bar{a}\bar{b}\bar{c}{]}^2=\frac{1}{4}+\frac{3\sqrt{3}-6}{8}+\frac{3\sqrt{3}-6}{8}$$

$$[\bar{a}\bar{b}\bar{c}{]}^2=\frac{1}{4}+\frac{6\sqrt{3}-12}{8}=\frac{1}{4}+\frac{3\sqrt{3}-6}{4}$$

$$[\bar{a}\bar{b}\bar{c}{]}^2=\frac{1+3\sqrt{3}-6}{4}=\frac{3\sqrt{3}-5}{4}$$

So, $|[\bar{a}\bar{b}\bar{c}]|=\sqrt{\frac{3\sqrt{3}-5}{4}}=\frac{\sqrt{3\sqrt{3}-5}}{2}$.

The volume of the tetrahedron is:

$$V=\frac{1}{6}|[\bar{a}\bar{b}\bar{c}]|=\frac{1}{6}\cdot \frac{\sqrt{3\sqrt{3}-5}}{2}=\frac{\sqrt{3\sqrt{3}-5}}{12}$$

The final answer is $\boxed{\frac{\sqrt{3\sqrt{3}-5}}{12}}$.