volume of tetrahedron whose edges are $\vec{a},\vec{b},\vec{c}$

Question

If $\vec{a},\vec{b},\vec{c}$ be three non coplanar unit vectors, each inclined with other at an angle of $30^\circ.$ Then volume of tetrahedron whose edges are $\vec{a},\vec{b},\vec{c}$ is

what i try

Volume of tetrahedron whose edges are $\vec{a},\vec{b},\vec{c}$ is

$$\displaystyle \frac{1}{6}|(\vec{a}\times \vec{b})\cdot \vec{c}|\; \text{cubic unit}$$

from $\displaystyle \vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \alpha\; \hat{n}=\sin \alpha \hat{n}=\frac{1}{2}\hat{n}$

and $$(\vec{a}\times \vec{b})\cdot \vec{c}=|\vec{a}\times\vec{b}||\vec{c}|\cos \beta = \frac{1}{2}\cos \beta$$

How do i solve it help me please

Answer 1

The volume of tetrahedron with edges $~\vec a, ~\vec b~$ and $~\vec c~$ is $~\dfrac 16~[\vec{a}~~\vec{b}~~\vec{c}]~$ cubic unit.

Given that $\vec{a},\vec{b},\vec{c}$ be three non coplaner unit vectors, hence $~\vec a \cdot \vec a = \vec b \cdot \vec b = \vec c \cdot \vec c = 1~$.

Also each inclined with other at an angle of $~30^{\circ}~$, so $~\vec a \cdot \vec c = \vec b \cdot \vec c = \vec c \cdot \vec a = \cos 30^{\circ}=\dfrac{\sqrt 3}{2}~$.

Now $$[\vec{a}~~\vec{b}~~\vec{c}]^2=\begin{vmatrix} \vec a \cdot \vec a & \vec a \cdot \vec b & \vec a \cdot \vec c \\ \vec b \cdot \vec a & \vec b \cdot \vec b & \vec b \cdot \vec c \\ \vec c \cdot \vec a & \vec c \cdot \vec b &\vec c \cdot \vec c \end{vmatrix}$$ $$~~~~~~~~~~~~~~=\begin{vmatrix} 1 & \dfrac{\sqrt 3}{2} & \dfrac{\sqrt 3}{2} \\ \dfrac{\sqrt 3}{2} & 1 & \dfrac{\sqrt 3}{2} \\ \dfrac{\sqrt 3}{2} & \dfrac{\sqrt 3}{2} & 1 \end{vmatrix}$$ $$=\dfrac{3\sqrt 3}{4}-\dfrac 54$$ Hence the volume of tetrahedron is $~~\dfrac 16~[\vec{a}~~\vec{b}~~\vec{c}]~=~\dfrac 16~\left[\sqrt{\dfrac{3\sqrt 3}{4}-\dfrac 54}\right]=~\dfrac 1{12}~\sqrt{{3\sqrt 3}- 5}~$ cubic unit.

Answer 2

A better (equivalent) expression for the volume is

$$V=\dfrac16 \det(A) \tag{1}$$

where $A=(a,b,c)$ is the matrix whose columns are the coordinates of $a,b,c$.

The key idea is to consider the Gram matrix associated with $A$ :

$$G=A^TA=\begin{pmatrix}a.a&a.b&a.c\\ b.a&b.b&b.c\\ c.a&c.b&c.c\end{pmatrix}=\begin{pmatrix}1&\cos \alpha&\cos \alpha\\ \cos \alpha&1&\cos \alpha\\ \cos \alpha&\cos \alpha&1\end{pmatrix} \tag{2}$$

('a line of $A^T$ times a column of $A$" is nothing else than a dot product : $G$ "records" all possible dot products of vectors $a,b,c$ with themselves)

From (1), on can deduce that

$$det(G)=\underbrace{\det(A^T)\det(A)}_{\det(A)^2}=1+2 \cos^3 \alpha-3 \cos^2\alpha \tag{1}$$

Knowing that $\cos \alpha= \dfrac{\sqrt{3}}{2}$, I leave you the end calculations...

Answer 3

The existing two answers make use of Gram determinant.Here I continue your own idea.How to find $\cos\beta$？The key idea is considering the projection vector of $\vec{c}$ on the plane $\alpha$ spanned by $\vec{a}$ and $\vec{b}$.

Firstly,we try to find two unit vectors $\boldsymbol{i}$ and $\boldsymbol{j}$ on the plane $\alpha$,and they are orthogonal to each other,that is ,$\{\boldsymbol{i},\boldsymbol{j}\}$ is a unit orthogonal base of plane $\alpha$ .$\boldsymbol{i}$ and $\boldsymbol{j}$ can be constructed by using Gram-Schimidt orthogonalization process as follows:$$ \boldsymbol{i}=\vec{a},\boldsymbol{j}=2\vec{b}-\sqrt{3}\vec{a}. $$ Denote the projection vector of $\vec{c}$ on the plane $\alpha$ by $\text{Proj}_{\alpha}\vec{c}$,then $$ \text{Proj}_{\alpha}\vec{c}=\left( \vec{c}\cdot \boldsymbol{i} \right)\boldsymbol{i}+\left( \vec{c}\cdot \boldsymbol{j} \right)\boldsymbol{j}=\left( 2 \sqrt{3}-3 \right)\left( \vec{a}+\vec{b} \right). $$ So $$ \sin\beta=\frac{\vec{c}\cdot \text{Proj}_{\alpha}\vec{c}}{|\vec{c}|\left|\text{Proj}_{\alpha}\vec{c}\right|}=\frac{3 \sqrt{2}-\sqrt{6}}{2}. $$ So $\cos\beta=\sqrt{3\sqrt{3}-5}$.So the volumn of the tetrahedron is $\dfrac{1}{12}\cos\beta=\dfrac{\sqrt{3 \sqrt{3}-5}}{12}$.