Question

A tetrahedron has vertices at $O(0,0,0),A(1,2,1),B(2,1,3)\&C(–1,1,2).$ Then the angle between the faces $OAB\&ABC$ will be?

Answer 1

Finding the angle between the faces $OAB\&ABC$ :

Given that tetrahedron has vertices at $O(0,0,0),A(1,2,1),B(2,1,3)\&C(–1,1,2).$

Vector perpendicular to $OAB,R1$ $=OAxOB$

$=\begin{array}{ccc}i& j& k\\ 1& 2& 1\\ 2& 1& 3\end{array}$

$R1=5i–j–3k$

Vector perpendicular to $ABCisR2=ABxAC$

$=\begin{array}{ccc}i& j& k\\ 1& -1& 2\\ -2& -1& 1\end{array}$

$=i–5j–3k$

angle between the faces $OAB\&ABC$ will be:

$\begin{array}{rcl}Cos\theta & =& R1/R2/\left[\right|R1\left|\right|R2\left|\right]\\ & =& \frac{19}{35}\\ \theta & =& {\cos }^{-1}\left(\frac{19}{35}\right)\end{array}$

Hence, the angle between the faces $OAB\&ABC$ is $\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\begin{array}{rcl}& & \frac{\mathbf{19}}{\mathbf{35}}\end{array}\mathbf{\right)}$.