# A tetrahedron has vertices $O(0,0,0),\:A(1,2,1)\:B(2,1,3)\:C(-1,1,2)$. The angle between the faces $OAB$ and $ABC$ is ?

$(a)\:\:90^{\circ}\:\:\:\qquad\:\:(b)\:\:30^{\circ}\:\:\:\qquad\:\:(c)\:\:cos^{-1}(19/35)\:\:\:\qquad\:\:(d)\:\:cos^{-1}(17/31)$

Let $\overrightarrow n_1$ and $\overrightarrow n_2$ be the normals to the face $OAB$ and $ABC$ respectively.
Given $O(0,0,0),\:A(1,2,1)\:C(2,1,3)\:C(-1,1,2)$ are vertices of tetrahedran.
$\overrightarrow {OA}=(1,2,1) \:and\:\:\overrightarrow {OB}=(2,1,3).$
Then $\overrightarrow n_1=\overrightarrow {OA}\times\overrightarrow {OB}=\left |\begin {array}{ccc} \hat i & \hat j & \hat k\\1 & 2 & 1\\ 2 & 1 & 3\end {array}\right |$
$\overrightarrow n_1=(5,-1,-3)$
$\overrightarrow {AB}=(1,-2,2)\:and\:\overrightarrow {AC}=(-2,-1,1)$
Then $\overrightarrow n_2=\overrightarrow {AB}\times\overrightarrow {AC}=\left |\begin {array}{ccc} \hat i & \hat j & \hat k\\1 & -2 & 2\\- 2 & -1 & 1\end {array}\right |$
$\overrightarrow n_2=(1,-5,-3)$
Angle between the faces $OAB$ and $ABC$ is same as the angle between their normals.
which is given by $cos\theta=\large\frac{\overrightarrow n_1.\overrightarrow n_2}{|\overrightarrow n_1||\overrightarrow n_2|}$
$=\large\frac{5+5+9}{\sqrt {35}.\sqrt {35}}=\frac{19}{35}$
$\therefore$ The required angle is $cos^{-1}\large\frac{19}{35}$