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Q.
A tetrahedron has vertices $O(0,0,0), A(1,2,1), B(2,1,3)$, $C(-1,1,2)$. If $\theta$ is the angle between the faces $O A B$ and $A B C$, then $\cos \theta=$
TS EAMCET 2020
Solution:
Equation of plane $O A B$ is given by,
$\begin{vmatrix}x-0 & y-0 & z-0 \\ 1-0 & 2-0 & 1-0 \\ 2-0 & 1-0 & 3-0\end{vmatrix}=0 \Rightarrow \begin{vmatrix}x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{vmatrix}=0$
$x(6-1)-y(3-2)+z(1-4)=0 $
$5 x-y-3 z=0$
Equation of plane $A B C$ is given by,
$\begin{vmatrix}x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1\end{vmatrix}=0$
$\begin{vmatrix}x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{vmatrix}=0$
$(x-1)(-1+2)-(y-2)(1+4)+(z-1)(-1-2)=0$
$x-1-5 y+10-3 z+3=0$
$x-5 y-3 z+12=0 \ldots $ (ii)
Then angle between planes represented by Eqs. (i) and (ii) is given by,
$\cos \theta=\frac{(5)(1)+(-1)(-5)+(-3)(-3)}{\sqrt{25+1+9} \sqrt{1+25+9}}=\frac{19}{35}$