Question

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 $OAB$ and $ABC$, then $\cos \theta =$

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{19}{35}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{17}{31}$

Answer 1

Answer: (B)

Equation of plane $OAB$ is given by,
$\left|\begin{array}{ccc}x-0& y-0& z-0\\ 1-0& 2-0& 1-0\\ 2-0& 1-0& 3-0\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x& y& z\\ 1& 2& 1\\ 2& 1& 3\end{array}\right|=0$
$x(6-1)-y(3-2)+z(1-4)=0$
$5x-y-3z=0$
Equation of plane $ABC$ is given by,
$\left|\begin{array}{ccc}x-1& y-2& z-1\\ 2-1& 1-2& 3-1\\ -1-1& 1-2& 2-1\end{array}\right|=0$
$\left|\begin{array}{ccc}x-1& y-2& z-1\\ 1& -1& 2\\ -2& -1& 1\end{array}\right|=0$
$(x-1)(-1+2)-(y-2)(1+4)+(z-1)(-1-2)=0$
$x-1-5y+10-3z+3=0$
$x-5y-3z+12=0\dots$ (ii)
Then angle between planes represented by Eqs. (i) and (ii) is given by,
$\cos \theta =\frac{(5)(1)+(-1)(-5)+(-3)(-3)}{\sqrt{25+1+9}\sqrt{1+25+9}}=\frac{19}{35}$