Question

A tetrahedron has vertices at $O ( 0,0,0 )$, $A ( 1,2,1 ) , B ( 2,1,3 )$ and $C ( - 1,1,2 )$ . Then the angle between the faces $O A B$ and $A B C$will be

• A. $\cos ^ { - 1 } \left( \frac { 19 } { 35 } \right)$
• B. $\cos ^ { - 1 } \left( \frac { 17 } { 31 } \right)$
• C. $30 ^ { \circ }$
• D. $90 ^ { \circ }$

Answer 1

Answer: (A)

$\cos ^ { - 1 } \left( \frac { 19 } { 35 } \right)$

Answer 2

Angle between two plane faces is equal to the angle between the normals $\mathbf { n } _ { 1 }$ and $\mathbf { n } _ { 2 }$ to the planes. $\mathbf { n } _ { 1 }$ the normal of face OAB is given by

$\overrightarrow { O A } \times \overrightarrow { O B } = \left| \begin{array} { c c c } \mathbf { i } & \mathbf { j } & \mathbf { k } \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array} \right| = 5 \mathbf { i } - \mathbf { j } - 3 \mathbf { k }$ .....(i)

$\mathbf { n } _ { 2 }$ the normal of face ABC is given by $\overrightarrow { A B } \times \overrightarrow { A C }$ $2 - 1,1 - 2,3 - 1$ and $- 1 - 1,1 - 2,2 - 1$i.e., $1 , - 1,2$ and $- 2 , - 1,1$

…..(ii)

If θ be the angle between $\mathbf { n } _ { 1 }$ and $\mathbf { n } _ { 2 }$, then

$\Rightarrow \theta = \cos ^ { - 1 } \left( \frac { 19 } { 35 } \right)$.