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 OAB and ABC will be

• A. $\cos ^ { - 1 } \left( \frac { 19 } { 35 } \right)$
• B. $\cos ^ { - 1 } \left( \frac { 17 } { 31 } \right)$
• C. 30⁰
• D. 90⁰

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 $n _ { 1 }$ and $n _ { 2 }$ to the planes. $\mathrm { n } _ { 1 }$, the normal to the face OAB is given by ……(i)

, the normal to the face ABC, is given by $\overrightarrow { A B } \times \overrightarrow { A C }$.

$\mathbf { n } _ { 2 } = \left| \begin{array} { c c c } \mathbf { i } & \mathbf { j } & \mathbf { k } \\ 1 & - 1 & 2 \\ - 2 & - 1 & 1 \end{array} \right| = \mathbf { i } - 5 \mathbf { j } - 3 \mathbf { k }$ ……(ii)

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

Then $\cos \theta = \frac { 19 } { 35 }$

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