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 th

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:

Accepted Answer

The angle between faces is the angle between their normal vectors.
Normal to face

O A B

:

n_{1} = O A \times OB = \hat{i} 12 \hat{j} 21 \hat{k} 13 = 5 \hat{i} - \hat{j} - 3 \hat{k}

.
Normal to face

A BC

:

n_{2} = A B \times A C = (OB - O A) \times (OC - O A) = (1, - 1, 2) \times (- 2, - 1, 1)

n_{2} = \hat{i} 1 - 2 \hat{j} - 1 - 1 \hat{k} 21 = \hat{i} - 5 \hat{j} - 3 \hat{k}

.
Angle

θ

:

cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣} = \frac{∣ ( 5 ) ( 1 ) + ( - 1 ) ( - 5 ) + ( - 3 ) ( - 3 ) ∣}{25 + 1 + 9 1 + 25 + 9} = \frac{5 + 5 + 9}{35} = \frac{19}{35}

.

θ = cos^{- 1} (\frac{19}{35})

.

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 $A BC$ will be: