Q.123. Vectors

Question 123.

Calculate an angle, θ, in a regular triangular pyramid

(tetrahedral) with 4 corners and 4 triangular faces. All

angles subtended by the edges at Z are equal and all the

corners are equidistant from Z.

Answer 123.
This is the same as the structure of methane, CH4. In that case C is at the position of Z of your figure and 4 hydrogen atoms are at A, B, C and T.

This figure can also be visualized as a cube ABCDA'B'C'D', where A, B, C, D, A' B' C' and D' are vertices of a cube such that AA', BB', CC' and DD' are the diagonals and Z can be taken as its centre of gravity which is the point of intesection of the diagonals.

A tetrahedral is formed by 4 of the 8 vertices of the cube such that any 2 of these 4 are diagonals of the 6 faces of the cube. These points can be A, C, B' and D'. If the length of the side of a cube is taken as 2, the coordinates of these four points with Z(0, 0, 0) as the origin can be taken as

A(-1, -1, -1), C(1, 1, -1), B'(-1, 1, 1) and D'(1, -1, 1)

The required angle is angle between vectors formed by any two of these 4 points.

Let us find the angle θ between ZA and ZC.
cos θ
= (ZA . ZC) / ( l ZA l * l ZC l )
= [(-1, -1, -1) . (1, 1, -1)] / [√3 . √3]
= (-1 -1 + 1) / 3 = - 1/3
θ =
arc cos (-1/3) =
180° - arc cos (1/3)
= 109° 28'

LINK to YA!