Regular Tetrahedron

Regular Tetrahedron

A regular tetrahedron is one in which all four faces are equilateral triangles.

Angle between any Edge and Face not Containing that Edge of Regular Tetrahedron:

Let us find the angle between edge OA and face OBC.

Since tetrahedron is regular and its face is an equilateral triangle, the required angle is the angle between vector $$\vec{a}$$
and vector along angle bisector of $$\angle BOC$$.

Let $$\vec{a},\vec{b},\vec{c}$$ be the vectors.

Angle bisector of $$\angle BOC$$ on $$\Delta OBC$$ is along vector $$\vec{b}+\vec{c}$$.

Therefore, required angle θ is the angle between vectors I.e., $$\vec{a}$$ and $$\vec{b}+\vec{c}$$,

$$\cos \theta =\frac{\vec{a}.\left( \vec{b}+\vec{c} \right)}{\left| {\vec{a}} \right|\left| \vec{b}+\vec{c} \right|}$$,

$$=\frac{\vec{a}.\vec{b}+\vec{a}.\vec{c}}{\left| \vec{b}+\vec{c} \right|}$$,

$${{\left| \vec{b}+\vec{c} \right|}^{2}}={{b}^{2}}+{{c}^{2}}+2\vec{b}.\vec{c}\times \cos \theta$$,

$$=1+1+2(1)(1)\cos {{60}^{o}}$$,

= 1+ 1 + 2 (½)

= 1 + 1 + 1 = 3

$$\left| \vec{b}+\vec{c}\right|=\sqrt{3}$$,

$$\cos \theta =\frac{\vec{a}.\vec{b}+\vec{a}.\vec{c}}{\left| \vec{b}+\vec{c} \right|}$$,

$$\cos \theta =\frac{1.1.\cos {{60}^{o}}+1.1.\cos {{60}^{o}}}{1\times \sqrt{3}}$$,

$$\cos \theta =\frac{1/2+1/2}{\sqrt{3}}$$,

$$\cos \theta =\frac{\frac{2}{2}}{\sqrt{3}}$$,

$$=\frac{1}{\sqrt{3}}$$.

$$\theta ={{\cos }^{-1}}\left( \frac{1}{\sqrt{3}} \right)$$.