The number of faces of a triangular pyramid or tetrahedron is _______.

Options

A. `cos^(-1)(1/3)`
B. `cos^(-1)(1/(sqrt(3))`
C. `cos^(-1)(2/3)`
D. none of these

Answer

A

Explanation

Let `OABC` be the regular tetradedron. Taking `O` as the origin let the position vectors of `A,B,C` be `veca, vecb` and `vecc` respectively.
We have
`OA=OB=OC=AB=BC=CA`
`implies|veca|=|vecb|=|vecc|` and `veca.vecb=vecb.vecc=vecc.veca`
Let `theta` be the acute between focus `OAB` and `OBC`. Then `theta` is also the angle between their normals.

i.e. `vecn_(1)=vecaxxvecb` and `vecn_(2)=vecbxxvecc`
`:. cos theta=|(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)|`
`implies cos theta=|(veca.{vecbxx(vecbxxvecc)})/(|vecaxxvecb||vecbxxvecc|)|`
i.e. `vecn_(1)=vecaxxvecb` and `vecn_(2)=vecbxxvecc`
`:.cos theta=|(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)|`
`impliescos theta=|(veca.{vecbxx(vecbxxvecc)})/(|vecaxxvecb||vecbxxvecc|)|`
`impliescos theta=|(veca.{(vecb. vecc)vecb-(vecb.vecb)vecc})/(|vecaxxvecb||vecbxxvecc|)|`
`impliescos theta=|({(veca.vecb)(vecb.vecc)-(vecb.vecb)(veca.vecc)})/(|vecaxxvecb||vecbxxvecc|)|`
Now,
`veca.vecb=vecb.vecc=vecc=veca=|veca||veca|"cos"(pi)/3=1/2|veca|^(2)`
and
`|vecaxxvecb|=|veca||vecb|"sin"(pi)/3=(sqrt(3))/2|veca|^(2)=|vecaxxvecc|=|veccxxveca|`
`:. cos theta=1/3impliestheta=cos^(-1)(1/3)^(2)`