1.

What is the angle substended by an edge of regular tetrahedron at its centre?

Answer»

`cos^(-1)((-1)/(2))`
`cos^(-1)((-1)/(sqrt(2)))`
`cos^(-1)((-1)/(3))`
`cos^(-1)((-1)/(sqrt(3)))`

Solution :
`veca,vecb,vecc`
are unit vector
`veca^^vecb=vecb^^vecc=vecc^^veca=(pi)/(3)`
centre p `((veco +veca+vecb+vecc)/(4))`
Now anglebetween `VEC(AP)&vec(BP)`
`costheta=(vec(AP)*vec(BP))/(|vec(AP)||vec(BP))=(((veca+vecb+vecc)/(4)-veca)*((veca+vecb+vecc)/(4)-vecb))/(|(veca+vecb+vecc)/(4)-veca||(veca+vecb+vecc)/(4)-vecb|)`
`=((vecb+vecc-3veca)*(veca+vecc-3VECB))/(|vecb+vecc-3veca|*|veca+vecc-3vecb|)`
`=(veca*vecb+vecb*vecc-3vec(b^(2))+veca*vecc+vec(c^(2))-3vecb*c-3vec(a^(2))-3veca*vecc+9veca*vecb)/((vec(b^(2))+vec(c^(2))+9vec(a^(2))+2vecb*vecc-6veca*vecc-6veca*vecb))`
`((1)/(2)+(1)/(2)-3+(1)/(2)+1-(3)/(2)-3-(3)/(2)+(9)/(2))/(1+1+9+1-3-3)`
`=(-5+3)/(6)=-(1)/(3)`
`theta=cos^(-1)(-(1)/(3))`


Discussion

No Comment Found

Related InterviewSolutions