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What is the angle substended by an edge of regular tetrahedron at its centre? |
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Answer» `cos^(-1)((-1)/(2))` `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))` |
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