KVPY SX SB Exam 20141 Question Paper

Question : 17 of 120

Marks: +1, -0

What is the angle subtended by an edge of a regular tetrahedron at its center?

Solution:

a↖{→}, b↖{→}, c↖{→}

are unit vectors

a↖{→} {\^}b↖{→} =b↖{→}{\^}c↖{→}=c↖{→}{\^}a↖{→}=π/3

centre p

({o↖{→}+a↖{→}+b↖{→}+ c↖{→}}/4)

Now angle between

\AP ↖{→}\ &\ \BP↖{→}

\cosθ={\AP↖{→}.\BP↖{→}}/{{|\AP ↖{→}|}{|\BP↖{→}|}}

={{({{a↖{→}+b↖{→}+ c↖{→}}/4-{a↖{→}}) .({{a↖{→}+b↖{→}+ c↖{→}}/4-{b↖{→}})}}}/ {|{a↖{→}+b↖{→}+ c↖{→}}/4-a↖{→}||{a↖{→}+b↖{→}+ c↖{→}}/4-b↖{→}|}

={(b↖{→}+ c ↖{→}-3a↖{→}).(a↖{→}+ c ↖{→}-3b↖{→})}/{|b↖{→}+ c ↖{→}-3a↖{→}|.|a↖{→}+ c ↖{→}-3b↖{→}|}

{a↖{→} . b↖{→}+ b↖{→} . c↖{→} -3b^2↖{→}+ a↖{→} · c↖{→} +c {^ 2↖{→}} - 3b↖{→} · c↖{→} -3a ^2 ↖{→}-3a ↖{→}. c↖{→}+ 9a ↖{→}.b↖{→}}/( b^2↖{→}+ c^2↖{→}+ 9a^2↖{→}+ 2b↖{→}. c↖{→} -6a↖{→}. c↖{→}- 6a↖{→} .b↖{→})

={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

θ=\cos^{-1}({-1}/3)

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