If OAB is a tetrahedronwith edges and hatp, hatq, hatr are unit vectors along bisectors of vec(OA), vec(OB):vec(OB), vec(OC):vec(OC), vec(OA) respectively and hata=(vec(OA))/(|vec(OA)|), vecb=(vec(OB))/(|vec(OB)|), vec c= (vec(OC))/(|vec(OC)|), then :

If OAB is a tetrahedronwith edges and hatp, hatq, hatr are unit vector

1.	If OAB is a tetrahedronwith edges and hatp, hatq, hatr are unit vectors along bisectors of vec(OA), vec(OB):vec(OB), vec(OC):vec(OC), vec(OA) respectively and hata=(vec(OA))/(\|vec(OA)\|), vecb=(vec(OB))/(\|vec(OB)\|), vec c= (vec(OC))/(\|vec(OC)\|), then :
Answer» `([hata HATB hatc])/([hatp hatq HATR])=(3sqrt(3))/(2)` `([hata+hatb" "hatb+hatc" "hatc +hata])/([hatp+hatq" "hatq+hatr" "hatr+hatp])=(3sqrt(3))/(4)` `([hata+hatb" "hatb+hatc" "hatc+hata])/([hatp hatq hatr])=(3sqrt(3))/(2)` `([hata hatb hatc])/([hatp +hatq" "hatq+hatr" "hatr+hatp])=(3sqrt(3))/(4)` Answer :A::D

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If OAB is a tetrahedronwith edges and hatp, hatq, hatr are unit vectors along bisectors of vec(OA), vec(OB):vec(OB), vec(OC):vec(OC), vec(OA) respectively and hata=(vec(OA))/(|vec(OA)|), vecb=(vec(OB))/(|vec(OB)|), vec c= (vec(OC))/(|vec(OC)|), then :

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