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Question:

||→OX × →OY|| = sin(P+R)sin(Q+R)sin(P+Q)sin2R

sin(P+R)

sin(Q+R)

sin2R

sin(P+Q)

Solution:

||→OX × →OY|| = ||→OX|| ||→OY|| sinθ, where θ is the angle between →OX and →OY. ||→OX|| = ||→OY|| = 1 since they are both unit vectors. Also since →OX ⊥ →QR and →OY ⊥ →RP, θ is also the angle between →QR and →RP, i.e., ∠R ∴ ||→OX × →OY|| = 1.1.sinR Now, since PQR is a triangle, P+Q+R=π ⇒ R=π−(P+Q) ∴ ||→OX × →OY|| = sin(π−(P+Q))=sin(P+Q) Hence, option D is correct.