Let O be the origin, and $\vec{OX}, \vec{OY},\vec{OZ}$ be three unit vectors in the directions of the sides $\vec{QR},\vec{RP},\vec{PQ}$ respectively, of a triangle PQR. Then, $|\vec{OX}×\vec{OY}|=$
Answer & explanation
Correct answer: option 1
$|\vec{OX}×\vec{OY}|$
As $\sin θ=\sin(\pi - θ)$
$\sin R=\sin(\pi - R)$
$=|\vec{OX}||\vec{OY}|\sin (π-R)$
$=\sin R = \sin (π-(P+Q)) = \sin (P+Q)$ $[∵P+Q+R=π]$
