Magnitude of Cross Product of Unit Vectors along Triangle Sides

To find the magnitude of the cross product $|\vec{OX} \times \vec{OY}|$, we first determine the angle between the vectors $\vec{OX}$ and $\vec{OY}$.

1. Definition of vectors:

   • $\vec{OX}$ is a unit vector in the direction of $\vec{QR}$, so $|\vec{OX}| = 1$.
   • $\vec{OY}$ is a unit vector in the direction of $\vec{RP}$, so $|\vec{OY}| = 1$.

2. Angle between $\vec{OX}$ and $\vec{OY}$: In $\triangle PQR$, the interior angle at the vertex $R$ is $R$. The vector $\vec{QR}$ represents motion towards vertex $R$, whereas $\vec{RP}$ represents motion away from vertex $R$. When both vectors are aligned to originate from the same point, the angle $\theta$ between the vectors $\vec{QR}$ and $\vec{RP}$ is the exterior angle at vertex $R$: $\theta = \pi - R$

3. Calculating the magnitude of the cross product: Using the definition of the cross product: $|\vec{OX} \times \vec{OY}| = |\vec{OX}| |\vec{OY}| \sin\theta$ Substituting the magnitudes and the angle $\theta$: $|\vec{OX} \times \vec{OY}| = (1)(1)\sin(\pi - R) = \sin R$

4. Relating to the angles of $\triangle PQR$: Since the sum of interior angles in a triangle is $\pi$, we have: $P + Q + R = \pi \implies R = \pi - (P + Q)$ Substituting $R$ into the trigonometric expression gives: $\sin R = \sin(\pi - (P + Q)) = \sin(P + Q)$

Thus, the magnitude of the cross product is: $|\vec{OX} \times \vec{OY}| = \sin(P + Q)$

Hence, the correct option is A.