Vector $PQ$ has magnitude 5 units and is inclined at $150^{\circ}$ to the $x$-axis. In the form $ai+bj$ it is:

1. Resolve into components: $a=|PQ|\cos\theta$ and $b=|PQ|\sin\theta$. 2. $\cos 150^{\circ}=-\dfrac{\sqrt3}{2}$, so $a=5\times\left(-\dfrac{\sqrt3}{2}\right)=-\dfrac{5\sqrt3}{2}$. 3. $\sin 150^{\circ}=\dfrac12$, so $b=5\times\dfrac12=\dfrac52$. 4. Hence $\vec{PQ}=-\dfrac{5\sqrt3}{2}i+\dfrac{5}{2}j$. 5. At $150^{\circ}$ the vector lies in the second quadrant, so $a$ must be negative and $b$ positive; any option breaking that sign pattern is wrong. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 3: Vectors_