UPSC NDA II 2023 Mathematics Solved Paper

Concept:Three vectors are collinear if the points they represent (as position vectors) lie on a single straight line.This means the vectors connecting any two points are parallel (scalar multiples of each other).Explanation:Let the given vectors be position vectors of points A, B, C:$\vec{A}=60\hat{i}+3\hat{j}$, $\vec{B}=40\hat{i}-8\hat{j}$, $\vec{C}=\beta\hat{i}-52\hat{j}$.Take A as the reference point. Then vector AB is from A to B:$\vec{AB} = \vec{B} - \vec{A} = (40-60)\hat{i} + (-8-3)\hat{j} = -20\hat{i} -11\hat{j}$.Vector AC is from A to C:$\vec{AC} = \vec{C} - \vec{A} = (\beta -60)\hat{i} + (-52-3)\hat{j} = (\beta -60)\hat{i} -55\hat{j}$.For collinearity, $\vec{AB}$ and $\vec{AC}$ must be parallel. Therefore, their components are proportional:$\frac{-20}{\beta-60} = \frac{-11}{-55}$.Simplify the right side: $\frac{-11}{-55} = \frac{1}{5}$.Thus, $\frac{-20}{\beta-60} = \frac{1}{5}$.Cross-multiply: $-20 \times 5 = 1 \times (\beta-60)$ → $-100 = \beta - 60$.Solve for $\beta$: $\beta = -100 + 60 = -40$.Answer:$\beta = -40$, which corresponds to option C.