UPSC NDA I 2024 Mathematics Paper

Concept:The given expression sums scalar-vector products where each term is a component of $\vec{r}$ times a cross product with a unit vector. The sum simplifies to zero due to cancellation of orthogonal components.Explanation:Let $\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}$ where $a,b,c$ are scalars.Compute each term separately.First term:$\vec{r} \cdot \hat{i} = a$$\vec{r} \times \hat{i} = (a\hat{i}+b\hat{j}+c\hat{k})\times\hat{i} = a(\hat{i}\times\hat{i}) + b(\hat{j}\times\hat{i}) + c(\hat{k}\times\hat{i}) = 0 + b(-\hat{k}) + c(\hat{j}) = c\hat{j} - b\hat{k}$Thus $(\vec{r}\cdot\hat{i})(\vec{r}\times\hat{i}) = a(c\hat{j} - b\hat{k}) = ac\hat{j} - ab\hat{k}$Second term:$\vec{r} \cdot \hat{j} = b$$\vec{r} \times \hat{j} = a(\hat{i}\times\hat{j}) + b(\hat{j}\times\hat{j}) + c(\hat{k}\times\hat{j}) = a\hat{k} + 0 + c(-\hat{i}) = a\hat{k} - c\hat{i}$Thus $(\vec{r}\cdot\hat{j})(\vec{r}\times\hat{j}) = b(a\hat{k} - c\hat{i}) = ab\hat{k} - bc\hat{i}$Third term:$\vec{r} \cdot \hat{k} = c$$\vec{r} \times \hat{k} = a(\hat{i}\times\hat{k}) + b(\hat{j}\times\hat{k}) + c(\hat{k}\times\hat{k}) = a(-\hat{j}) + b\hat{i} + 0 = -a\hat{j} + b\hat{i}$Thus $(\vec{r}\cdot\hat{k})(\vec{r}\times\hat{k}) = c(-a\hat{j} + b\hat{i}) = -ac\hat{j} + bc\hat{i}$Now add all three terms:$(ac\hat{j} - ab\hat{k}) + (ab\hat{k} - bc\hat{i}) + (-ac\hat{j} + bc\hat{i})$The $\hat{j}$ terms: $ac\hat{j} - ac\hat{j} = 0$The $\hat{k}$ terms: $-ab\hat{k} + ab\hat{k} = 0$The $\hat{i}$ terms: $-bc\hat{i} + bc\hat{i} = 0$So the sum is $\vec{0}$ for any $\vec{r}$.Answer:A. $\vec{0}$