UPSC NDA I 2023 Mathematics Solved Paper

Concept:The angle between two vectors is found using the dot product formula: $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos\theta$, where $\theta$ is the angle between them.Explanation:Given vectors: $\vec{a} = 3\hat{i} + 3\hat{j} + 3\hat{k}$, $\vec{b} = 5\hat{i} + 2\hat{j} + 2\hat{k}$, and $\vec{c} = \hat{j} - \hat{k}$.First, find $\vec{a} + \vec{b} = (3+5)\hat{i} + (3+2)\hat{j} + (3+2)\hat{k} = 8\hat{i} + 5\hat{j} + 5\hat{k}$.Let $\theta$ be the angle between $(\vec{a}+\vec{b})$ and $\vec{c}$.Compute the dot product: $(\vec{a}+\vec{b})\cdot \vec{c} = (8)(0) + (5)(1) + (5)(-1) = 0 + 5 - 5 = 0$.Magnitude of $(\vec{a}+\vec{b})$: $|\vec{a}+\vec{b}| = \sqrt{8^2 + 5^2 + 5^2} = \sqrt{64+25+25} = \sqrt{114}$.Magnitude of $\vec{c}$: $|\vec{c}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0+1+1} = \sqrt{2}$.Using the dot product formula: $0 = (\sqrt{114})(\sqrt{2})\cos\theta \Rightarrow \cos\theta = 0$.Thus, $\theta = \frac{\pi}{2}$ (since $\cos\theta = 0$ for $\theta$ in $[0,\pi]$).Answer:The angle between $(\vec{a}+\vec{b})$ and $\vec{c}$ is $\frac{\pi}{2}$.Option A is correct.