UPSC NDA I 2016 Mathematics Solved Paper

Concept:The dot product of two unit vectors equals the cosine of the angle between them.The half‑angle identity $1+\cos\theta = 2\cos^{2}(\theta/2)$ simplifies the expression.Explanation:Since $\hat{a}$ and $\hat{b}$ are unit vectors, their magnitudes are $1$.Their dot product is $\hat{a}\cdot\hat{b} = |\hat{a}||\hat{b}|\cos\theta = (1)(1)\cos\theta = \cos\theta$.Start from the magnitude square of the sum: $|\hat{a}+\hat{b}|^{2} = |\hat{a}|^{2} + |\hat{b}|^{2} + 2\hat{a}\cdot\hat{b}$.Substitute the known values: $|\hat{a}+\hat{b}|^{2} = 1^{2} + 1^{2} + 2(\cos\theta) = 2 + 2\cos\theta$.Factor out $2$: $|\hat{a}+\hat{b}|^{2} = 2(1+\cos\theta)$.Apply the identity $1+\cos\theta = 2\cos^{2}(\theta/2)$: $|\hat{a}+\hat{b}|^{2} = 2 \times 2\cos^{2}(\theta/2) = 4\cos^{2}(\theta/2)$.Rearrange to isolate $\cos^{2}(\theta/2)$: $\cos^{2}(\theta/2) = \frac{|\hat{a}+\hat{b}|^{2}}{4}$.Take the positive square root (the magnitude is non‑negative): $\cos(\theta/2) = \frac{|\hat{a}+\hat{b}|}{2}$.Answer:Option B: $\frac{ | \hat{a}+\hat{b} | }{2}$.