UPSC NDA I 2024 Mathematics Paper

Concept:Given two linear equations in direction cosines, find the ratios of $l,m,n$, then use $l^2+m^2+n^2=1$ to determine the required expression.Explanation:We have the equations:$l + 2m + n = 0$$2l - 2m + 3n = 0$Solve for ratios using cross‑multiplication:$\frac{l}{2 \cdot 3 - (-2) \cdot 1} = \frac{m}{1 \cdot 3 - 2 \cdot 1}?$ Wait, careful: standard formula for two equations $a_1l+b_1m+c_1n=0$, $a_2l+b_2m+c_2n=0$ gives $\frac{l}{b_1c_2-b_2c_1} = \frac{m}{c_1a_2-c_2a_1} = \frac{n}{a_1b_2-a_2b_1}$.Here $a_1=1, b_1=2, c_1=1$;$a_2=2, b_2=-2, c_2=3$Thus:$l : m : n = (2 \cdot 3 - (-2) \cdot 1) : (1 \cdot 3 - 2 \cdot 1) : (1 \cdot (-2) - 2 \cdot 2)$$= (6+2) : (3-2) : (-2-4)$$= 8 : 1 : -6$Wait: sign of m? Actually compute precisely: For m, denominator is $c_1a_2 - c_2a_1 = 1 \cdot 2 - 3 \cdot 1 = 2-3 = -1$. But ratio includes sign: $\frac{l}{8} = \frac{m}{-1} = \frac{n}{-6}$.So let common ratio be $k$:$l = 8k,\; m = -k,\; n = -6k$Since $l,m,n$ are direction cosines, we have $l^2+m^2+n^2 = 1$:$(8k)^2 + (-k)^2 + (-6k)^2 = 1$$64k^2 + k^2 + 36k^2 = 1$$101k^2 = 1 \implies k^2 = \frac{1}{101}$Now compute $l^2 + m^2 - n^2$:$l^2+m^2-n^2 = (l^2+m^2+n^2) - 2n^2 = 1 - 2n^2$$n^2 = (-6k)^2 = 36k^2 = 36 \cdot \frac{1}{101} = \frac{36}{101}$Thus $l^2+m^2-n^2 = 1 - 2 \cdot \frac{36}{101} = 1 - \frac{72}{101} = \frac{101-72}{101} = \frac{29}{101}$Answer:$\frac{29}{101}$, which corresponds to option B.