UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The line of intersection of two planes is perpendicular to the normals of both planes. Its direction ratios are given by the cross product of the normals.Explanation:Let the given planes be $x + y + z = 1$ and $2x - y - 2z = 3$.Their normals are $\vec{n}_1 = \langle 1, 1, 1 \rangle$ and $\vec{n}_2 = \langle 2, -1, -2 \rangle$.Compute the cross product $\vec{n}_1 \times \vec{n}_2$ using the determinant:$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & -2 \end{vmatrix}$.Expand: $\hat{i}((1)(-2) - (1)(-1)) - \hat{j}((1)(-2) - (1)(2)) + \hat{k}((1)(-1) - (1)(2))$.Simplify each term: $\hat{i}(-2 + 1) - \hat{j}(-2 - 2) + \hat{k}(-1 - 2)$.This gives $\hat{i}(-1) - \hat{j}(-4) + \hat{k}(-3) = \langle -1, 4, -3 \rangle$.The direction ratios can be multiplied by a scalar; multiplying by $-1$ gives $\langle 1, -4, 3 \rangle$.Thus, the required direction ratios are $\langle 1, -4, 3 \rangle$.Answer:Option C: $< 1, -4, 3 >$