JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

Since MM^T = I, M is an orthogonal matrix. The columns of an orthogonal matrix form an orthonormal basis for R ^3. The given vectors u , v , w are exactly the columns of M. Therefore, u , v , w are mutually orthogonal unit vectors. For (P): The rows of M are also orthonormal. The third row is bmatrix & & bmatrix , so ^2 + ^2 + ^2 = 1. The third column w is a unit vector, so | w |^2 = (- 1 2 )^2 + ( 1 3 )^2 + ^2 = 1. 1 2 + 1 3 + ^2 = 1 ^2 = 1 6 Substituting ^2 into the row equation gives ^2 + ^2 = 1 - 1 6 = 5 6 . Thus, (P) (5). For (Q): Given x u + y v + z w = j . Taking the dot product with u on both sides gives x( u u ) + y( v u ) + z( w u ) = j u . Since the vectors are orthonormal, u u = 1 and v u = w u = 0. Thus, x = j u . From the definition of u , the j component is 1 3 , so x = 1 3 . Thus, (Q) (4). For (R): The scalar triple product u ( v w ) represents the determinant of the matrix formed by these column vectors, which is M. Since M is orthogonal, (M) = 1. Thus, | u ( v w )| = | (M)| = 1. Thus, (R) (2). For (S): Using the vector triple product expansion, u ( v w ) = ( u w ) v - ( u v ) w . Since u , v , w are mutually orthogonal, u w = 0 and u v = 0. Thus, u ( v w ) = 0 , and its magnitude is 0. Thus, (S) (1). Answer: (P) (5), (Q) (4), (R) (2), (S) (1)