Let the vectors u = x i + ^2 x j + k and v = 2 ^2 x i - j - 2 x k be given. The number of values of x in the interval [0, 4 ] for which u and v are orthogonal is

Step-by-step solution

For the vectors u and v to be orthogonal, their dot product must be zero. u v = ( x)(2 ^2 x) + ( ^2 x)(-1) + (1)(-2 x) = 0 2 ^3 x - ^2 x - 2 x = 0 Substituting ^2 x = 1 - ^2 x , we get: 2 ^3 x - (1 - ^2 x) - 2 x = 0 2 ^3 x + ^2 x - 2 x - 1 = 0 Factoring by grouping: ^2 x (2 x + 1) - 1(2 x + 1) = 0 ( ^2 x - 1)(2 x + 1) = 0 This gives ^2 x = 1 or x = - 1 2 . Case 1: ^2 x = 1 x = 1 or x = -1 In the interval [0, 4 ] , x = 1 gives x = 2 , 5 2 (2 solutions). x = -1 gives x = 3 2 , 7 2 (2 solutions). Case 2: x = - 1 2 In