The value of x so that the volume of the parallelopiped formed by the vectors i + x j + k , j + x k and x i + k is minimum, is

Step-by-step solution

The volume of the parallelopiped formed by the vectors a , b , and c is given by the absolute value of their scalar triple product, V = |[ a , b , c ]| . Given vectors: a = i + x j + k b = 0 i + j + x k c = x i + 0 j + k Calculating the scalar triple product: [ a , b , c ] = vmatrix 1 & x & 1 0 & 1 & x x & 0 & 1 vmatrix Expanding the determinant along the first row: = 1(1 1 - x 0) - x(0 1 - x x) + 1(0 0 - 1 x) = 1(1) - x(-x^2) + 1(-x) = x^3 - x + 1 So, the volume is V = |x^3 - x + 1| . Let f(x) = x^3 - x + 1 . To f