GATE 2018 | MATHS | Q-51

Question

Consider \( \mathbb{R}^3 \) with the usual inner product. If \( d \) is the distance from \( (1, 1, 1) \) to the subspace ${(1, 1, 0), (0, 1, 1)}$ of \( \mathbb{R}^3 \), then \( 3d^2 \) is given by

Answer 1

vector (1,-1,1) is orthogonal to both of these vectors

(1,1,1) = 2/3(1,1,0) + 2/3(0,1,1) + 1/3(1,-1,1)

hence the orthogonal projection is 1/3(1,-1,1) and so d is the length of this vector. Then d^2 = 1/3 and 3d^2 = 1