M346 Section 6.4 Number 5

# Problem 6.4 Number 5
# Problem givens:
b1 = [1; 1; 1]
b2 = [-2; 1; 1]
b3 = [0; 1; -1]
d1 = b1/norm(b1)
d2 = b2/norm(b2)
d3 = b3/norm(b3)
L(x) = [x[1]; x[2]; -x[3]]

# Using formula from current section 6.4: $A_{ij} = \bk{d_i}{L(d_j)}$
LD = [dot(d1, L(d1)) dot(d1, L(d2)) dot(d1, L(d3)); dot(d2, L(d1)) dot(d2, L(d2)) dot(d2, L(d3)); dot(d3, L(d1)) dot(d3, L(d2)) dot(d3, L(d3))]

# Verification using older technique
# conversion matrix from basis D to standard basis E
PED = hcat(d1, d2, d3)
# Matrix in standard basis is quite intuitive, just reverse x_3 in three standard basis vectors.
LE = [1 0 0; 0 1 0; 0 0 -1]
LD2 = inv(PED) * LE * PED