Our vectors are
#A = langle -3, 8, -1 rangle,#
#B = langle 0, -4, 2 rangle.#
Firstly, it's important to understand how the norm #|| cdot ||# is related to the inner product. By definition,
#||A||^2 = A cdot A.#
Therefore,
#A cdot B - ||A|| ||B|| = A cdot B - sqrt((A cdot A) (B cdot B)).#
Calculating #A cdot B#, #A cdot A# and #B cdot B#, using the definition of the inner product in three dimensions, where #A_{i}# is the #i#-th component of the vector #A = langle A_{1}, A_{2}, A_{3} rangle#,
#A cdot B = sum_{i=1}^{3} A_{i} B_{i},#
#A cdot B = -3 cdot 0 + 8 cdot (-4) + (-1) cdot 2 = -34,#
#A cdot A = (-3)^2 + 8^2 + (-1)^2 = 74,#
#B cdot B = 0^2 + (-4)^2 + 2^2 = 20.#
Back to our expression,
#A cdot B - sqrt((A cdot A) (B cdot B))
= -34 - sqrt(74 cdot 20)#
#A cdot B - sqrt((A cdot A) (B cdot B))
= -34 - 2 cdot sqrt(37 cdot 10)#
#A cdot B - sqrt((A cdot A) (B cdot B))
approx -72,5.#
Therefore,
#A cdot B - ||A|| ||B|| = -34 - 2 cdot sqrt(37 cdot 10).#
Geometrically, this is a measure of how disaligned the two vectors are, since
#(A cdot B)/(||A|| ||B||) - (||A|| ||B||)/(||A|| ||B||) = cos(theta) - 1,#
where #theta# is the angle between the vectors, and, therefore, the closer #A cdot B - ||A|| ||B||# is to #0#, the more aligned are the vectors.
