Our vectors are
A=⟨−3,8,−1⟩,
B=⟨0,−4,2⟩.
Firstly, it's important to understand how the norm ||⋅|| is related to the inner product. By definition,
||A||2=A⋅A.
Therefore,
A⋅B−||A||||B||=A⋅B−√(A⋅A)(B⋅B).
Calculating A⋅B, A⋅A and B⋅B, using the definition of the inner product in three dimensions, where Ai is the i-th component of the vector A=⟨A1,A2,A3⟩,
A⋅B=3∑i=1AiBi,
A⋅B=−3⋅0+8⋅(−4)+(−1)⋅2=−34,
A⋅A=(−3)2+82+(−1)2=74,
B⋅B=02+(−4)2+22=20.
Back to our expression,
A⋅B−√(A⋅A)(B⋅B)=−34−√74⋅20
A⋅B−√(A⋅A)(B⋅B)=−34−2⋅√37⋅10
A⋅B−√(A⋅A)(B⋅B)≈−72,5.
Therefore,
A⋅B−||A||||B||=−34−2⋅√37⋅10.
Geometrically, this is a measure of how disaligned the two vectors are, since
A⋅B||A||||B||−||A||||B||||A||||B||=cos(θ)−1,
where θ is the angle between the vectors, and, therefore, the closer A⋅B−||A||||B|| is to 0, the more aligned are the vectors.
