Question

If `A= <-3 ,8 ,-1 >` and `B= <0 ,-4 ,2 >`, what is `A*B -||A|| ||B||`?

Answer 1

`A cdot B - ||A|| ||B|| = -34 - 2 cdot sqrt(37 cdot 10) approx 72,5`

Explanation:

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.