Question

Show that the dot product of any two unit vectors is the sum of the product of the components?

Answer 1

We have:

`bb(ul(A)) = A_1bb(ul(i)) + A_2bb(ul(j)) + A_3bb(ul(k))`
`bb(ul(B)) = B_1bb(ul(i)) + B_2bb(ul(j)) + B_3bb(ul(k))`

So then by definition the dot product is

`bb(ul(A)) * bb(ul(B)) = (A_1bb(ul(i)) + A_2bb(ul(j)) + A_3bb(ul(k))) * (B_1bb(ul(i)) + B_2bb(ul(j)) + B_3bb(ul(k)))`

Using the distributive property of the dot product:

`bb(ul(A)) * bb(ul(B)) = A_1bb(ul(i)) * (B_1bb(ul(i)) + B_2bb(ul(j)) + B_3bb(ul(k))) +`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ A_2bb(ul(j)) * (B_1bb(ul(i)) + B_2bb(ul(j)) + B_3bb(ul(k))) +`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ A_3bb(ul(k)) * (B_1bb(ul(i)) + B_2bb(ul(j)) + B_3bb(ul(k)))`

`\ \ \ \ \ \ \ \ = A_1bb(ul(i)) * B_1bb(ul(i)) + A_1bb(ul(i)) * B_2bb(ul(j)) + A_1bb(ul(i)) * B_3bb(ul(k)) +`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ A_2bb(ul(j)) * B_1bb(ul(i)) + A_2bb(ul(j)) * B_2bb(ul(j)) + A_2bb(ul(j)) * B_3bb(ul(k)) +`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ A_3bb(ul(k)) * B_1bb(ul(i)) + A_3bb(ul(k)) * B_2bb(ul(j)) + A_3bb(ul(k)) * B_3bb(ul(k))`

`\ \ \ \ \ \ \ \ = A_1B_1bb(ul(i)) * bb(ul(i)) + A_1B_2bb(ul(i)) * bb(ul(j)) + A_1B_3bb(ul(i)) * bb(ul(k)) +`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ A_2B_1bb(ul(j)) * bb(ul(i)) + A_2B_2bb(ul(j)) * bb(ul(j)) + A_2B_3bb(ul(j)) * bb(ul(k)) +`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ A_3B_1bb(ul(k)) * bb(ul(i)) + A_3B_2bb(ul(k)) * bb(ul(j)) + A_3B_3bb(ul(k)) * bb(ul(k))`

We now use the property that the dot product of a unit vector with itself is unity whilst the dot product of the unit vector with any other perpendicular vector is zero:

`bb(ul(A)) * bb(ul(B)) = A_1B_1(1) + A_1B_2(0) + A_1B_3(0) +`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ A_2B_1(0) + A_2B_2(1) + A_2B_3(0) +`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ A_3B_1(0) + A_3B_2(0) + A_3B_3(1)`

`\ \ \ \ \ \ \ \ = A_1B_1 + A_2B_2 + A_3B_3` QED