Question

If `vec a, vec b and vec c` are unit vectors, then prove that `|vec a - vec b|^2 + |vec b - vec c|^2 + |vec c - vec a|^2` ≤ 9.

Answer 1

If `vec a, vec b and vec c` are unit vectors, their magnitudes are:

`|vec a| = |vec b| = |vec c| = 1`

Using the identity `|vec x - vec y|^2 = |vec x|^2 + |vec y|^2 - 2(vec x . vec y)`, we expand each term:

`|vec a - vec b|^2 = |vec a|^2 + |vec b|^2 - 2(vec a . vec b)`

= `1 + 1 - 2(vec a . vec b)`

= `2 - 2(vec a . vec b)`   ...(i)

`|vec b - vec c|^2 = |vec b|^2 + |vec c|^2 - 2(vec b . vec c)`

= `1 + 1 - 2(vec b . vec c)`

= `2 - 2(vec b . vec c)`   ...(ii)

`|vec c - vec a|^2 = |vec c|^2 + |vec a|^2 - 2(vec c . vec a)`

= `1 + 1 - 2(vec c . vec a)`

= `2 - 2(vec c . vec a)`   ...(iii)

Adding (i), (ii) and (iii):

`|vec a - vec b|^2 + |vec b - vec c|^2 + |vec c - vec a|^2 = 2 - 2(vec a . vec b) + 2 - 2(vec b . vec c) + 2 - 2(vec c . vec a)`

= `(2 + 2 + 2) - 2(vec a . vec b) - 2(vec b . vec c) - 2(vec c . vec a)`

= `6 - 2(vec a . vec b + vec b . vec c + vec c . vec a)`   ...(iv)

Consider the magnitude of the sum of the vectors squared, which is always greater than or equal to zero:

`|vec a  + vec b + vec c|^2 ≥ 0`

Expanding this expression gives:

`|vec a|^2 + |vec b|^2 + |vec c|^2 + 2(vec a . vec b + vec b . vec c + vec c . vec a) ≥ 0`

`1 + 1 + 1 + 2(vec a . vec b + vec b ⋅ vec c + vec c ⋅ vec a) ≥ 0`

`3 + 2(vec a . vec b + vec b ⋅ vec c + vec c ⋅ vec a) ≥ 0`

Rearranging to find a limit for the dot products:

`- 2(vec a . vec b + vec b ⋅ vec c + vec c ⋅ vec a) ≤ 3`   ...(v)

Substitute inequality (iv) into equation (v):

`|vec a - vec b|^2 + |vec b - vec c|^2 + |vec c - vec a|^2 = 6 - 2(vec a . vec b + vec b . vec c + vec c . vec a)`

`|vec a - vec b|^2 + |vec b - vec c|^2 + |vec c - vec a|^2 ≤ 6 + 3`

`|vec a - vec b|^2 + |vec b - vec c|^2 + |vec c - vec a|^2 ≤ 9`

Hence proved.