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प्रश्न
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.
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उत्तर
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.
