Prove i×(a×i)+j×(a×j)+k×(a×k)=2a

Short answer

Use i×(a×i)=a-(a·i)i and analogous for j, k. Summing gives 3a - (a₁i+a₂j+a₃k) = 2a.

Prove that i×(a×i) + j×(a×j) + k×(a×k) = 2a, where i, j, k are mutually perpendicular unit vectors along the coordinate axes.

Use the vector triple product A × (B × C) = B(A · C) − C(A · B).

With A = i, B = a, C = i:

i × (a × i) = a(i · i) − i(i · a) = a − (a₁) i, since i · i = 1 and i · a = a₁ (the x-component of a).

Similarly:

j × (a × j) = a − (a₂) j
k × (a × k) = a − (a₃) k

i×(a×i) + j×(a×j) + k×(a×k) = 3a − (a₁i + a₂j + a₃k)
                              = 3a − a
                              = 2a

Hence i × (a × i) + j × (a × j) + k × (a × k) = 2a.

✓ Answer checked by Claude Opus 5.5 (high effort)

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