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Vectors a, b and c satisfy the scalar triple product [a b c] = 1, with c = lambda*(a x b), |a| = sqrt(2), |b| = sqrt(3), and |c| = 1/sqrt(3). Find the angle between vectors a and b.

1. pi/6
2. pi/4
3. pi/3
4. pi/2

Correct answer: pi/3

Solution

Since c = lambda*(a x b), we get [a b c] = c.(a x b) = lambda*(a x b).(a x b) = lambda*|a x b|² = 1, so lambda = 1/|a x b|². Also |c| = |lambda|*|a x b| = 1/sqrt(3). So |a x b| = 1/(|lambda|*sqrt(3)). Combined: lambda * |a x b|² = 1 and |a x b| = lambda^(-1/2)... solving: |a x b| = sqrt(3)/1 = sqrt(3). Wait: let S = |a x b|. Then lambda * S² = 1 and (1/S²)*S = 1/sqrt(3) => 1/S = 1/sqrt(3) => S = sqrt(3). Then sin(theta) = S/(|a||b|) = sqrt(3)/(sqrt(2)*sqrt(3)) = 1/sqrt(2), so theta = pi/4. Re-check |c|: lambda = 1/S² = 1/3. |c| = lambda*S = (1/3)*sqrt(3) = 1/sqrt(3). Correct! So theta = pi/4.

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