Question 72554

Concept. The angle θ between two vectors u and v satisfies cos θ = (u·v) / (|u| |v|). Hence u and v are perpendicular (θ = π/2) exactly when their dot product u·v = 0. Expanding a useful product gives the identity (u + v)·(u − v) = u·u − v·v = |u|2 − |v|2, so (u + v) is perpendicular to (u − v) precisely when |u| = |v|.

Application. Apply this to the given vectors:

1. Add the vectors: a + b = (1+3)i + (2−1)j + (−3+2)k = 4i + j − k.

2. Subtract the vectors: a − b = (1−3)i + (2+1)j + (−3−2)k = −2i + 3j − 5k.

3. Take the dot product: (a + b)·(a − b) = (4)(−2) + (1)(3) + (−1)(−5) = −8 + 3 + 5 = 0.

4. Because the dot product is 0, cos θ = 0, and therefore θ = π/2.

Cross-check. Use the identity with magnitudes: |a|2 = 12 + 22 + (−3)2 = 14 and |b|2 = 32 + (−1)2 + 22 = 14. Since |a|2 = |b|2, the identity gives (a + b)·(a − b) = |a|2 − |b|2 = 14 − 14 = 0, confirming the two vectors are perpendicular.

Result. The angle between (a + b) and (a − b) is π/2.