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Vector between two points in cylindrical

Short answer

R_AB = B−A = −2 a_x + 6 a_y − 14 a_z. At A, ρ=5, φ=233.13°. Transform: R_ρ = −3.6, R_φ = −5.2, R_z = −14. Hence R_AB = −3.6 a_ρ − 5.2 a_φ − 14 a_z.

Find the vector that extends from A(−3, −4, 6) to B(−5, 2, −8) and express it in cylindrical co-ordinate system.

Given: A(−3, −4, 6), B(−5, 2, −8). The vector must be written in cylindrical components. A vector in cylindrical coordinates depends on the point where it is located, so we take its location as the starting point A.

Step 1: Vector from A to B (Cartesian)

R_AB = B − A = (−5 + 3) a_x + (2 + 4) a_y + (−8 − 6) a_z
     = −2 a_x + 6 a_y − 14 a_z

Step 2: Cylindrical coordinates of A

ρ = √(3² + 4²) = 5
φ = tan⁻¹(−4/−3): third quadrant → φ = 180° + 53.13° = 233.13° (−126.87°)
cosφ = −3/5 = −0.6,  sinφ = −4/5 = −0.8

Step 3: Transform (Cartesian to cylindrical)

R_ρ = R_x cosφ + R_y sinφ = (−2)(−0.6) + (6)(−0.8) = 1.2 − 4.8 = −3.6
R_φ = −R_x sinφ + R_y cosφ = −(−2)(−0.8) + (6)(−0.6) = −1.6 − 3.6 = −5.2
R_z = −14

Check: (−3.6)² + (−5.2)² + (−14)² = 12.96 + 27.04 + 196 = 236 = (−2)² + 6² + (−14)² ✓

Answer: R_AB = −3.6 a_ρ − 5.2 a_φ − 14 a_z (components valid at A, where φ = 233.13°). Its magnitude is √236 = 15.36.

✓ Answer checked by Claude Opus 5.5 (high effort)

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