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Transform vector to spherical coordinates

Short answer

At P, r = √29 ≈ 5.385, θ = 42.03°, φ = 236.31°. Using transformation matrix: A_r = −3.343, A_θ = 2.266, A_φ = 4.438. Thus A = −3.34 a_r + 2.27 a_θ + 4.44 a_φ.

Transform the vector A = 4a_x − 2a_y − 4a_z into spherical co-ordinates at a point P(x = −2, y = −3, z = 4).

Given: A = 4a_x − 2a_y − 4a_z at P(x = −2, y = −3, z = 4). Find A_r, A_θ, A_φ.

Transformation (Cartesian to spherical)

| A_r |   | sinθ cosφ   sinθ sinφ   cosθ  | | A_x |
| A_θ | = | cosθ cosφ   cosθ sinφ  −sinθ  | | A_y |
| A_φ |   | −sinφ       cosφ        0     | | A_z |

Step 1: Location of P in spherical coordinates

r = √(x² + y² + z²) = √(4 + 9 + 16) = √29 = 5.385
θ = cos⁻¹(z/r) = cos⁻¹(4/5.385) = 42.03°
φ = tan⁻¹(y/x) : x < 0, y < 0 (third quadrant) → φ = 180° + tan⁻¹(3/2) = 236.31° (= −123.69°)

So sinθ = √13/√29 = 0.6695, cosθ = 4/√29 = 0.7428, cosφ = −2/√13 = −0.5547, sinφ = −3/√13 = −0.8321.

Step 2: Components

A_r = A_x sinθ cosφ + A_y sinθ sinφ + A_z cosθ
    = 4(0.6695)(−0.5547) + (−2)(0.6695)(−0.8321) + (−4)(0.7428)
    = −1.4856 + 1.1142 − 2.9712 = −3.343

A_θ = A_x cosθ cosφ + A_y cosθ sinφ − A_z sinθ
    = 4(0.7428)(−0.5547) + (−2)(0.7428)(−0.8321) − (−4)(0.6695)
    = −1.6481 + 1.2362 + 2.6780 = 2.266

A_φ = −A_x sinφ + A_y cosφ
    = −4(−0.8321) + (−2)(−0.5547)
    = 3.3284 + 1.1094 = 4.438

Check: |A|² = 16 + 4 + 16 = 36, and (−3.343)² + (2.266)² + (4.438)² = 11.18 + 5.13 + 19.70 = 36.0. ✓

Answer: A = −3.34 a_r + 2.27 a_θ + 4.44 a_φ (at P, r = 5.385, θ = 42.03°, φ = 236.3°)

✓ Answer checked by Claude Opus 5.5 (high effort)

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