Motion in three dimensions

A particle in space is located by r(t) = x(t) î + y(t) ĵ + z(t) k̂, measured from a chosen origin in a chosen frame. The three coordinate functions share a single parameter, t, and the points they trace form the trajectory — a curve in space. Displacement between two instants is the vector difference Δr = r₂ − r₁, and its magnitude is the straight-line separation, never larger than the distance travelled along the curve. Nothing in this description privileges an axis. Which coordinates happen to stay constant is a statement about the axes you picked, not about the motion.

In a fixed Cartesian frame, î, ĵ and k̂ have constant magnitude and constant direction, so their derivatives vanish and d/dt passes straight through to the coefficients: v = ẋ î + ẏ ĵ + ż k̂ and a = ẍ î + ÿ ĵ + z̈ k̂. Velocity is tangent to the path at every point, and speed is |v| = √(ẋ² + ẏ² + ż²) in m s⁻¹. This is the whole reason three-dimensional kinematics is three one-dimensional problems sharing one clock. The property belongs to the fixed basis alone: in cylindrical or spherical coordinates the unit vectors turn as the particle moves, and differentiating them produces the extra terms that carry the centripetal and Coriolis pieces.

Given a(t), integrate componentwise: v(t) = v₀ + ∫a dt′ and r(t) = r₀ + ∫v dt′. Each integration introduces one constant per component, so a complete solution needs six scalars — three for r₀ and three for v₀. For constant a this collapses to r(t) = r₀ + v₀t + ½at², the same statement as in one dimension with vectors in place of signed numbers. Arc length is the one quantity that refuses to decompose: s = ∫|v| dt must be built from the magnitude, because √(ẋ² + ẏ² + ż²) is not the sum of the separate pieces.

Read r(t) − r₀ = v₀t + ½at² as a recipe: every displacement from the launch point is v₀ times a number plus a times a number. Two fixed vectors span a plane, so the whole trajectory lies in the plane through r₀ containing v₀ and a — at every t, for any launch angle. If v₀ × a = 0 the two are parallel and the path degenerates to a straight line. A ball thrown with v₀ = (3.0 î + 4.0 ĵ + 12.0 k̂) m s⁻¹ under a = −9.81 k̂ m s⁻² is therefore not a three-dimensional problem. Rotate the horizontal axes to put one along 3î + 4ĵ and the launch reads 5.0 m s⁻¹ horizontally with 12.0 m s⁻¹ vertically: ordinary projectile motion, rising 12.0²/(2 × 9.81) = 7.34 m and returning to launch height after 2 × 12.0/9.81 = 2.45 s.

Two conditions cover most planar cases. Constant acceleration, as above. And central acceleration, with a always along the line to one fixed point: then d(r × v)/dt = v × v + r × a = 0, so r × v is fixed and r stays perpendicular to it. Orbits are planar for that reason, not by assumption. For a general path the exact test is zero torsion, (v × a)·(da/dt) = 0 at every instant where v × a ≠ 0; fail it and the trajectory leaves every plane. A charge moving partly along a uniform magnetic field is the standard failure: the perpendicular velocity circles while the parallel velocity is untouched, giving a helix. A proton in a 0.50 T field turns at ω = |q|B/m = 4.8 × 10⁷ rad s⁻¹, so v⊥ = 2.0 × 10⁶ m s⁻¹ and v∥ = 1.0 × 10⁶ m s⁻¹ give a radius v⊥/ω = 4.2 cm and a pitch 2πv∥/ω = 13 cm.

When a reduction exists, make it explicit rather than assumed. If the motion is planar, the plane normal is n̂ = (v₀ × a)/|v₀ × a| for constant acceleration, or the direction of r × v for a central force; align k̂ with n̂, put the origin on the plane, and the z equation becomes z = 0. If the motion is genuinely three-dimensional, look instead for a direction along which the equations decouple: put k̂ along B for the helix and z(t) = z₀ + v∥t separates completely from the circular x–y motion, leaving one 2D problem plus one trivial 1D problem. Dropping a dimension is a claim about geometry, so state the claim — the trajectory lies in the plane containing v₀ and a — rather than quietly deleting a coordinate.