Distance Point-Line

Fastest with the cross product formula d = |(q⃗ − p⃗) × u⃗|/|u⃗|. Procedure: form the connection vector from the support point of the line to the point, compute the cross product with the direction vector, divide its magnitude by |u⃗|. Example: g: x⃗ = (1|0|2) + t·(2|1|2) and Q(3|4|2): q⃗ − p⃗ = (2|4|0), cross product (8|−4|−6), magnitude √(64 + 16 + 36) = √116 ≈ 10.77; divided by |u⃗| = 3 gives d ≈ 3.59 LU. The formula automatically measures the shortest, i.e. perpendicular distance. Important: use (q⃗ − p⃗), not the position vector q⃗ alone, and do not forget the final division.

You look for the point F on the line where the perpendicular from Q lands. Ansatz: F = p⃗ + t·u⃗, and the vector QF⃗ must be perpendicular to u⃗, i.e. QF⃗·u⃗ = 0. Solving yields t = (q⃗ − p⃗)·u⃗/(u⃗·u⃗). Example: g: x⃗ = (1|1|0) + t·(1|0|1) and Q(2|1|3): t = ((1|0|3)·(1|0|1))/2 = 4/2 = 2, so F(3|1|2) and d = |QF⃗| = |(1|0|−1)| = √2 ≈ 1.41 LU. The added value over the cross product formula: you obtain the foot point F itself and can continue working with it, e.g. construct the reflected point Q′ = Q + 2·QF⃗.

Ask yourself what the task really requires. If only the distance number is required, the cross product formula is the shortest route: one cross product, two magnitudes, one division, done. If instead you need the foot point itself, a reflected point, the contact point of a perpendicular or a justification via orthogonality, there is no way around the foot point method; it delivers F and the distance at once. In exams the combination is also strong: compute with the foot point and check with the cross product formula in one line (or vice versa). Both methods must give exactly the same value; deviations immediately reveal a calculation error.

In the plane there is the Hesse normal form of the line: for g: ax + by = c and the point Q(q₁|q₂), d = |a·q₁ + b·q₂ − c|/√(a² + b²). Example: g: 3x + 4y = 10 and Q(5|5): d = |15 + 20 − 10|/√(9 + 16) = 25/5 = 5 LU. If the line is given as y = mx + t, first bring it into the form mx − y = −t. The cross product is not available in 2D (it needs three dimensions); alternatively the foot point method works with the normal slope −1/m. The 2D formula is the exact counterpart of the point-plane formula in space, just with two instead of three coordinates.

Skew lines do not intersect and are not parallel; their distance is the length of the common perpendicular. The formula uses both direction vectors: d = |(q⃗ − p⃗)·(u⃗ × v⃗)|/|u⃗ × v⃗|, where p⃗ and q⃗ are the support vectors. The difference from the point-line formula: instead of the magnitude of a cross product, the numerator holds a scalar triple product, i.e. the projection of the connection vector onto the common normal direction u⃗ × v⃗. If the lines are parallel (u⃗ × v⃗ = 0⃗), this formula fails; then you pick a point of one line and compute point-line as usual. So first check the relative position, then the matching formula.