Perpendicular feet, distances and angles

A nearest point must satisfy two conditions

A perpendicular foot is a point on the target line or plane. Its displacement from the original point must also be perpendicular to that target. A point satisfying only one condition is not enough: many points lie on a line, and many perpendicular displacements start from points elsewhere.

Use an orthonormal Cartesian frame throughout. Recall that a zero dot product tests perpendicularity, and that a nonzero normal to a plane is perpendicular to every direction within it.

Find a point-to-line foot before its distance

For an infinite line r = a + td with d ≠ 0, the foot H from P must have position h = a + td. The perpendicular condition is

Expanding the dot product and solving gives

The denominator is the squared magnitude of d. Dividing only by |d| would calculate a signed scalar component, not the parameter multiplying the original direction vector.

Take P = (0,2,3) and the line r = (1,0,0) + t(1,1,0). Here p-a = (-1,2,3), so the numerator is -1 + 2 + 0 = 1 and the denominator is 2. Thus t = 1/2 and

The difference from H to P is (-3/2,3/2,3). Its dot product with (1,1,0) is zero. The foot is on the line because the same parameter 1/2 gives all three coordinates. Now take the length:

The answer t = 1/2 locates the foot but is not the distance. Likewise, 27/2 is the squared distance, not the distance.

Why does the perpendicular point give a minimum? Every other line point has form q = h + sd. Since p-h is perpendicular to d,

The extra term is nonnegative and vanishes only at s = 0. This proves the foot is the unique nearest point on the infinite line.

An alternative uses the area of the parallelogram with sides p-a and d. Its area is base times perpendicular height, giving distance

Here the vector product is (-3,3,-3), so the distance is 3 square root of 3/square root of 2 = 3 square root of 6/2 again. This method gives the distance directly; it does not give the foot’s coordinates.

A point-to-plane foot moves along the normal

Let the plane be n · r = c, with n ≠ 0. The line from P perpendicular to the plane follows n, so write

Requiring H to lie in the plane gives n · p-λ|n|² = c. Therefore

The numerator is the plane equation’s residual at P. Its sign tells you which way to move along the chosen normal; it is not yet a distance. The perpendicular distance is |λ||n|, or

For the plane x + 2y + 2z = 9 and P at the origin, n = (1,2,2) has magnitude 3. The residual is -9, so λ = -1 and h = 0-(-1)n = (1,2,2). The distance is 9/3 = 3.

Check the two conditions separately: H gives 1 + 4 + 4 = 9 in the plane equation, and p-h = (-1,-2,-2) is parallel to the normal. The length from P to H is 3.

If P already lies in the plane, the residual and distance are zero and H is P itself. A zero distance is valid; a zero normal would not define a plane. Scaling the whole equation by a nonzero constant changes neither the foot nor the distance: the residual and normal scale together, with the absolute value preserving a nonnegative length.

Choose the geometrical objects before the inverse function

An angle between two vectors lies between 0° and 180°. Keep the sign in cos θ = (u · v)/(|u||v|).

A line can be represented by either orientation of its direction vector. For the smaller angle between lines, remove that arbitrary orientation:

This gives 0° ≤ θ ≤ 90°: parallel lines give zero and perpendicular directions give ninety degrees. The rule also gives the angle between skew lines by comparing their directions; it does not calculate a shortest distance between them.

For example, (1,0,0) and (-1,1,0) have vector angle 135°, but the smaller angle between lines in those directions is 45°. Reversing one direction changes the vector angle, while the lines themselves stay unchanged.

For the smaller angle between planes, use the same cosine formula with their nonzero normals. Parallel normals give parallel plane orientations and angle zero, even when the planes have different offsets. If a question identifies a particular obtuse region, use that region and its supplementary angle rather than automatically reporting the smaller angle.

A line-to-plane angle is complementary to a normal angle

Let d be the line direction and n a normal to the plane. The dot product first compares d with n. The smaller angle φ to the normal line and the inclination θ to the plane satisfy θ + φ = 90°. Hence

Use inverse sine for the inclination, or find the smaller normal angle by inverse cosine and subtract its unrounded value from ninety degrees. Round once at the end, in the requested unit.

The solid rising line and its dashed projection enclose θ. Together with φ to the normal, it fills the right angle. The schematic shows the relationship, not values to measure from the drawing.

The limiting cases help select the formula. If d · n = 0, the line direction is parallel to the plane and θ = 0°, whether the line lies in the plane or is disjoint from it. If d is parallel to n, the line is perpendicular to the plane and θ = 90°. A zero direction or zero normal cannot define either angle.

Before using a remembered formula, identify whether you need a point, a length or an angle. For a foot, check membership and perpendicularity. For an angle, identify the line directions or plane normals and the intended region. Point-to-line and point-to-plane distances do not provide a shortest-distance method between skew lines.