Lines and Planes

01 / UNDERSTAND & APPLY

1. A line needs a direction; a plane needs a normal

A line in three-dimensional space is specified by a point A and a nonzero direction vector d. Its vector equation is r = a + td, with real parameter t. A plane is specified by a point A and a nonzero normal vector n. Its equation is n·(r − a) = 0. The line direction lies along the line; the plane normal points perpendicular to the plane. Confusing these roles reverses many angle and parallelism conditions.

The line parameter t is a coordinate along the line, not necessarily a distance. If d is a unit vector, changing t by one changes distance by one; otherwise the distance change is |d| times the parameter change. Multiplying d by any nonzero scalar describes the same geometric line with a different parametrization.

For a plane written Ax + By + Cz = D, the normal is (A,B,C), which must be nonzero. Multiplying every coefficient, including D, by the same nonzero scalar leaves the plane unchanged. If all three normal coefficients are zero, the equation is either an identity or a contradiction, not an ordinary plane.

Three-dimensional space differs crucially from a flat page: two nonparallel lines can fail to intersect. Such lines are skew. A diagram drawn on paper may make them seem to cross, but their coordinates reveal different heights or depths at the apparent crossing. Treat a spatial sketch as a guide and verify incidence algebraically.

02 / UNDERSTAND & APPLY

2. Parametric form safely includes zero direction components

For d = (l,m,n), the line r = a + td becomes x = a₁ + lt, y = a₂ + mt, z = a₃ + nt. If all direction components are nonzero, one may combine these as (x−a₁)/l = (y−a₂)/m = (z−a₃)/n. If a component is zero, do not divide by it: keep the corresponding coordinate fixed and use parametric form.

A line through distinct points A and B has direction b − a. Any nonzero multiple is acceptable, and simplifying common factors often makes later arithmetic easier. To test whether P lies on the line, one parameter value must satisfy all three coordinate equations. Matching only two coordinates is insufficient.

Two lines with parallel direction vectors are either the same line or disjoint parallel lines. Check whether a point of one lies on the other. Nonparallel directions alone do not establish intersection. Use independent parameters t and s for the two lines, solve the coordinate equations, and verify all three. Using one shared parameter can falsely exclude or invent intersections.

03 / UNDERSTAND & APPLY

3. Three noncollinear points determine a plane

Given A,B,C, form the two in-plane displacements u = b − a and v = c − a. Their cross product n = u×v is perpendicular to both, so n·(r − a) = 0 is the plane equation if n ≠ 0. If the cross product is zero, the points are collinear or repeated and do not determine a unique plane.

A plane containing a line r = a + td and an additional point P not on the line has two in-plane directions d and p − a. Their cross product is a normal. A plane containing two intersecting nonparallel lines similarly uses their two directions. Before applying this construction, verify the specified lines actually intersect or are otherwise coplanar.

Two planes with nonparallel normals intersect in a line whose direction is n₁×n₂. To locate that line, solve the two plane equations for at least one point and combine it with the direction. If the normals are parallel, the planes are either identical or disjoint; proportionality of the full equations decides which.

The intercept form x/a + y/b + z/c = 1 applies when the plane has finite nonzero intercepts a,b,c on all three axes. It is not a universal form for a plane through the origin or parallel to an axis. Normal form avoids these special-case restrictions.

04 / UNDERSTAND & APPLY

4. Solve incidence and keep the exceptional cases

For a line r = a + td and plane n·r = c, substitution gives n·a + t(n·d) = c. If n·d ≠ 0, there is exactly one intersection parameter t = (c − n·a)/(n·d). If n·d = 0, the line direction is parallel to the plane. Then either n·a = c and the entire line lies in the plane, or n·a ≠ c and there is no intersection.

For two nonparallel lines, the scalar triple product (b − a)·(d₁×d₂) tests coplanarity. A zero value means they are coplanar and therefore intersect; a nonzero value means they are skew. This criterion assumes the directions are nonparallel. If their cross product is zero, handle parallel or coincident lines separately.

When solving coordinate equations for an intersection, do not discard an inconsistent third equation. It is exactly the evidence that two nonparallel lines are skew. Likewise, when a parameter-dependent denominator vanishes, return to the original equations rather than declaring the problem undefined; that parameter may create a parallel or coincident configuration.

For three planes, the associated linear system may have one point, a line, a plane, or no common point. A zero coefficient determinant rules out a unique point but does not by itself distinguish infinitely many solutions from inconsistency. Row reduction or explicit compatibility checks are needed.

05 / UNDERSTAND & APPLY

5. Use the right vectors for the requested angle

The acute or zero angle θ between two unoriented lines with nonzero directions d₁,d₂ satisfies cos θ = |d₁·d₂|/(|d₁||d₂|). Without the absolute value, the formula gives the angle between the chosen oriented vectors, which may be obtuse. Reversing a line direction should not change the angle between the geometric lines.

The acute angle between two planes equals the acute angle between their normal lines, so use the same absolute dot-product formula with n₁,n₂. If the question requests a particular dihedral angle, there may be a supplementary choice depending on the specified half-planes. State the convention rather than silently choosing one.

The angle α between a line and a plane is complementary to the acute angle between the line and the plane’s normal. Consequently sin α = |d·n|/(|d||n|). A direction parallel to the plane gives α = 0; a direction parallel to the normal gives α = π/2. This endpoint check is an easy way to remember why the formula uses sine.

Parallelism and perpendicularity follow immediately. A line is parallel to a plane’s direction when d·n = 0, but it may lie in the plane. A line is perpendicular to the plane when d is parallel to n. Two planes are perpendicular when their nonzero normals have zero dot product.

06 / UNDERSTAND & APPLY

6. Point-to-plane distance, foot and reflection

For a plane n·r = c and point P with position p, the signed normal displacement is (n·p − c)/|n|. Its absolute value is the perpendicular distance. This follows by choosing a point A in the plane and projecting p − a onto the unit normal n/|n|. Tangential parts do not contribute to the dot product.

The foot H lies along the normal through P, so h = p − λn. Requiring n·h = c gives λ = (n·p − c)/|n|². Therefore h = p − [(n·p − c)/|n|²]n. Reflection in the plane goes the same distance beyond H: p_ref = p − 2[(n·p − c)/|n|²]n.

These formulas are unchanged if every plane coefficient is multiplied by a nonzero constant. The signed distance changes sign if the normal orientation is reversed, but the unsigned distance, foot and reflected point do not. This is a useful consistency check when two equivalent equations look different.

When two planes are parallel, rewrite them with the same normal before subtracting their constants. If they are n·r = c₁ and n·r = c₂, the distance is |c₂ − c₁|/|n|. Subtracting constants from unnormalized equations with different normal scales produces a wrong answer.

07 / UNDERSTAND & APPLY

7. A closest point on a line comes from a projection

For line r = a + td, with d ≠ 0, minimize the distance from P by requiring the connecting displacement p − (a + td) to be perpendicular to d. The dot-product equation gives t₀ = [(p − a)·d]/|d|². The foot is h = a + t₀d, and the distance is |p − h|.

The same distance can be written |(p − a)×d|/|d|. The cross-product magnitude is the area of a parallelogram with base |d| and perpendicular height equal to the point-line distance. This second derivation is valuable when only the distance, rather than the foot, is requested.

The closest point on an infinite line is not necessarily on a specified segment. If the problem asks for distance to segment AB, compute the line parameter with d = b − a. If t₀ lies in [0,1], the foot is inside the segment. If t₀ < 0, the nearest segment point is A; if t₀ > 1, it is B. A line formula used without this restriction can underestimate the segment distance.

A spatial diagram should show a perpendicular connector from P to H and label d along the line. The point A can be anywhere on that line; changing the chosen starting point changes t₀ but not H or the distance. This explains why equivalent parametrizations must give the same geometric answer.

08 / UNDERSTAND & APPLY

8. Shortest distance between nonparallel lines

For lines L₁: r = a + td₁ and L₂: r = b + sd₂ with nonparallel directions, n = d₁×d₂ is a nonzero common perpendicular direction. The component of b − a along its unit direction gives the separation of the two parallel planes that contain the lines and are perpendicular to n. The shortest distance is |(b − a)·n|/|n|.

To see why the starting points do not matter, replace a or b by another point on its line. The difference b − a changes by a combination of d₁ and d₂, both perpendicular to n, so its dot product with n stays the same. If the numerator is zero, the nonparallel lines are coplanar and intersect, giving distance zero.

If the closest points themselves are required, let Q₁ = a + td₁ and Q₂ = b + sd₂. Their connector Q₂ − Q₁ must be perpendicular to both directions. This produces two linear equations in t,s. Solve them and verify both perpendicularity conditions. The distance formula alone does not tell you the endpoints of the shortest segment.

Do not use the cross-product denominator for parallel lines; it vanishes there. The parallel case has a different formula and a different family of closest-point pairs. Treating 0/0 as zero would incorrectly claim that all parallel lines coincide.

09 / UNDERSTAND & APPLY

9. Parallel cases and an independent minimization check

For parallel lines with common nonzero direction d, use the distance from a point on one line to the other: |(b − a)×d|/|d|. It is zero exactly when the starting-point difference is also parallel to d, meaning the lines coincide. There are infinitely many shortest connecting segments related by translation along d.

A direct squared-distance calculation gives another route. For a point and a line, |p − a − td|² is a quadratic in t with positive leading coefficient |d|². Completing the square or differentiating gives the same minimizing t₀ as projection. For two nonparallel lines, the analogous two-variable quadratic gives the pair of perpendicularity equations.

This minimization viewpoint clarifies why the perpendicular connector is the shortest. If any connector has a component along an allowed line direction, moving its endpoint to remove that component reduces its squared length. At the minimum, the connector must be perpendicular to all directions in which the endpoints are free to move.

For rays, segments, or bounded portions of planes, boundary points can become the minimum instead. The infinite-line and infinite-plane formulas solve the unconstrained geometric problem. Read the requested object carefully before assuming those unconstrained feet lie in the allowed region.

10 / UNDERSTAND & APPLY

10. Original practice with classification and checks

Before doing arithmetic, identify whether each requested object is a line, plane, segment, direction, angle or distance. The formulas are sensitive to those distinctions.

11 / UNDERSTAND & APPLY

11. A spatial answer needs incidence, direction and scope

The most reliable workflow is to write each line as a point plus a direction and each plane as a nonzero normal plus a constant. Then classify before calculating: parallel or nonparallel, contained or disjoint, intersecting or skew. The classification determines whether a denominator is safe and which distance formula applies.

Check a proposed intersection by substituting into every original equation. Check a foot by verifying both membership and perpendicularity. Check a reflection by verifying that its midpoint with the original point is the foot. Check a shortest connector between lines by dotting it with both directions. These independent geometric checks are stronger than merely repeating arithmetic.

The common mistakes are using the plane normal as an in-plane direction, using cosine instead of sine for a line-plane angle, dividing by a zero direction component, giving both lines the same parameter, and treating a zero determinant or cross product as a complete classification. Each error comes from skipping one of the geometric distinctions above.

Finally, keep the scope of the object visible. A closest point on an infinite line can lie outside a segment. A parallel line may lie inside a plane. A diagrammatic crossing may be skew in space. A signed normal component is not an unsigned distance until its absolute value is taken. Correct formulas work only when matched to the correct geometric question.