Difficulty: Intermediate to Advanced | Prerequisites: §12.1–12.5 (vectors, dot product, cross product, equations of lines and planes)
This material asks you to work with lines in three-dimensional space: writing their parametric equations, finding the closest approach between two moving objects or two static paths, and determining whether a line intersects a sphere. These problems combine vector arithmetic, distance minimisation (often via calculus or the cross-product method), and the geometry of spheres. If the cross product or the formula for distance between skew lines is not fresh, review §12.4 first.
The locus of points equidistant from three given points is a line (the intersection of two perpendicular bisector planes). The closest distance between two simultaneously moving objects is found by minimising a distance-squared function of time. The closest distance between two line trajectories (ignoring timing) uses the cross product of their direction vectors. Whether a line of sight clears a sphere is checked by substituting the parametric line into the sphere equation and examining the discriminant.
Equidistant locus
The set of all points that are the same distance from two or more given objects. The locus equidistant from two points is a plane (the perpendicular bisector plane). The locus equidistant from three non-collinear points is a line (the intersection of two such planes). Think of it as the place where you would stand to be equally far from each target.
Parametric line in 3D
r(t) = P + t·d, where P is a point on the line and d is the direction vector. In component form: x = x₀ + at, y = y₀ + bt, z = z₀ + ct. Think of t as a dial: each value places you at a different point along the line.
Direction vector
A vector that points along a line or trajectory. For the line r(t) = ⟨2, −1, 3⟩ + t⟨4, 2, 8⟩, the direction vector is ⟨4, 2, 8⟩. Two lines are parallel when their direction vectors are scalar multiples.
Closest distance between two lines (skew lines)
The minimum distance between two lines that do not intersect and are not parallel. Computed using the cross product of their direction vectors and a vector connecting a point on each line. In simple terms, it is the length of the shortest bridge you could build between the two lines.
Line-sphere intersection
Substituting a parametric line into a sphere equation produces a quadratic in t. If the discriminant is positive, the line passes through the sphere at two points. If zero, the line is tangent. If negative, the line misses the sphere entirely.
To find all points equidistant from three points A, B, C, find the perpendicular bisector plane of AB and the perpendicular bisector plane of AC. Their intersection is a line.
Perpendicular bisector plane of A = (2, 1, 1) and B = (−1, −1, 10): the midpoint is (1/2, 0, 11/2). The normal vector is AB = ⟨−3, −2, 9⟩. The plane is −3(x − 1/2) − 2(y) + 9(z − 11/2) = 0.
Repeat for A and C = (1, 3, −4): midpoint (3/2, 2, −3/2), normal AC = ⟨−1, 2, −5⟩.
Solve the two plane equations simultaneously to get the parametric equation of the line.
Given r₁(t) = ⟨2, −1, 3⟩ + t⟨4, 2, 8⟩ and r₂(t) = ⟨−1, −3, 1⟩ + t⟨1, 4, −2⟩, the distance between the jets at time t is |r₁(t) − r₂(t)|.
Compute the difference: r₁(t) − r₂(t) = ⟨3 + 3t, 2 − 2t, 2 + 10t⟩.
The squared distance is D²(t) = (3 + 3t)² + (2 − 2t)² + (2 + 10t)².
Expand: 9 + 18t + 9t² + 4 − 8t + 4t² + 4 + 40t + 100t² = 113t² + 50t + 17.
Minimise by setting dD²/dt = 226t + 50 = 0, giving t = −25/113.
The minimum distance is √(D²(−25/113)).
The trajectories are lines, but they may not be traversed at the same time. The distance between two skew lines uses a different formula.
Direction vectors: d₁ = ⟨4, 2, 8⟩ and d₂ = ⟨1, 4, −2⟩.
Cross product: d₁ × d₂ = ⟨(2)(−2) − (8)(4), (8)(1) − (4)(−2), (4)(4) − (2)(1)⟩ = ⟨−36, 16, 14⟩.
Vector connecting any point on line 1 to any point on line 2: w = ⟨−1 − 2, −3 − (−1), 1 − 3⟩ = ⟨−3, −2, −2⟩.
Distance = |w · (d₁ × d₂)| / |d₁ × d₂| = |(−3)(−36) + (−2)(16) + (−2)(14)| / √(36² + 16² + 14²) = |108 − 32 − 28| / √(1296 + 256 + 196) = 48/√1748.
The line from (0, −2, 1) to (1, 0, 2): direction = ⟨1, 2, 1⟩, so r(t) = ⟨0, −2, 1⟩ + t⟨1, 2, 1⟩ for t in [0, 1].
To check whether this line is blocked by a sphere centred at (0, −1, 1) with radius 2/3:
Substitute into the sphere equation: (t)² + (−2 + 2t + 1)² + (1 + t − 1)² = 4/9.
Simplify: t² + (2t − 1)² + t² = 4/9, i.e. t² + 4t² − 4t + 1 + t² = 4/9.
This gives 6t² − 4t + 1 = 4/9, or 6t² − 4t + 5/9 = 0.
Discriminant: 16 − 4(6)(5/9) = 16 − 40/3 = 8/3 > 0.
Two solutions for t exist. If both lie in [0, 1], the line passes through the sphere and the view is blocked. The two t-values give the points on the sphere visible to each person.
\text{Parametric line: } \mathbf{r}(t) = \mathbf{P} + t\,\mathbf{d} = \langle x_0 + at,\; y_0 + bt,\; z_0 + ct \rangle\text{Distance between skew lines: } D = \frac{|\mathbf{w} \cdot (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|}where w is any vector from a point on line 1 to a point on line 2.
\text{Sphere: } (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2Substituting a parametric line into the sphere equation yields a quadratic in t. The discriminant determines intersection (positive = two points, zero = tangent, negative = miss).
The closest-distance-between-lines calculation is used in air traffic control to determine whether two aircraft will pass within a safe separation. The line-sphere intersection test is fundamental in ray tracing (computer graphics), where you check whether a ray of light hits a spherical object to determine shading and reflections.
The shortest distance between two jets at the same time t is different from the shortest distance between their trajectories (the lines themselves, ignoring timing). Students often conflate the two. The first is a calculus problem (minimise D²(t)); the second uses the cross-product formula for skew lines.
When finding the equidistant line from three points, students sometimes try to find one plane equidistant from all three. The correct approach is to find two perpendicular bisector planes and intersect them.
In the line-sphere intersection problem, finding that the quadratic has real solutions is necessary but not sufficient. You must also check that the t-values fall within the relevant range (e.g. t in [0, 1] for a line segment between two specific points).
Students sometimes forget to normalise when computing the skew-line distance. The denominator is the magnitude of the cross product, not the cross product itself.
⚠️ The distance between skew lines formula (using the cross product) is a classic exam question. Memorise the formula and practise computing the cross product quickly.
⚠️ The distinction between "closest distance at the same time" and "closest distance between the trajectories" appears frequently. Be prepared for both in a single problem.
⚠️ Line-sphere intersection is a standard computational technique. If you see a problem involving visibility, line of sight, or ray-object intersection, this is the method.
True or false: the set of points equidistant from two distinct points in 3D is a plane.
Fill in the blank: if two lines have direction vectors d₁ and d₂, and d₁ × d₂ = 0, the lines are ____.
True or false: to find the closest distance between two jets flying at the same time, you minimise |r₁(t) − r₂(t)|².
Fill in the blank: a line intersects a sphere at two points when the discriminant of the resulting quadratic is ____.
True or false: two skew lines always have a unique shortest distance between them.
Answers: 1. True. 2. Parallel (or the same line). 3. True. 4. Positive. 5. True.
Q: Two jets travel as r₁(t) = ⟨2, −1, 3⟩ + t⟨4, 2, 8⟩ and r₂(t) = ⟨−1, −3, 1⟩ + t⟨1, 4, −2⟩. Find the time of closest approach.
A: D²(t) = (3 + 3t)² + (2 − 2t)² + (2 + 10t)² = 113t² + 50t + 17. Setting dD²/dt = 226t + 50 = 0 gives t = −25/113 seconds.
Q: Find the shortest distance between the trajectories (lines) of the two jets, ignoring timing.
A: d₁ × d₂ = ⟨−36, 16, 14⟩. Vector w = ⟨−3, −2, −2⟩. Distance = |w · (d₁ × d₂)| / |d₁ × d₂| = |108 − 32 − 28| / √1748 = 48/√1748.
Q: A line goes from (0, −2, 1) to (1, 0, 2). Is it blocked by a sphere centred at (0, −1, 1) with radius 2/3?
A: Parametrise: r(t) = ⟨t, −2 + 2t, 1 + t⟩. Substitute into (x)² + (y + 1)² + (z − 1)² = 4/9: t² + (2t − 1)² + t² = 4/9. This gives 6t² − 4t + 5/9 = 0. Discriminant = 16 − 40/3 = 8/3 > 0, so the line does intersect the sphere. Check that both roots fall in [0, 1] to confirm the view is blocked.
Q: What is the perpendicular bisector plane of the points (2, 1, 1) and (−1, −1, 10)?
A: Midpoint: (1/2, 0, 11/2). Normal vector: ⟨−3, −2, 9⟩. Equation: −3(x − 1/2) − 2(y) + 9(z − 11/2) = 0, which simplifies to −3x − 2y + 9z = 48.
The skew-line distance formula relies on the cross product from §12.4 and reappears in §13.2 when computing the torsion and binormal of a space curve. The line-sphere intersection technique is a special case of the general method of substituting parametric equations into implicit surface equations, which extends to every quadric surface in §12.6. The perpendicular bisector plane is the foundational idea behind Voronoi diagrams in computational geometry.
Tags: lines in 3D, parametric line, equidistant locus, perpendicular bisector plane, skew lines, closest distance, cross product distance formula, line-sphere intersection, discriminant, ray tracing, MATH 215, Calculus III, §12.5, §13.1, University of Michigan