Difficulty: Intermediate | Prerequisites: Cross products, dot products, basic vector operations
Big picture: Lines and planes are the fundamental geometric objects of three-dimensional space, and nearly every MATH 126 exam includes at least one problem asking you to find a plane equation or determine how two lines relate to each other. The key tools are parametric equations for lines, the cross product for generating normal vectors, and scalar projection for computing distances. These ideas return throughout the course whenever you need to describe surfaces, set up integrals over regions, or work with tangent planes.
A line in 3D is described by a point plus a direction vector. A plane is described by a point plus a normal vector (perpendicular to the plane). To find a plane containing a line and an external point, take the cross product of the line's direction vector with a vector from the line to the point. Two lines in 3D can intersect, be parallel, or be skew (neither intersecting nor parallel); the distance between skew lines uses a scalar projection onto their shared normal.
Parametric equation of a line
A line through point P₀ = (x₀, y₀, z₀) with direction vector v = ⟨a, b, c⟩ is written as r(t) = ⟨x₀ + at, y₀ + bt, z₀ + ct⟩, where t is a real parameter.
Think of it as: t slides you along the line. At t = 0 you are at P₀; increasing or decreasing t moves you in the direction of v or its opposite.
Direction vector
A vector parallel to a line. If the line passes through two points A and B, then AB = B − A is a direction vector.
In simple terms, this means: the direction vector tells you which way the line runs and how fast.
Normal vector (to a plane)
A vector n perpendicular to every vector lying in the plane. A plane is completely determined by one point in it and one normal vector.
Think of it as: the normal vector sticks straight out of the plane, like a flagpole on a flat field.
Scalar equation of a plane
Given a normal vector n = ⟨A, B, C⟩ and a point P₀ = (x₀, y₀, z₀) on the plane, the equation is A(x − x₀) + B(y − y₀) + C(z − z₀) = 0, which simplifies to Ax + By + Cz = D.
In simple terms, this means: the coefficients A, B, C come directly from the normal vector, and D is found by plugging in any known point.
Skew lines
Two lines in 3D that are neither parallel nor intersecting. They pass each other without meeting and point in different directions. This cannot happen in 2D.
Think of it as: one line on the ground floor and another on the first floor, running at an angle to each other.
Scalar projection (component)
The scalar projection of a onto b is comp_b a = (a · b) / |b|. It gives the signed length of the shadow of a onto the direction of b.
In simple terms, this means: how much of a points in the direction of b, measured as a number (positive if roughly the same direction, negative if opposite).
Given the line r(t) = ⟨1 + t, 3 − 4t, 5t⟩ and the point P(1, 2, 3), find the equation of the plane containing both.
First check whether P is on the line. Setting x = 1 gives t = 0, but then y = 3 ≠ 2 and z = 0 ≠ 3. So P is not on the line.
Extract a point on the line: at t = 0, the point is (1, 3, 0).
Extract the direction vector of the line: v₁ = ⟨1, −4, 5⟩.
Form a second vector in the plane by connecting the point on the line to P: v₂ = ⟨1 − 1, 2 − 3, 3 − 0⟩ = ⟨0, −1, 3⟩.
The normal vector is n = v₁ × v₂:
i: (−4)(3) − (5)(−1) = −12 + 5 = −7
j: −((1)(3) − (5)(0)) = −3
k: (1)(−1) − (−4)(0) = −1
n = ⟨−7, −3, −1⟩
Using the point (1, 3, 0): −7(x − 1) − 3(y − 3) − (z − 0) = 0
Expanding: −7x + 7 − 3y + 9 − z = 0, so −7x − 3y − z = −16
Final answer: −7x − 3y − z = −16 (equivalently 7x + 3y + z = 16)
Given L₁ through A(1, 0, 0) and B(0, 2, 1), and L₂ through C(0, 0, 2) and D(0, a, 0):
Write L₁ parametrically: direction v₁ = B − A = ⟨−1, 2, 1⟩, so r₁(t) = ⟨1 − t, 2t, t⟩.
Equivalently: x = 1 − t, y = 2t, z = t.
Write L₂ parametrically: direction v₂ = D − C = ⟨0, a, −2⟩, so r₂(s) = ⟨0, as, 2 − 2s⟩.
Equivalently: x = 0, y = as, z = 2 − 2s.
Set equal and solve the system:
x: 1 − t = 0, so t = 1
z: t = 2 − 2s, so 1 = 2 − 2s, giving s = 1/2
y: 2t = as, so 2(1) = a(1/2), giving a = 4
The lines intersect when a = 4.
When a ≠ 4, the lines are skew. To find the values of a where the distance between L₁ and L₂ is exactly 1:
Compute the cross product n = v₁ × v₂ = ⟨a + 4, 2, a⟩. This is the vector perpendicular to both lines.
Pick a vector CA connecting a point on L₂ to a point on L₁: CA = A − C = ⟨1, 0, −2⟩.
The distance between the skew lines is the absolute value of the scalar projection of CA onto n:
d = |CA · n| / |n|
CA · n = (1)(a + 4) + (0)(2) + (−2)(a) = a + 4 − 2a = 4 − a
|n| = √((a + 4)² + 4 + a²) = √(2a² + 8a + 20)
Setting d = 1: |4 − a| / √(2a² + 8a + 20) = 1
Squaring: (4 − a)² = 2a² + 8a + 20
Expanding: 16 − 8a + a² = 2a² + 8a + 20
Rearranging: a² + 16a + 4 = 0
Quadratic formula: a = (−16 ± √(256 − 16)) / 2 = (−16 ± √240) / 2 = −8 ± 2√15
\text{Line: } \vec{r}(t) = \vec{r}_0 + t\vec{v}\text{Plane: } A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \quad \text{where } \vec{n} = \langle A, B, C \rangle\text{Distance between skew lines} = \frac{|\vec{P_1P_2} \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}Students often try to find a plane equation without first checking whether the given point is on the line. If it is, the problem is under-determined (infinitely many planes contain the line), and the exam would not ask the question that way. The check also picks up easy marks.
When finding line intersections, students sometimes use the same parameter for both lines. You must use different parameters (e.g. t and s) because the two lines are traversed independently.
Students confuse parallel lines with skew lines. Parallel lines have proportional direction vectors but never meet. Skew lines have non-proportional direction vectors and still never meet. The distinction matters because the distance formula for skew lines uses the cross product of the direction vectors, which would be zero for parallel lines.
When computing the distance between skew lines, students sometimes forget to take the absolute value of the scalar projection. The sign depends on which direction you measure, but distance is always non-negative.
⚠️ Plane equations appear on virtually every MATH 126 exam. The recipe is always the same: find two vectors in the plane, cross them to get the normal, plug in a point.
⚠️ The skew-lines distance formula is a classic multi-step problem worth significant marks. On this midterm it was worth 4 out of 10 points on Problem 5 and required setting up a quadratic equation.
⚠️ Always check your plane equation by verifying that all given points satisfy it. This is a free error-check that takes ten seconds.
True or false: two lines in 3D that do not intersect must be parallel. (False; they could be skew.)
Fill in the blank: the normal vector to a plane Ax + By + Cz = D is ______. (⟨A, B, C⟩)
True or false: you need two distinct vectors lying in a plane to find its normal vector. (True)
Fill in the blank: to find the distance between skew lines, you project a connecting vector onto the ______ of their direction vectors. (cross product)
True or false: if v₁ × v₂ = 0, the two lines are skew. (False; it means the lines are parallel.)
Q: Find the equation of the plane through the points (1, 0, 0), (0, 1, 0), and (0, 0, 1).
A: Form vectors ⟨−1, 1, 0⟩ and ⟨−1, 0, 1⟩. Their cross product is ⟨1, 1, 1⟩. Using (1, 0, 0): 1(x − 1) + 1(y) + 1(z) = 0, so x + y + z = 1.
Q: Lines L₁ and L₂ have direction vectors ⟨2, 1, −1⟩ and ⟨1, 3, 2⟩. Are they parallel, intersecting, or skew?
A: The direction vectors are not proportional (2/1 ≠ 1/3), so the lines are not parallel. To determine whether they intersect or are skew, you would need to set up a system of equations using their parametric forms. Without the specific points, you cannot yet decide.
Q: The distance between two skew lines is |a · n| / |n|, where n = v₁ × v₂. What is the geometric meaning of n?
A: n is the vector perpendicular to both lines. The two skew lines lie in parallel planes whose common normal direction is n, and the distance between those parallel planes equals the distance between the lines.
Q: If a plane has equation 3x − 2y + z = 7, what is a normal vector to this plane?
A: ⟨3, −2, 1⟩. The coefficients of x, y, z directly form the normal vector.
Plane equations are the foundation for tangent planes to surfaces, which appear later in MATH 126. The normal vector concept returns when you compute surface integrals and flux. The distance-between-lines formula uses the same scalar projection idea that appears in the tangential component of acceleration (see the Vector-Valued Functions study notes).
lines in 3D, parametric equations, plane equation, normal vector, skew lines, parallel lines, line intersection, scalar projection, distance between lines, direction vector, cross product for planes, MATH 126, Calculus III, multivariable geometry, Ax + By + Cz = D