Source: Homework Set 1, §12.5
Tags: parametric line, direction vector, skew lines, plane equation, normal vector, intercept form, parametric plane, coordinate planes
Difficulty: Intermediate | Prerequisites: §12.1-12.4 (vectors, dot product, cross product). You should be comfortable with vector arithmetic and the idea of a normal vector.
Big Picture
§12.5 moves from individual vectors to the geometric objects they define: lines and planes in three-dimensional space. A line in R³ needs a point and a direction; a plane needs a point and a normal (or equivalently, a point and two direction vectors). This section brings together dot products, cross products, and projections into a unified geometric toolkit. The homework tests whether you can prove lines are skew, manipulate plane equations into different forms, and navigate a plane using parametric coordinates, all skills that recur throughout the rest of multivariable calculus.
A line in 3D is written as r(t) = r₀ + tv (point plus direction). Two lines in R³ can be parallel, intersecting, or skew (neither parallel nor intersecting). A plane is ax + by + cz + d = 0, where ⟨a, b, c⟩ is the normal, or equivalently r₀ + tv₁ + sv₂ (point plus two independent direction vectors). Converting between these forms is a core skill.
Parametric line equation
The representation r(t) = ⟨x₀, y₀, z₀⟩ + t⟨a, b, c⟩, where (x₀, y₀, z₀) is a known point and ⟨a, b, c⟩ is the direction vector.
Think of it as: a starting point plus a scaled direction that sweeps out every point on the line as t varies.
Direction vector
The vector that determines which way a line runs. Two lines are parallel precisely when their direction vectors are scalar multiples of each other.
Skew lines
Two lines in R³ that are neither parallel nor intersecting. This cannot happen in R², only in three or more dimensions.
In simple terms, skew lines pass each other like two roads at different levels of an overpass.
Plane equation (standard form)
ax + by + cz + d = 0, where the vector ⟨a, b, c⟩ is normal (perpendicular) to the plane.
Intercept form of a plane
x/x₀ + y/y₀ + z/z₀ = 1, where x₀, y₀, z₀ are the plane's intercepts with the coordinate axes. Valid only when the plane does not pass through the origin and is not parallel to any coordinate axis.
Parametric plane equation
⟨x₀, y₀, z₀⟩ + tv₁ + sv₂, where v₁ and v₂ are two non-parallel vectors lying in the plane. The parameters t and s act as coordinates on the plane.
Think of it as: a 2D coordinate system laid flat on the plane, centred at a reference point.
Showing two lines are skew (Q3a)
Given r₁(t) = ⟨1, -1, 2⟩ + t⟨-1, 4, 3⟩ and r₂(s) = ⟨-1, -3, 1⟩ + s⟨1, 4, -2⟩:
Step 1, not parallel: check whether the direction vectors are scalar multiples. ⟨-1, 4, 3⟩ and ⟨1, 4, -2⟩ are not multiples of each other (the ratio -1/1 ≠ 4/4 is fine but 3/(-2) ≠ -1, so they fail), confirming the lines are not parallel.
Step 2, do not intersect: set r₁(t) = r₂(s) component by component and solve. You get a system of three equations in two unknowns. If the system is inconsistent (the third equation contradicts the first two), the lines do not intersect.
Both conditions together mean the lines are skew.
Plane intercept form (Q3b)
Starting from ax + by + cz + d = 0 with d ≠ 0 (the plane does not pass through the origin):
Rearrange to ax + by + cz = -d.
Divide through by -d: (a/-d)x + (b/-d)y + (c/-d)z = 1.
Set x₀ = -d/a, y₀ = -d/b, z₀ = -d/c. These are the axis intercepts (found by setting the other two variables to zero).
Result: x/x₀ + y/y₀ + z/z₀ = 1.
If the plane passes through the origin, d = 0 and you cannot divide by d. If the plane is parallel to a coordinate axis, one of a, b, or c is zero, and the corresponding intercept does not exist, so the intercept form breaks down.
Parametric description of a plane (Q4)
Consider the plane x + y - 4z = 2 with reference point (1, 1, 0):
You need two orthogonal unit vectors lying in the plane.
v₁ = ⟨1/√2, -1/√2, 0⟩ is given. It lies in the plane because (1/√2) + (-1/√2) - 4(0) = 0 (it satisfies the homogeneous form of the plane equation, which direction vectors must satisfy).
To find v₂: it must satisfy four conditions: unit length, orthogonal to v₁ (dot product zero), lies in the plane (its components satisfy the normal equation), and all components positive.
A direction vector lies in the plane when it is perpendicular to the normal ⟨1, 1, -4⟩, that is, the components satisfy a + b - 4c = 0.
Solve the system: v₁ · v₂ = 0 and ⟨1, 1, -4⟩ · v₂ = 0, then normalise, then pick the sign so all components are positive.
Navigating the plane (Q4b)
To reach an arbitrary point (a, b, c) on the plane from (1, 1, 0):
Write ⟨a - 1, b - 1, c⟩ = t v₁ + s v₂.
Solve for t and s using dot products: t = ⟨a - 1, b - 1, c⟩ · v₁ and s = ⟨a - 1, b - 1, c⟩ · v₂ (this works because v₁ and v₂ are orthonormal).
Parametric line:
\mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t\,\mathbf{v}Standard plane equation:
a x + b y + c z + d = 0 \quad\text{with normal } \mathbf{n} = \langle a, b, c \rangleIntercept form (plane not through origin, not parallel to any axis):
\frac{x}{x_0} + \frac{y}{y_0} + \frac{z}{z_0} = 1Parametric plane:
\langle x_0, y_0, z_0 \rangle + t\,\mathbf{v}_1 + s\,\mathbf{v}_2Parametric line equations are the basis for ray tracing in computer graphics: every pixel on your screen is drawn by casting a parametric ray from the camera and testing where it hits objects.
Plane equations are used constantly in engineering and architecture. Determining whether a structural beam intersects a wall, or computing the angle between two surfaces, reduces to the algebra of plane normals and line-plane intersection.
Students often assume two non-parallel lines must intersect. In R³ they can be skew. Always check for intersection explicitly.
When converting to intercept form, forgetting the condition d ≠ 0. If the plane passes through the origin, the intercept form does not exist.
Confusing a direction vector of a line with the normal vector of a plane. A direction vector lies in the object; a normal vector is perpendicular to it.
When finding direction vectors for a plane, students sometimes forget that the vector must satisfy the homogeneous equation (normal · v = 0), not the full plane equation with the constant.
⚠️ Proving lines are skew is a two-part argument: show not parallel AND show no intersection. Leaving out either part loses marks.
⚠️ The intercept form derivation is a common "show that" proof question. Be ready to handle the edge cases (origin, parallel to axis).
⚠️ Parametric plane problems test your ability to combine dot product, cross product, and normalisation in one problem. They are longer but earn more credit.
⚠️ The orthonormality shortcut (t = displacement · v₁) works only when v₁ and v₂ are orthogonal unit vectors. If they are not orthonormal, you must solve a system instead.
True or false: two non-parallel lines in R³ always intersect.
False. They may be skew.
Fill in the blank: the intercept form x/x₀ + y/y₀ + z/z₀ = 1 requires that the plane does not pass through the ___.
Origin.
True or false: a direction vector of a plane is perpendicular to the plane's normal vector.
True.
Fill in the blank: to show lines are skew, show they are not ___ and do not ___.
Parallel; intersect.
True or false: in a parametric plane equation, the parameters t and s act as coordinates on the plane.
True.
Q: Are the lines r₁(t) = ⟨0, 0, 0⟩ + t⟨1, 1, 1⟩ and r₂(s) = ⟨1, 0, 0⟩ + s⟨2, 2, 2⟩ skew, parallel, or intersecting?
A: The direction vectors ⟨1, 1, 1⟩ and ⟨2, 2, 2⟩ are scalar multiples, so the lines are parallel. (Check: they do not share a common point, so they are parallel and distinct.)
Q: Convert 2x + 3y + 6z = 12 to intercept form.
A: Divide by 12: x/6 + y/4 + z/2 = 1. The intercepts are x₀ = 6, y₀ = 4, z₀ = 2.
Q: Why does the intercept form fail for the plane x + y + z = 0?
A: This plane passes through the origin (d = 0 in ax + by + cz + d = 0), so you cannot divide by d to isolate 1 on the right-hand side.
Q: Find a direction vector lying in the plane 3x - y + 2z = 5.
A: Any vector ⟨a, b, c⟩ with 3a - b + 2c = 0. For example, set a = 1, c = 0, then b = 3, giving ⟨1, 3, 0⟩.
Q: If v₁ and v₂ are orthonormal vectors spanning a plane, and the displacement from the reference point to a target is d, how do you find the coordinates t and s?
A: t = d · v₁ and s = d · v₂. This works because orthonormality means the dot product extracts the component along each basis vector.
Lines and planes tie directly to the projection material (§12.2-12.3): the distance from a point to a line or plane is computed via projection, and the parametric plane uses orthogonal direction vectors found by the same decomposition.
The cross product (§12.4) provides the quickest route to a plane's normal: given two direction vectors in the plane, their cross product is the normal.
These ideas are foundational for later work on surfaces, tangent planes, and the geometry of curves in R³.
parametric line, vector equation of a line, direction vector, skew lines, parallel lines, plane equation, normal vector, intercept form, parametric plane, coordinate planes, orthonormal basis, plane through three points, point-normal form, Calculus III §12.5, MATH 215