Lines, planes and intersections
Build line and plane equations, verify intersections and interpret parameter restrictions.
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A line needs a point and a nonzero direction
A point fixes a location; a nonzero vector fixes a direction. Together they describe an infinite line:
Every coordinate uses the same parameter. For a = (1,2,3) and d = (2,-1,0),
Eliminating t gives the Cartesian description
The third direction component is zero, so z is constant. A fraction with zero denominator is not another way to state that condition.
A line has many vector equations. You may choose another point on it and multiply its direction by any nonzero scalar. For example, (5,0,3) is on this line at t = 2, and (-4,2,0) = -2d. Hence
describes the same line. The parameter labels need not agree: here t = 2-2s. To compare lines, compare their points and directions rather than their parameter names.
Two independent Cartesian equalities also identify the line. You can split a three-part ratio chain into two equalities. For the line above, the ratio equality and z = 3 are the two conditions; a fixed coordinate must not become division by zero.
Check your understanding 1 Keep the whole line
Give another pair of Cartesian equalities for r = (1,2,3) + t(2,-1,0). Would x + 2y = 5 and 2x + 4y = 10 be enough?
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One valid pair is x + 2y = 5 and z = 3. Substituting x = 1 + 2t and y = 2-t gives x + 2y = 5 for every t, while z stays three.
The proposed pair repeats the same condition: its second equation is twice its first. Together they describe a plane, not the required line. A different nonzero scaling is allowed, but the two equalities must remain independent and keep the line’s location as well as its direction.
Two line parameters must describe one common point
For nonparallel lines, equate the coordinates and solve for both parameters. Then test the proposed pair in every coordinate equation.
Consider
The x- and y-coordinates give 1 + 2t = 3 + s and t = 1-s. Substituting s = 1-t into the first equation gives t = 1, hence s = 0. The z-coordinate must still agree: 2-1 = 1 + 2(0). Thus the lines intersect at (3,1,1).
Now replace m by
The x- and y-equations give the same candidate parameters, but the z-values are now 1 and 4. There is no common point. The directions are not proportional, so these lines are skew: they neither meet nor run parallel. A drawing projected onto a flat page can make skew lines look as though they cross.
For parallel directions, first test whether one line’s anchor point lies on the other line. If it does, the lines coincide; otherwise they are distinct parallel lines. Two parallel lines are coplanar. Two nonparallel lines are coplanar exactly when they intersect; skew lines cannot lie in one plane.
Check your understanding 2 The direction does not fix the location
Compare the line r = (1,2,3) + t(2,-1,0) with each of r = (5,0,3) + s(-4,2,0) and r = (5,0,4) + s(-4,2,0). State whether each pair is coincident, distinct parallel, intersecting at one point or skew. Justify the distinction.
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Both new directions are proportional to (2,-1,0). The first pair coincides because (5,0,3) lies on the original line at t = 2. The second pair is distinct and parallel: its anchor has z = 4, while every point on the original has z = 3. Neither pair is skew; parallel directions alone do not decide coincidence.
A plane needs two independent directions
A plane through A can be written
where λ,μ∈R and u, v are nonzero and nonparallel. Varying the two parameters moves in two independent directions. Parallel directions would describe only a line.
For a plane through three noncollinear points A, B and C, use the displacements u = b-a and v = c-a. Crossing the position vectors b and c instead uses directions from O, which need not be directions within ABC. Those position vectors can even be parallel when A, B and C are noncollinear.
Take A = (1,-1,2), B = (2,0,2) and C = (1,0,4). Their displacements from A are (1,1,0) and (0,1,2), so
A normal is perpendicular to both plane directions. The vector product gives
Check n · (1,1,0) = 0 and n · (0,1,2) = 0. Every displacement r-a within the plane is therefore perpendicular to n:
Since n · a = 6, the Cartesian equation is 2x-2y + z = 6. Substituting A, B and C checks both orientation and offset. A normal vector alone fixes only the orientation; you still need a point or constant to locate the plane.
You can reverse the process. In 2x-2y + z = 6, choose x = s, y = t and obtain z = 6-2s + 2t. Thus another vector equation for the same plane is
Its anchor satisfies the Cartesian equation, and its two independent directions are perpendicular to (2,-2,1).
A line and plane give one parameter equation
Substitute a line r = a + td into a plane n · r = c:
If n · d ≠ 0, this has exactly one parameter value and one intersection point. If n · d = 0, travelling along the line never changes the left-hand side. Then compare n · a with c: equality means the entire line lies in the plane; inequality means no intersection.
An identity such as 0 = 0 means every line parameter works. It does not mean t = 0 is the only solution. A contradiction such as 0 = 4 means no parameter works; dividing by zero cannot recover one.
Check your understanding 3 Substitute the whole parameterised point
Find where r = (1,0,0) + t(1,2,-1) meets the plane x + y + z = 5. Check your point against both original equations.
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Substitution gives 1 + t + 2t-t = 5, so t = 2. The point is (3,4,-2). Its coordinate sum is 5, and all three line coordinates use the same parameter value 2. A parameter alone is not the intersection point.
Check your understanding 4 Decide what the equation permits
For each real k, classify the relationship between r = (0,0,k) + t(1,1,0) and the plane 2x-2y + z = 6. Is there any k for which they have exactly one intersection point?
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Substitution gives 2t-2t + k = 6, or k = 6, with no t term. When k = 6, every line point lies in the plane. For every other k, the line is parallel to the plane and disjoint from it. No k gives exactly one intersection point: changing the anchor cannot change the direction’s zero dot product with the normal.
Two plane normals identify the possible relationships
Proportional nonzero normals give parallel orientations. Scale the whole equation before comparing constants. The plane 4x-4y + 2z = 12 is the same as 2x-2y + z = 6; the plane 4x-4y + 2z = 13 is distinct and parallel to it.
Nonparallel normals give an intersection line. Solve the two Cartesian equations together with one free parameter. For
subtracting the second equation from the first gives x-3y = 3. Set y = s, so x = 3 + 3s and z = -4s. The intersection is
Its anchor satisfies both plane equations. Its direction has zero dot product with both normals: 6-2-4 = 0 and 3 + 1-4 = 0. These checks confirm a common point and a direction lying in both planes; checking the direction alone would not locate their intersection.
Try it yourself 1
Interpret the moving point
Problem
A point moves at constant velocity with position r(t) = (0,0,2) + t(1,1,1), where coordinates are in metres and t is time in seconds. The model applies for t ≥ 0. Does the point reach the plane x + y + z = 1 during the modelled motion? Explain the meaning of any intersection you calculate.
Write your answer, then compare it with the solution
Hints
Hint 1
Check the meaning of the parameter
An algebraic intersection also needs to satisfy the time interval in the model.
Show full solution
Test the algebraic intersection
Method
Substitute the position coordinates into the plane equation.Reason
The point must satisfy all coordinate conditions at the same time.Working
Answer
t + t + (2 + t) = 1 gives t = -1/3 and point (-1/3,-1/3,5/3).Apply the model's time interval
Method
The computed time is outside t ≥ 0.Reason
The infinite line intersects the plane, but the model includes only positions from the start time onwards.Working
Answer
For t ≥ 0, the coordinate sum is 2 + 3t ≥ 2, so it never equals 1.
Quick check
The algebraic intersection belongs to the backward extension of the path, one third of a second before the start. Compare your interpretation as well as your arithmetic: a unique line–plane intersection does not establish that a time-restricted path reaches it.
When you classify an intersection, state the objects, any parameter restrictions and the conditions you checked. Orientation decides which relationships are possible; a common point or matching offset decides which one actually occurs.
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