CEGEP Linear Algebra Lines and Planes in Space Free · no sign-up

CEGEP Linear Algebra — Lines and Planes in Space Worksheet

Q: Is this for 201-NYC-05 or 201-SN4-RE?

Both — they are the same course under two numbering systems, the first the legacy code and the second the current one. The content is written against ministerial competency 0M04 in the Sciences de la nature programme, so it matches whichever code your college prints on the outline.

Q: What are skew lines, and why did they never come up before?

Skew lines have different directions and no common point, and no single plane contains them both. They cannot happen in the plane: two lines in ℝ² that do not meet must be parallel, which is the rule most students bring with them. The third dimension creates the extra case, so in space the directions and the intersection have to be checked separately — neither answer implies the other.

Q: Which form of a line's equation should I use?

Whichever makes the next step easy, and expect to convert. The vector and parametric forms carry a parameter, so they are what you substitute into a plane's equation or set equal to another line. The symmetric form has no parameter, which makes it a quick test of whether a point is on the line. The devis asks for all three, so a question can name any of them.

Q: Why does a zero dot product mean the line is parallel to the plane?

Because the vector the plane gives you is its normal, which points away from the plane rather than along it. A direction orthogonal to the normal is a direction lying flat with respect to the plane — so it is parallel to it, and the line either lies inside the plane or misses it entirely. A non-zero dot product is the case where the line crosses.

Q: Why is the angle between a line and a plane a sine when the others are cosines?

Because the dot product can only compare the line's direction with the plane's normal, and the angle to the normal is the complement of the angle to the plane. The usual quotient gives the cosine of the angle to the normal, which is the sine of the angle you were asked for. Two lines, or two planes, are compared through vectors of the same kind, so no complement is involved and the cosine is read directly.

Q: Do I really need row reduction here? Isn't this a geometry topic?

You need it. Three planes meeting is a linear system, and the type of its solution set is the geometric answer: one point, a common line, a common plane, or nothing in common. This set is where the matrix half of the course and the vector half meet, which is one reason it sits at the end.

Q: Where do determinants come into this?

In two places. The coefficient determinant of a three-plane system tells you whether the three normals are coplanar, and so whether a unique common point exists. And the scalar triple product — itself a determinant — measures whether two directions and the gap between two lines lie in a common plane, which is exactly the coplanar-versus-skew question.

By Marius — private math tutor in Montreal. Updated 30 August 2026

The last block of the course, and the one that puts everything before it to work. Writing a line three ways and a plane three ways, and converting between the forms; deciding whether a point is on a line or in a plane; sorting two lines into parallel, intersecting or skew, and a line and a plane into contained, parallel or crossing; finding intersections — a line meeting a plane by substitution, three planes by row reduction, with the type of the solution set read back as a picture; distances from a point to a plane, a point to a line and between parallel planes; and angles between lines, between planes, and between a line and a plane, where the formula switches from a cosine to a sine for a reason. Have a look on this page, then print the free PDF when you want to write on it.

Practice worksheet — free PDF

7 pages 8 questions Letter size, print-ready

Download the worksheet Read the questions here

No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 7 harder problems come with the CEGEP Linear Algebra bundle.

All 8 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

Q1Vector, Parametric and Symmetric Equations of a Line

A surveyor sights along a straight guy wire that passes through the points $A (2, - 1, 3)$ and $B (5, 1, - 1)$ , coordinates in metres.
1. Give a direction vector $\vec{d}$ for the line $A B$ , then write a vector equation of the line.
2. Write parametric equations of the line.
3. Write symmetric equations of the line.
4. Decide, showing your reasoning, whether $P (11, 5, - 9)$ and $R (8, 3, - 6)$ lie on the line.
Q2Vector, Parametric and Cartesian Equations of a Plane

Three anchor bolts of a solar panel sit at $A (1, 0, 2)$ , $B (3, 1, 2)$ and $C (2, - 1, 4)$ . The panel lies in the plane $π$ through these three points.
1. Write a vector equation of $π$ .
2. Write parametric equations of $π$ .
3. Find a normal vector to $π$ and hence its cartesian equation in the form $a x + b y + c z + d = 0$ .
4. Decide whether $E (4, 2, 1)$ and $F (0, 1, 0)$ lie in $π$ .
Q3The Relative Position of Two Lines

Determine the relative position of each pair of lines. Where they meet, give the point of intersection; where they do not, say why not.
1. $ℓ_{1} : (x, y, z) = (1, 0, 2) + t (2, 1, - 1)$ and $ℓ_{2} : (x, y, z) = (5, 3, - 1) + s (- 4, - 2, 2)$ .
2. $ℓ_{3} : (x, y, z) = (2, - 1, 3) + t (1, 2, - 1)$ and $ℓ_{4} : (x, y, z) = (1, 2, 1) + s (2, - 1, 1)$ .
Q4The Relative Position of Lines and Planes

Let $π : 2 x - y + 3 z - 4 = 0$ . For each line below, determine its position relative to $π$ — contained in $π$ , parallel to $π$ and disjoint from it, or meeting $π$ in exactly one point. State the test you used, and give the point of intersection where there is one.
1. $ℓ_{1} : (x, y, z) = (1, 0, 1) + t (1, 2, 0)$
2. $ℓ_{2} : (x, y, z) = (2, 3, 1) + t (1, 2, 0)$
3. $ℓ_{3} : (x, y, z) = (0, 2, 0) + t (1, - 1, 1)$
Q5Intersections of Lines and Planes
1. A laser beam travels along $ℓ : (x, y, z) = (4, - 2, 1) + t (- 1, 3, 1)$ and strikes the screen lying in the plane $π : x + y + z - 6 = 0$ . Find the point where it meets the screen.
2. Three flat panels meet along the edges of a corner bracket: $π_{1} : x + y + z - 6 = 0, π_{2} : 2 x - y + z - 3 = 0, π_{3} : x + 2 y - z - 2 = 0 .$ Using row reduction on the augmented matrix, find all points common to the three planes, and say what the type of solution set means geometrically.
Q6Distances Between Points, Lines and Planes

Compute each of the following exactly; all coordinates are in metres.
1. The distance from the sensor at $P (3, - 2, 1)$ to the wall $π : 2 x - y + 2 z + 5 = 0$ .
2. The distance from the point $Q (2, 2, 4)$ to the line $ℓ : (x, y, z) = (1, 0, 2) + t (2, 1, - 2)$ .
3. The distance between the parallel planes $π_{1} : x - 2 y + 2 z - 3 = 0$ and $π_{2} : x - 2 y + 2 z + 6 = 0$ .
Q7Angles Between Lines and Planes

Give each angle to three decimal places where it is not exact, and take every angle to be the acute (or right) one.
1. Find the angle between the lines of directions $\vec{d_{1}} = (1, - 2, 2)$ and $\vec{d_{2}} = (2, 2, - 1)$ . Note that $\vec{d_{1}} \cdot \vec{d_{2}} < 0$ and explain what you do about that.
2. Find the angle between the line of direction $\vec{u} = (2, - 1, 2)$ and the plane $π : x + 2 y + 2 z - 7 = 0$ .
3. Find the angle between the planes $π : x + 2 y + 2 z - 7 = 0$ and $σ : 2 x - 2 y + z + 3 = 0$ .
4. The formula in (a) and (c) uses a cosine, the one in (b) a sine. Explain why, in one or two sentences.
Q8Synthesis — drawing on several topics in this set

A triangular sail is stretched between the mast fittings $A (1, 0, 0)$ , $B (0, 2, 0)$ and $C (0, 0, 3)$ , coordinates in metres; the sail lies in the plane $π$ through those three points. A guy rope runs along $ℓ : (x, y, z) = (0, 0, 4) + t (1, - 1, - 2)$ .
1. Find the cartesian equation of $π$ in the form $a x + b y + c z + d = 0$ .
2. Find the point where the rope's line meets the plane of the sail.
3. Find the acute angle between the rope and the sail, to three decimal places.
4. Find the distance from the mast foot at the origin to the plane of the sail.

The 7 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

Which stream is this for? This is CEGEP Linear Algebra — 201-NYC-05 under the legacy numbering, 201-SN4-RE under the current one, both the same course — written against ministerial competency 0M04, Analyser des problèmes par l'utilisation de concepts de l'algèbre linéaire et de la géométrie vectorielle, in the Sciences de la nature programme. The full title of the competency is linear algebra and vector geometry, and this set is the geometry half arriving at its destination. Scope follows the devis, whose précisions enumerate the forms exactly: a line has vector, parametric and symmetric equations; a plane has vector, parametric and cartesian equations. That enumeration is why converting between forms is treated here as a skill in its own right rather than as notation.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

This is the payoff set, not a new topic

Almost nothing here is a new idea. A line in space is a base point and a direction vector; a plane is a base point and a normal vector; and every question is one of those two objects handled with tools the course has already built. What is new is that you now have to choose the tool, and the choice is the thing being marked.

Where each earlier block reappears: the cross product builds a normal from two directions, and its norm gives a point-to-line distance; the dot product tests perpendicularity and produces every angle; the scalar triple product — a determinant — decides whether two lines and the gap between them are coplanar; row reduction finds where planes meet, and the type of the solution set is the geometric answer, not a step towards it. If a technique in this set feels shaky, the repair is usually in the set that introduced it.

Three forms for a line, and how to move between them

Q1 gives two points on a line and asks for all three forms in turn, then asks whether two given points lie on it. Take the parts in order — each one is produced from the one above it by a single mechanical step, and that is the whole conversion skill.

From two points to symmetric equations

Each step is the previous line rewritten, not a new calculation.

1
Direction vector
Subtract the two points: d = AB = B − A. Any non-zero multiple of it does the same job, so tidy the components if it helps.
2
Vector equation
(x, y, z) = A + t·d, with t ∈ ℝ — a base point plus a parameter times the direction. Say what the parameter ranges over.
3
Parametric equations
Read off the three components separately. Nothing is computed here; the vector equation already contains them.
4
Symmetric equations
Isolate t in each parametric equation and set the three expressions equal. That is the step that divides by each component of d — which is why it is a step with a condition attached, not a formula to memorise.

(x, y, z) = A + t·d → x = a₁ + t·d₁, y = a₂ + t·d₂, z = a₃ + t·d₃ → (x − a₁)/d₁ = (y − a₂)/d₂ = (z − a₃)/d₃

Part (d) is the test the rest of the set keeps reusing: a point lies on the line exactly when the vector from the base point to it is a scalar multiple of the direction — and one single scalar has to work for all three components at once. Two components agreeing is not a result. Check the third every time.

Three forms for a plane, and the normal that unlocks the third

Q2 does the same tour for a plane through three given points: vector equation, parametric equations, then a normal vector and the cartesian equation ax + by + cz + d = 0, and finally a test of whether two points lie in it.

A plane needs two directions, a line needs one. From three points, take two vectors out of the same corner — that gives the vector and parametric forms directly. The cartesian form is a different description: it is built from the normal, and the normal is where the cross product enters, because the one direction orthogonal to both of those vectors is their cross product.

n = AB × AC, then n · ((x, y, z) − A) = 0 expands to ax + by + cz + d = 0

Two habits that cost nothing and save marks. First, substitute the other given points back into your cartesian equation — a normal computed with one sign slip is caught instantly this way, and the check is two lines of arithmetic. Second, notice that the cartesian form makes the point-in-plane test trivial: put the coordinates in and see whether you get zero. Comparing the forms is not busywork; each form makes a different question easy.

The case that only exists in space: skew lines

Q3 gives two pairs of lines and asks for their relative position — the point of intersection where there is one, and the reason where there is not. This is the question students carry a wrong instinct into, straight from geometry in the plane.

In space, two lines that do not meet need not be parallel. They can be skew: different directions, no common point, and no plane containing them both. That case has no analogue in the plane — in ℝ² two lines that fail to intersect really are parallel — and it is created by the third dimension. So "the system has no solution, therefore they are parallel" is not an argument in this course. Parallel is a statement about the directions, and it has to be checked separately.

Sorting two lines in space

Directions first, points second. The order matters.

1
Are the directions proportional?
If one direction is a scalar multiple of the other the lines are parallel. Then test whether one line's base point satisfies the other line — the same line written twice, or two distinct parallel lines.
2
Not proportional: set the coordinates equal
Three equations in the two parameters t and s. Solve two of them, then test the pair in the third — the equation you did not use is the whole decision.
3
Consistent → they intersect
Substitute back into both lines and confirm you get the same point. That is the check that catches an arithmetic slip in the solve.
4
Inconsistent → skew
Not parallel and no common point leaves exactly one possibility. Name it: an over-determined system with no solution is what skewness looks like in algebra.

A line and a plane: one dot product, then possibly one substitution

Q4 fixes a plane and gives three lines, and asks you to state the test you used as well as the answer. The test has two stages, and the second stage is the one that gets skipped.

Stage one — compare the line's direction with the plane's normal. A non-zero dot product means the line cuts the plane in exactly one point; substitute the parametric coordinates into the cartesian equation, solve for the parameter, and report the point. Stage two — a zero dot product means the line is parallel to the plane, and that is not yet an answer: test one point of the line in the equation. Satisfied, and the line lies in the plane; not satisfied, and it misses the plane entirely. Two very different pictures share the same first stage.

Note the shape of that test, because it is the recurring surprise of this set: a zero dot product with the normal means the line is parallel to the plane. The normal is perpendicular to the plane, so being perpendicular to the normal is being parallel to the plane. Reading "orthogonal" as "crossing" here gets every part of the question backwards.

Intersections, where row reduction becomes a picture

Q5 has two halves. Part (a) is one line meeting one plane, solved by substituting the parametric coordinates into the cartesian equation — a single linear equation in the parameter. Part (b) is three planes at once, and it asks for the augmented matrix to be row reduced and for the type of the solution set to be interpreted geometrically.

Each solution type is a configuration: a unique solution means the three planes share exactly one point, like the three faces at the corner of a box; a solution with one free variable means they share a whole line, a pencil of planes through a common edge; two free variables mean the three equations describe a single plane; and an inconsistent system means no common point at all, which is a configuration and not a failure. Answer the geometry, not just the algebra — the sentence naming the picture is what the question is asking for.

Two supports worth using. The coefficient determinant settles the unique-solution case on its own — non-zero means three normals that are not coplanar, hence one common point — so it is a fast sanity check on a reduction. And the row operations are far easier to trust if you write each one down beside the matrix, in the notation the earlier sets used; a reduction with unlabelled steps cannot be marked, and cannot be debugged by you either.

Three distances, three different formulas

Q6 asks for a point-to-plane distance, a point-to-line distance, and the gap between two parallel planes, all exactly. They look alike and are not, so keep them apart by what each one divides by.

point → plane: |ax₀ + by₀ + cz₀ + d| / ‖n‖ point → line: ‖AQ × d‖ / ‖d‖

Point to plane substitutes the point into the left-hand side of the cartesian equation, takes the absolute value, and divides by the norm of the normal. Point to line uses a cross product, not a dot product, and divides by the norm of the direction. Two parallel planes: write both with the same normal first — otherwise nothing you compute afterwards is comparable — then take any convenient point of one plane and measure its distance to the other.

The absolute value is doing real work in the first formula: without it you get a signed quantity that tells you which side of the plane the point is on, which is sometimes useful and is never a distance. Each of these also has a free check available. If the vector from the line to the point turns out to be already perpendicular to the direction — dot product zero — then the point-to-line distance is simply the norm of that vector, and it should agree with the cross-product formula exactly.

Angles: cosine, cosine, sine — and the complement that explains the third

Q7 asks for the angle between two lines, between a line and a plane, and between two planes, and then part (d) asks you to explain why two of those use a cosine and one uses a sine. Get part (d) right and the other three stop being formulas to memorise.

The angle between a line and a plane is the complement of the angle between the line's direction and the plane's normal. The dot product can only compare two vectors, and the only vector the plane offers is its normal — which points away from the plane, not along it. So the usual quotient gives cos θ for the direction-to-normal angle θ, while the angle φ you were actually asked for satisfies φ = 90° − θ. Since sin φ = cos θ, that same quotient is read as a sine. Feed it to arccos instead and you report the complement of the right answer, with no arithmetic error anywhere in your work.

two lines: cos θ = |d₁ · d₂| / (‖d₁‖‖d₂‖) line and plane: sin φ = |u · n| / (‖u‖‖n‖)

Two lines are compared through their directions and two planes through their normals; in both of those cases the two vectors are of the same kind, the angle between the vectors is the angle asked for, and a cosine is correct. Take the absolute value in every case: a line has no preferred direction and a plane no preferred side, so reversing a vector must not change the answer, and the absolute value is what selects the acute angle. Q7(a) is built on exactly that point — it hands you two directions whose dot product is negative and asks what you do about it.

The synthesis question, and why it is last

Q8 is a single physical set-up — a triangular sail on three fittings, a guy rope along a given line — carrying four parts that are four different techniques from this set applied to the same two objects: build the plane's cartesian equation from three points, find where the line meets it, find the angle between the line and the plane, and find a distance from a point to the plane. Nothing in it is new, which is the point: on a final, questions do not arrive labelled with the section they came from. Work part (a) carefully and check it, because the remaining parts all consume its normal vector, and a sign slip there propagates through everything after it.

Getting the most out of it

Write down what each object is before you compute

A line is a base point and a direction; a plane is a base point and a normal. Extract those first and name them, and most questions in this set collapse to choosing which product to apply to which pair. Starting from the equations exactly as printed, without naming the vectors inside them, is what makes these questions feel long.

Directions before points, always

Every relative-position question — two lines, a line and a plane, two planes — is decided in two stages, and the first is a comparison of directions or normals. Only when that stage returns "parallel" does a point test decide between coincident, contained and disjoint. Running the stages in the other order produces a lot of algebra and no conclusion.

Use the third equation you did not use

Setting two lines equal gives three equations in two parameters. Solve two, then substitute into the third: that is not a check bolted on at the end, it is the step that separates intersecting from skew. An answer that never touches the third equation has not answered the question.

Substitute your answers back

Cross products, determinants and row reductions all fail the same way — one sign, silently. But every answer in this set is verifiable in a couple of lines: put the given points into your cartesian equation, put an intersection point into both original objects, put the parameter back into both lines. The devis asks for rigorous mathematical reasoning, and a stated check is part of what that looks like on a CEGEP final.

Want the solutions, or something more challenging?

The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the CEGEP Linear Algebra Solutions Bundle, which is what keeps the rest of the series free.

What else exists for Lines and Planes in Space

Three PDFs · 16 pages · all three are in the bundle below.

Answer key — 4 pages. All 8 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
Challenge problems — 7 pages, 7 problems. A separate sheet at exam-plus difficulty covering the same 7 concepts. Harder than anything on the free sheet.
Challenge answer key — 5 pages. Every challenge problem worked to the same standard, with the checks shown.
PDF, letter size, print-ready.

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Common questions

Is this worksheet really free?

Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete CEGEP Linear Algebra Solutions Bundle, which covers every set at this level.

Is this for 201-NYC-05 or 201-SN4-RE?

Both — they are the same course under two numbering systems, the first the legacy code and the second the current one. The content is written against ministerial competency 0M04 in the Sciences de la nature programme, so it matches whichever code your college prints on the outline.

What are skew lines, and why did they never come up before?

Skew lines have different directions and no common point, and no single plane contains them both. They cannot happen in the plane: two lines in ℝ² that do not meet must be parallel, which is the rule most students bring with them. The third dimension creates the extra case, so in space the directions and the intersection have to be checked separately — neither answer implies the other.

Which form of a line's equation should I use?

Whichever makes the next step easy, and expect to convert. The vector and parametric forms carry a parameter, so they are what you substitute into a plane's equation or set equal to another line. The symmetric form has no parameter, which makes it a quick test of whether a point is on the line. The devis asks for all three, so a question can name any of them.

Why does a zero dot product mean the line is parallel to the plane?