CEGEP Linear Algebra Vectors in the Plane and in Space Free · no sign-up

CEGEP Linear Algebra — Vectors in the Plane and in Space

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

This is where the geometry half of the course begins. Everything up to here — matrices, systems, inverses, determinants — was algebra you could do without drawing anything; from here on there is a picture behind every calculation. What a vector actually is, and why the same vector can be drawn in two places; components from two points, head minus tail; norm, direction angles and the identity the three direction cosines must satisfy; addition and scalar multiplication done componentwise, and a vector equation solved like a scalar one; the tip-to-tail and parallelogram pictures on a grid; linear independence decided by solving a homogeneous system rather than by looking; a basis, and why the components of a vector in it are unique; and normalisation, including the one vector that cannot be normalised. 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 5 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.

Q1Geometric and Algebraic Vectors

Points are given in the plane and in space.
1. In the plane, $A (- 2, 5)$ , $B (4, 1)$ and $C (0, - 3)$ . Write $\vec{A B}$ and $\vec{B C}$ in component form.
2. Find the point $D$ for which $\vec{C D} = \vec{A B}$ .
3. The segments $A B$ and $C D$ are different segments of the plane. Explain, in one sentence, in what sense $\vec{A B}$ and $\vec{C D}$ are nevertheless the same vector.
4. In space, $P (1, - 2, 4)$ and $Q (- 3, 0, 5)$ . Write $\vec{P Q}$ and $\vec{Q P}$ in component form, and state the relation between them.
Q2Norm, Direction and Direction Angles

Give every angle in degrees, to three decimals.
1. For $\vec{u} = (- 5, 12)$ , find $‖ \vec{u} ‖$ and the direction angle $θ \in [0^{\circ}, 360^{\circ})$ that $\vec{u}$ makes with the positive $x$ -axis.
2. For $\vec{v} = (2, - 3, 6)$ , find $‖ \vec{v} ‖$ , the three direction cosines, and the direction angles $α$ , $β$ , $γ$ .
3. Verify that your three direction cosines satisfy $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$ , and say why that had to happen.
Q3Vector Operations and Linear Combinations

Let $\vec{u} = (3, - 1, 2)$ , $\vec{v} = (- 2, 4, 0)$ and $\vec{w} = (1, 1, - 5)$ .
1. Compute $2 \vec{u} - 3 \vec{v} + \vec{w}$ .
2. Compute $‖ \vec{u} + \vec{v} ‖$ , leaving the answer in exact form.
3. Solve $3 \vec{x} - \vec{u} = 2 (\vec{x} + \vec{v})$ for the vector $\vec{x}$ .
Q4Representing Vectors in the Plane and in Space

Use the grid below, with the origin at its centre and one square as one unit.
1. Draw $\vec{u} = (4, 1)$ and $\vec{v} = (- 2, 3)$ as arrows starting at the origin.
2. Draw $\vec{u} + \vec{v}$ starting at the origin, and outline the parallelogram whose sides at the origin are $\vec{u}$ and $\vec{v}$ .
3. Draw the arrow that starts at the tip of $\vec{v}$ and ends at the tip of $\vec{u}$ . State its components and say which vector it represents.
4. A classmate says the arrow from the tip of $\vec{u}$ to the tip of $\vec{v}$ represents $\vec{u} - \vec{v}$ just as well. Is that right? Justify in one sentence.
A blank Cartesian grid for this question is on the printable PDF.
Q5Linear Independence and Dependence

For each set, decide whether it is linearly independent by determining all solutions of $c_{1} {\vec{u}}_{1} + c_{2} {\vec{u}}_{2} + \dots = \vec{0}$ . Where a set is dependent, write down an explicit dependence relation.
1. $\vec{u} = (1, 2, - 1)$ , $\vec{v} = (3, 1, 4)$ .
2. $\vec{u} = (1, 0, 2)$ , $\vec{v} = (2, - 1, 1)$ , $\vec{w} = (4, - 1, 5)$ .
3. $\vec{u} = (1, 1, 0)$ , $\vec{v} = (0, 1, 1)$ , $\vec{w} = (1, 0, 1)$ .
Q6Bases and Components
1. Show that $ℬ = {{\vec{b}}_{1}, {\vec{b}}_{2}}$ , with ${\vec{b}}_{1} = (2, 1)$ and ${\vec{b}}_{2} = (- 1, 3)$ , is a basis of $ℝ^{2}$ , then find the components of $\vec{x} = (7, 7)$ in $ℬ$ .
2. Show that $𝒞 = {{\vec{c}}_{1}, {\vec{c}}_{2}, {\vec{c}}_{3}}$ , with ${\vec{c}}_{1} = (1, 0, 1)$ , ${\vec{c}}_{2} = (0, 1, 1)$ and ${\vec{c}}_{3} = (1, 1, 0)$ , is a basis of $ℝ^{3}$ , then find the components of $\vec{y} = (4, 2, 6)$ in $𝒞$ .
3. Explain why a vector cannot have two different sets of components in the same basis.
Q7Unit Vectors and Normalization
1. Find the unit vector in the direction of $\vec{u} = (6, - 3, 2)$ .
2. Find the vector of magnitude $21$ pointing in the sense opposite to $\vec{u}$ .
3. A vector $\vec{v}$ in the plane has magnitude $20$ and direction angle $150^{\circ}$ . Write $\vec{v}$ in exact component form.
4. Explain why $\frac{\vec{u}}{‖ \vec{u} ‖}$ is undefined when $\vec{u} = \vec{0}$ , and what that says about the direction of the zero vector.
Q8Synthesis — drawing on several topics in this set

A courier drone flies in a straight line from a depot at $D (2, - 1, 0)$ to a rooftop at $R (8, 5, - 3)$ . Coordinates are in hundreds of metres.
1. Write the displacement $\vec{d} = \vec{D R}$ in component form and find $‖ \vec{d} ‖$ . How many metres does the drone travel?
2. Find the unit vector $\hat{d}$ in the direction of the flight, and the direction angles of $\vec{d}$ to three decimals of a degree.
3. The drone must report its position when it is $600$ m from the depot along the straight route. Find the coordinates of that point.
4. Two fixed corridor directions are $\vec{p} = (4, 2, - 1)$ and $\vec{q} = (2, 4, - 2)$ . Express $\vec{d}$ as a linear combination of $\vec{p}$ and $\vec{q}$ , then state whether ${\vec{p}, \vec{q}, \vec{d}}$ is linearly independent, with a reason.

The 5 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 that competency names vector geometry alongside linear algebra, and this set is where the geometry half starts — the half a calculus-trained student finds least familiar, because nothing in it looks like a function. Scope follows the devis rather than any one college's outline, which fixes two things students often expect and will not find here: there are no abstract vector spaces — no axioms, no subspaces, no dimension theory — and no eigenvalues. Independence and bases appear in their concrete ℝ² and ℝ³ form, which is exactly the form the devis names.

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.

Why this set has no dot product in it

Read the questions and you will notice something missing. There is no dot product here, no cross product, no projection and no triple product — those are the whole of the next set. So an angle is found from direction angles and plane trigonometry, and linear independence is settled by solving a system. That is not an oversight; it is the order the subject is best learnt in.

The products are shortcuts, and a shortcut is only useful once you know what it is short for. Perpendicularity found by comparing two direction angles, and coplanarity found by solving c₁u + c₂v + c₃w = 0, are the definitions themselves. Meet the dot product first and it is a formula you apply; meet it after this set and it is a computation that replaces work you have already done by hand. Everything you build here is what the next set makes fast.

A vector is fixed by three things, and where you drew it is not one of them

Q1 starts in the plane with three points, asks for two vectors between them, then asks you to find a fourth point D making one vector equal to another. Part (c) is the question the whole set rests on: AB and CD are literally different segments of the plane, so in what sense are they the same vector?

A vector carries a norm, a direction and a sense — and nothing else. Two arrows with the same components are the same vector, however far apart they are drawn. That is why part (b) is answered by adding a displacement to a point rather than by solving anything, and it is the single idea that makes the rest of the course work: a direction vector for a line, a normal vector for a plane, none of them have a location.

components of AB = (head) − (tail), so BA = −AB

Part (d) repeats the whole thing in ℝ³ with PQ and QP and asks for the relation between them. Everything in this set that works in the plane works in space with one more coordinate; the point of putting both in one question is that you stop treating them as two topics.

Norm and direction angles: the identity is not optional

Q2 asks for a norm and a direction angle in the plane, then a norm, three direction cosines and three direction angles in space, all in degrees to three decimals. Two habits are being trained, and both are places marks are lost.

A calculator's arctan does not know which quadrant you are in. Compute the reference angle from the absolute values, then place the angle yourself using the signs of the components — a vector with a negative first component and a positive second one is in the second quadrant whatever the calculator returned. In space the same information arrives differently: a negative direction cosine is what pushes its direction angle past 90°, and that sign is telling you the vector points backwards along that axis.

cos²α + cos²β + cos²γ = 1

Part (c) asks you to verify that identity and to say why it had to hold, which is the real question. The three direction cosines are the components of the unit vector in the same direction, and the squared components of any vector sum to the square of its norm — here, to 1. So they are not three independent pieces of data: fix two and the third is determined up to sign. That is worth holding on to, because it is what makes several later questions answerable at all.

Componentwise operations, and a vector equation you may rearrange

Q3 is three parts on one triple of vectors: a linear combination, a norm of a sum left in exact form, and an equation solved for an unknown vector. The third is the interesting one.

You may collect terms in a vector equation exactly as in a scalar one — but the reason is worth stating rather than assuming. Addition and scalar multiplication act componentwise, so each coordinate is an ordinary linear equation and the familiar rearrangement is three ordinary rearrangements done at once. Solve for the vector symbolically first and substitute components at the end; expanding into coordinates immediately triples the arithmetic and buries the structure.

Part (b) says exact form, which means a surd, not a decimal. Norms rarely come out whole, and a decimal at this stage is an approximation carried into every line that follows. Leave the radical.

The grid question is not a colouring exercise

Q4 gives you a grid and asks for two vectors drawn from the origin, their sum, and the parallelogram they span. Then it asks for the arrow running from the tip of one to the tip of the other, and what vector that arrow represents. The picture is where the algebra becomes memorable.

Tip-to-tail and the parallelogram rule give the same arrow — the sum is one diagonal of the parallelogram, and the difference is the other. For the difference, the order matters and the sense is the thing to get right: the arrow representing u − v is the one that ends at the tip of u. Swap the two endpoints and you have drawn a vector of the same length and direction but the opposite sense, which is a different vector. Part (d) is exactly that check, so read the endpoints before you name the arrow.

Linear independence is not something you can eyeball

Q5 gives you three sets of vectors and one instruction: decide independence by determining all solutions of the homogeneous vector equation, and where a set is dependent, write down an explicit dependence relation. The instruction is the lesson.

Independence is a question about an equation, not an impression of a picture. The set {u, v, w} is independent when

c₁u + c₂v + c₃w = 0 forces c₁ = c₂ = c₃ = 0

and that is a homogeneous linear system in the unknowns c₁, c₂, c₃ — the machinery from earlier in the course, put to a geometric purpose. Only the trivial solution: independent. A free variable, so a non-trivial solution: dependent, and any non-zero choice of the free parameter hands you the explicit relation the question asks for. Nothing about "they look like they point different ways" is an argument.

Two shortcuts are legitimate, and worth knowing precisely because they follow from that system rather than replacing it. Two vectors are dependent exactly when one is a scalar multiple of the other, which is the whole of the two-vector case. And with three vectors in ℝ³ the coefficient matrix is square, so the determinant settles it in one line — zero determinant, non-trivial solutions, dependent. Part (c) is placed where it is so you can see the determinant do in a line what parts (a) and (b) did by elimination.

What dependence means geometrically: two dependent vectors are collinear — they lie on one line through the origin. Three dependent vectors in ℝ³ are coplanar — one of them is a combination of the other two, so all three, drawn from the origin, fit in a single plane. Independent vectors do not collapse like that, and three independent vectors in ℝ³ open out a genuine box.

A basis, and why the components in it are unique

Q6 asks you to show that a given pair is a basis of ℝ², find the components of a vector in it, then do the same with a triple in ℝ³. Showing a set is a basis has two halves, and the determinant is what lets you do both at once.

Showing a set is a basis, and using it

In ℝ² with two vectors, or ℝ³ with three. Work down the list.

1
Count first
A basis of ℝⁿ has exactly n vectors. Two vectors can never span ℝ³, and four vectors in ℝ³ can never be independent, so a wrong count ends the question before any arithmetic.
2
Take the determinant of the matrix carrying them
Non-zero means the homogeneous system has only the trivial solution, so the set is independent — and the same non-zero determinant makes the non-homogeneous system solvable for every right-hand side, so the set spans. One computation, both halves.
3
Components are the solution of a system
The components of x in the basis are the coefficients with c₁b₁ + c₂b₂ + … = x. Set up that system and solve it. The answer is an ordered tuple relative to that basis, and it is not the same tuple as the standard components of x.
4
Check by rebuilding the vector
Multiply your coefficients back through the basis vectors and add. Thirty seconds, and it catches every sign slip.

Part (c) asks why a vector cannot have two different sets of components in the same basis, and the proof is three lines you should be able to reproduce: suppose two expansions, take the difference, and you have a combination of the basis vectors equal to the zero vector. Independence forces every coefficient of such a combination to vanish, so the two expansions had the same coefficients all along. Uniqueness of components is precisely what independence buys you — it is what independence is for.

Normalisation, and the vector that has no direction

Q7 asks for a unit vector, then a vector of a prescribed magnitude in the opposite sense, then a plane vector reconstructed from its magnitude and direction angle in exact form. Three questions, one move.

Normalise, then scale. Dividing by the norm strips out the length and leaves the direction; multiplying by any real number then sets the length, and a negative multiplier reverses the sense at the same time. Building the vector in two steps like that is why part (b) needs no new idea. Part (c) runs the machine backwards: magnitude times (cos θ, sin θ) rebuilds a plane vector from its polar description, and the direction angle determines the direction cosines, which are the components of the unit vector.

The zero vector cannot be normalised, and part (d) asks you to say what that means rather than just to notice the division by zero. Its norm is 0, so the quotient is undefined — but the formula is not being awkward. The zero vector has no direction and no sense to record, so it has no direction angles either; there is nothing for the calculation to return. Every non-zero vector, by contrast, has exactly one unit vector in its direction.

The synthesis question, and why it is last

Q8 is a straight-line flight between two points in space, with coordinates in hundreds of metres, and it uses the whole set in order. Part (a) is components from two points and a norm, plus a unit conversion — read the units line, because the norm is not yet the answer in metres. Part (b) is the unit vector and the direction angles, and the sign of one of them should match the fact that the flight descends; if it does not, you have an arithmetic error rather than a surprise. Part (c) is the reason normalisation was worth learning: a point a given distance along a route is the starting point plus that distance times the unit vector, which is the whole idea of a direction vector doing the steering while a scalar does the walking. Part (d) then hands you two fixed directions, asks you to express the flight as a combination of them, and asks whether the three together are independent — with a reason. Solve the system, and let the coefficients answer both questions at once.

Getting the most out of it

Head minus tail, every time

Almost every error in this set that is not a sign slip is a subtraction done the wrong way round. The components of the vector from P to Q are Q − P. Write the letters down in the order the question names them, subtract in that order, and if the answer looks like the reverse of what you expected, check the order before you check the arithmetic.

Decide independence by solving, not by looking

Set up c₁u + c₂v + … = 0, turn it into a homogeneous system, and read off the solution set. Only the trivial solution means independent; a free variable means dependent, and it hands you the relation. With three vectors in ℝ³ the determinant is a legitimate one-line version of the same test — but it is the same test, and knowing that is what stops you reaching for a determinant in a case where the matrix is not square.

Keep exact values exact

Norms are surds and direction cosines are fractions. Carry them as they are and round only at the last step — the angle questions ask for three decimals of a degree, and rounding a cosine first is how a third decimal goes wrong. Where a question says exact form, a decimal is not an answer.

Say what the numbers mean

Several parts here ask for a sentence, not a value: in what sense two segments are the same vector, why the direction cosines had to satisfy that identity, why components in a basis are unique, what a dependence relation means geometrically. The devis asks for a just interpretation of linear independence and dependence, and an interpretation is written in words. Answer those parts in a full sentence — they are the ones a marker reads closely.

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 Vectors in the Plane and in Space

Three PDFs · 12 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 — 5 pages, 5 problems. A separate sheet at exam-plus difficulty covering the same 7 concepts. Harder than anything on the free sheet.
Challenge answer key — 3 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.

Why is there no dot product or cross product here?

Because they are the next set. Angles here come from direction angles, and independence from solving a homogeneous system — the definitions themselves rather than the shortcuts. The products are much easier to use well once you have built the geometry by hand, and everything on this sheet is what they will later make fast.

How do I tell whether a set of vectors is linearly independent?

Solve c₁u + c₂v + … = 0. If the only solution is all coefficients zero, the set is independent; if there is a non-zero solution, it is dependent, and that solution is the dependence relation. It is a homogeneous system, not a judgement call — and with three vectors in ℝ³ the determinant of the matrix carrying them is a one-line form of the same test.

No two of my vectors are parallel. Doesn't that make the set independent?

No, and this is the commonest wrong instinct in the topic. Parallel is a condition on a pair; dependence is a condition on the whole set. Three vectors can have no two of them parallel and still be dependent, because one of them is a combination that genuinely uses both of the others — geometrically, all three lie in one plane through the origin.

Why can't I choose all three direction angles freely?

Because the direction cosines are the components of a unit vector, so their squares must sum to 1. Fix two of the angles and the identity determines the third up to sign — which is why some combinations of angles are simply impossible, and why others correspond to two vectors rather than one.

Does this cover vector spaces, subspaces and eigenvalues?

No. Competency 0M04 names none of them, so the course stays concrete in ℝ² and ℝ³: no axioms, no subspaces, no dimension theory, no eigenvalues. Independence and bases appear here in their concrete form, which is the form the devis asks for and the form your final will test.

Can teachers use this in class?

Yes. Print and photocopy it for your own classes freely — I just ask that the tutorinmontreal.ca footer stays on the page.

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Yes — through one-on-one tutoring, in Montreal or online. Get in touch to arrange a session, or see the current rates.

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