MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: geometric vectors and lines in space (MATH 133)

This sheet is not a summary of section 4.1 of Nicholson: you already have the notes. It answers one question only, what makes students lose marks on geometric vectors and lines in MATH 133 at McGill University, and which precise gesture avoids each loss.

Almost every trap on this chapter comes from one fact: the same line can be written in infinitely many ways. So the sheet is built around two tests, the direction test and the point test, applied in that order every time two lines appear, and around the rule that each line keeps its own parameter.

The thread of the chapter

A line has infinitely many equations, because neither its point nor its direction is unique: two lines are never compared by reading their equations, they are compared by a DIRECTION test on all three components, then a POINT test, with one independent parameter per line and the third equation always checked.

This chapter is part of MATH 133, Linear Algebra and Geometry (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Geometric vectors in space
Parallelogram rule
Norm of a vector
Unit vector
Distance between two points
Parallel vectors
Midpoint and section formula
Centroid of a triangle
Vector equation of a line
Parametric and symmetric equations
Skew lines
Intersection of two lines

The essentials

A vector is a move, a point is a place

• $\overrightarrow{AB} = B - A$ , the END minus the START. Reversed, it is the same length pointing the other way: $\overrightarrow{BA} = -\overrightarrow{AB}$ .
• Chasles: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ . Two moves in a row are one move.
• On the parallelogram built on $\mathbf{u}$ and $\mathbf{v}$ from one point, the diagonal leaving that point is $\mathbf{u} + \mathbf{v}$ , and the other diagonal, from the tip of $\mathbf{v}$ to the tip of $\mathbf{u}$ , is $\mathbf{u} - \mathbf{v}$ .
• $\|(a, b, c)\| = \sqrt{a^2 + b^2 + c^2}$ and $\|c\mathbf{v}\| = |c|\,\|\mathbf{v}\|$ . The distance $AB$ is $\|\overrightarrow{AB}\|$ : subtract first, square second.
• $\mathbf{u}$ and $\mathbf{v}$ , both non-zero, are parallel when $\mathbf{v} = c\,\mathbf{u}$ with ONE scalar $c$ valid on every component.
• $ABCD$ , in that order, is a parallelogram when $\overrightarrow{AB} = \overrightarrow{DC}$ , which is the same as $A + C = B + D$ .

Both diagonals come from the same two vectors:

\mathbf{u} + \mathbf{v}

leaves the common origin

O

, and

\mathbf{u} - \mathbf{v}

goes from the tip of

\mathbf{v}

to the tip of

\mathbf{u}

A vector proof is one equality between two moves. If your proof uses the figure to conclude, it is not a proof yet.

A line is one point plus one direction, and neither is unique

• Vector form $\mathbf{x} = \mathbf{p} + t\mathbf{d}$ ; parametric $x = x_0 + at$ , $y = y_0 + bt$ , $z = z_0 + ct$ .
• Symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ , only with $a$ , $b$ , $c$ non-zero. A zero component becomes a separate equation, such as $y = y_0$ .
• In symmetric form, each numerator must read coordinate MINUS number, with coefficient $1$ on the coordinate: $\frac{1 - y}{2} = \frac{y - 1}{-2}$ and $\frac{2z - 4}{6} = \frac{z - 2}{3}$ .
• Any point of the line can replace $\mathbf{p}$ , and any non-zero multiple of $\mathbf{d}$ can replace $\mathbf{d}$ : the same line has infinitely many correct equations.
• Section formula: $AP : PB = m : n$ gives $P = \frac{nA + mB}{m + n}$ . Midpoint $\frac{1}{2}(A + B)$ , centroid $\frac{1}{3}(A + B + C)$ .

A point is on a line when ONE value of $t$ gives all three coordinates. Three equations, one unknown, and the third coordinate is checked, never assumed.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Two lines in space: what the two tests say

Direction test first, then the point test or the system in $t$ and $s$ . The last column is the conclusion; the red row is a test that does not exist.

Directions	Point test or system	Position
proportional	common point	coincident
Example: $(1, 0, -1) + t(2, -1, 3)$ and $(5, -2, 5) + s(-4, 2, -6)$ : factor $-2$ , and $(5, -2, 5)$ is the first line at $t = 2$ .
proportional	no common point	parallel, distinct
Example: Same first line and $(3, 1, 2) + s(4, -2, 6)$ : factor $2$ , but $x = 3$ needs $t = 1$ , which gives $y = -1 \ne 1$ .
not proportional	solution, all 3 equations	one point
Example: $(1, 2, 0) + t(1, -1, 2)$ and $(4, 2, 3) + s(1, 2, -1)$ : $t = 2$ , $s = -1$ , third equation $4 = 4$ , point $(3, 0, 4)$ .
not proportional	third equation fails	skew
Example: $(1, 2, 0) + t(1, -1, 2)$ and $s(2, 1, 1)$ : the first two give $t = s = 1$ , the third gives $2 \ne 1$ .
not checked	same letter t in both	no conclusion not a test
Example: $1 + t = 4 + t$ gives $1 = 4$ , yet the two lines of the third row meet at $(3, 0, 4)$ . What to do: Rename the second parameter $s$ before writing a single equation, then solve two equations and check the third.

No solution to the system means parallel OR skew. Only the direction test separates them, which is why it comes first.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. The same parameter letter for both lines

the whole question

What not to write

“ $1 + t = 4 + t$ gives $1 = 4$ , so the lines do not intersect.”

What to write

“Let $s$ be the parameter of $L_2$ . Then $1 + t = 4 + s$ , $2 - t = 2 + 2s$ , $2t = 3 - s$ give $t = 2$ , $s = -1$ , and the lines meet at $(3, 0, 4)$ .”

Why: The parameter says where you are on ONE line. Using one letter for both asks whether two particles are at the same place at the same time, a stronger question that is usually false even when the lines cross.

2. Solving two coordinate equations and stopping

2 to 3 marks, and every later part

What not to write

“From $1 + t = 2s$ and $2 - t = s$ , $t = s = 1$ , so the lines meet at $(2, 1, 2)$ .”

What to write

“The third equation gives $2t = 2$ but $s = 1$ , and $2 \ne 1$ : no common point. The directions are not proportional, so the lines are skew.”

Why: Two equations in two unknowns almost always have a solution. In $\mathbb{R}^3$ the third equation is the actual test; announcing a point without checking it is announcing a guess.

3. Proportional directions read as the same line

2 marks

What not to write

“ $(4, -2, 6) = 2(2, -1, 3)$ , so the two lines are the same.”

What to write

“The directions are parallel. The point $(3, 1, 2)$ needs $t = 1$ on the first line, which gives $y = -1 \ne 1$ : the lines are parallel and distinct.”

Why: Proportional directions only say parallel OR equal. The point test is what separates the two, and it takes one line to write.

4. Calling two lines that never meet parallel

the classification mark

What not to write

“The system has no solution, so the lines are parallel.”

What to write

“The system has no solution and the directions are not proportional, so the lines are skew.”

L_1

runs along the top front edge,

L_2

across the floor: they never meet and do not point the same way. No solution plus non-proportional directions means skew.

Why: In the plane, lines that never meet are parallel. In space they can also be skew, which is in fact the typical case: two edges of a box, one on the top face and one on the floor, never meet and are not parallel.

5. Dividing by a zero component in the symmetric form

1 mark

What not to write

“ $\frac{x - 2}{4} = \frac{y + 1}{0} = \frac{z - 3}{-2}$ .”

What to write

“ $\frac{x - 2}{4} = \frac{z - 3}{-2}$ and $y = -1$ .”

Why: A zero component means that coordinate never changes along the line. It is not divided by anything, it is written as its own equation.

6. Reading the point and direction off a symmetric form that is not in standard shape

the whole line, and every question built on it

What not to write

“ $\frac{x + 2}{3} = \frac{1 - y}{2} = \frac{2z - 4}{6}$ has point $(-2, 1, 4)$ and direction $(3, 2, 6)$ .”

What to write

“ $\frac{1 - y}{2} = \frac{y - 1}{-2}$ and $\frac{2z - 4}{6} = \frac{z - 2}{3}$ , so the point is $(-2, 1, 2)$ and the direction $(3, -2, 3)$ .”

Why: The form must read coordinate minus number over a component, with coefficient $1$ on the coordinate. A minus in front of $y$ goes into the denominator, a factor in front of $z$ is divided out.

7. The weight on the wrong end of a ratio

2 marks

What not to write

“ $AP : PB = 2 : 1$ , so $P = \frac{1}{3}(2A + B)$ .”

What to write

“ $P = A + \frac{2}{3}\overrightarrow{AB} = \frac{1}{3}(A + 2B)$ : $P$ is closer to $B$ , so $B$ gets the weight $2$ .”

Three equal parts:

P

sits after two of them, next to

B

, and is

\frac{1}{3}(A + 2B)

. The swapped weights give the point one third of the way, next to

A

Why: The ratio counts equal parts from $A$ . The point after two parts out of three is two thirds of the way, and the heavier weight goes on the endpoint it is near. With $A(1, 2, -1)$ and $B(7, -1, 5)$ the right answer is $(5, 0, 3)$ and the wrong one $(3, 1, 1)$ .

8. The order of the letters in a parallelogram

1 to 2 marks

What not to write

“ $A(1, 1)$ , $B(5, 2)$ , $C(2, 4)$ : the fourth vertex of $ABDC$ is $A + C - B = (-2, 3)$ .”

What to write

“In $ABDC$ the diagonal from $A$ is $AD$ , so $D = A + \overrightarrow{AB} + \overrightarrow{AC} = (6, 5)$ , and $\overrightarrow{CD} = (4, 1) = \overrightarrow{AB}$ .”

Why: Three points give three parallelograms, one for each vertex that can be opposite the missing one. The order of the letters in the statement picks one, and the check $\overrightarrow{AB} = \overrightarrow{DC}$ in that order confirms it.

Which method to choose

Two lines are given: which test, in which order

Look at the two direction vectors before writing any system

The four outcomes, left to right: proportional directions with a common point, proportional without one, non-proportional with a solution, non-proportional with the third equation failing.

If directions proportional, same factor on all three components → test ONE point of the second line on the first

Example: $(-4, 2, -6) = -2(2, -1, 3)$ , and $(5, -2, 5)$ is on the first line at $t = 2$ : coincident

point on it: coincident; point off it: parallel and distinct
If directions not proportional → rename one parameter, solve two coordinate equations, then check the THIRD

Example: $t = 2$ , $s = 1$ from two equations, then $5 \ne 2$ in the third: skew

third equation true: one point, computed from BOTH lines; false: skew
If one direction has a zero component → compare components with a ratio only where it is defined, and require the zero to match

Example: $(2, 0, 3)$ and $(4, 0, 6)$ are parallel; $(2, 0, 3)$ and $(4, 1, 6)$ are not
If the lines are given in symmetric form → convert both to vector form first, then run the same two tests

Example: $\frac{x + 2}{3} = \frac{1 - y}{2} = \frac{2z - 4}{6}$ becomes $(-2, 1, 2) + t(3, -2, 3)$

The distance between two skew lines needs the cross product, in the next chapters. On this chapter, skew is a conclusion, not a starting point for a computation.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Deciding the position of two lines

When to use it: The statement gives two lines and asks whether they intersect, or asks for their relative position

1 Write both lines with DIFFERENT parameters, $t$ for the first and $s$ for the second, saying so in words.
2 Direction test: compare the direction vectors component by component and write the factor, or write that no single factor exists.
3 If parallel: test one point of the second line on the first, with all three coordinates.
4 If not parallel: equate the coordinates, solve two equations for $t$ and $s$ , and substitute into the THIRD, writing the numbers on both sides.
5 Conclude with the name of the position, and if the lines meet, compute the point from both lines.

Concluding sentence

“The directions are not proportional and the values $t = 2$ , $s = -1$ satisfy all three equations, so the lines intersect at $(3, 0, 4)$ .”

The trap: Writing the intersection point after two equations. The third equation is the one the marker looks for first.

Marking: Typically 1 mark for the direction test, 2 for the system with the third equation checked, 1 for the named conclusion and 1 for the point.

Writing a vector proof

When to use it: Prove that a quadrilateral is a parallelogram, that points are collinear, or that segments bisect each other

1 Write every point you need as a combination of the given ones: midpoints $\frac{1}{2}(A + B)$ , section points, centroid.
2 Write the two vectors whose equality proves the claim, each as end minus start.
3 Simplify both to the same expression, for example $\frac{1}{2}(C - A)$ on each side.
4 Conclude with the definition you used, word for word.

Concluding sentence

“ $\overrightarrow{PQ} = \frac{1}{2}\overrightarrow{AC} = \overrightarrow{SR}$ , so $PQRS$ has a pair of opposite sides equal as vectors and is a parallelogram.”

The trap: Proving the sides parallel instead of equal. Parallel with different lengths is a trapezoid.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The intersection point from both lines

Compute the point with $t$ on the first line AND with $s$ on the second. If the two triples differ, the system was solved wrong, whatever it looked like.

$(1, 2, 0) + 2(1, -1, 2) = (3, 0, 4)$ and $(4, 2, 3) - (1, 2, -1) = (3, 0, 4)$ : confirmed.
A second point on a converted line

After converting a symmetric form, pick $t = 1$ in your vector form and put the point back into the ORIGINAL symmetric equations: every fraction must give the same value.

$(1, -1, 5)$ in $\frac{x + 2}{3} = \frac{1 - y}{2} = \frac{2z - 4}{6}$ gives $1 = 1 = 1$ . Getting $1$ , $-1$ , $1$ means a lost minus sign.
A section point by two lengths

Compute $AP$ and $PB$ as norms. Their ratio must be the one in the statement, in the right order.

$\|(4, -2, 4)\| = 6$ and $\|(2, -1, 2)\| = 3$ : the ratio is $2 : 1$ from $A$ , as asked.
A unit vector has norm 1

Square and add the components of the vector you found. Anything other than $1$ means the division by the norm was skipped or done on one component only.

$\left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)$ : $\frac{4 + 1 + 4}{9} = 1$ .

The typical problem, taken apart

Two lines written with the same letter

Decide the relative position of $L_1 : (x, y, z) = (2, -1, 1) + t(1, 1, -2)$ and $L_2 : (x, y, z) = (5, -4, 5) + t(1, -2, 3)$ , and give the point of intersection if there is one.

Step 1

Rename the parameter of $L_2$ : $L_2 : (x, y, z) = (5, -4, 5) + s(1, -2, 3)$ .

Why

With one letter the first equation would read $2 + t = 5 + t$ , that is $2 = 5$ , and the lines would be declared disjoint although they meet. Renaming costs nothing and is the first mark.

Step 2

Direction test: $(1, -2, 3)$ against $(1, 1, -2)$ . The first components give the factor $1$ , the second $-2$ : no single factor, so the lines are not parallel.

Why

This settles the family of the answer before any system: intersecting or skew, never parallel. It also tells you what no solution will mean later.

Step 3

System: $2 + t = 5 + s$ , $-1 + t = -4 - 2s$ , $1 - 2t = 5 + 3s$ . Subtracting the second from the first: $3 = 9 + 3s$ , so $s = -2$ , then $t = 3 + s = 1$ .

Why

Two equations are enough to fix the two unknowns, and a subtraction that eliminates $t$ is faster than a substitution. Nothing is concluded yet.

Step 4

Third equation: $1 - 2 \times 1 = -1$ on the left and $5 + 3 \times (-2) = -1$ on the right. It holds.

Why

This is the actual test in $\mathbb{R}^3$ . Writing both sides with their numbers is what the marker checks, and it is where skew lines are caught.

Step 5

Point from both lines: $L_1$ at $t = 1$ gives $(3, 0, -1)$ , and $L_2$ at $s = -2$ gives $(5 - 2, -4 + 4, 5 - 6) = (3, 0, -1)$ .

Why

Computing the point twice is the verification: a wrong $t$ or $s$ gives two different triples, and you see it before the marker does.

The conclusion, written out

“The directions are not proportional and $t = 1$ , $s = -2$ satisfy all three equations, so $L_1$ and $L_2$ intersect at $(3, 0, -1)$ .”

The classic mistake on this problem: Keeping the letter $t$ in both lines: $2 + t = 5 + t$ gives $2 = 5$ , and the paper concludes that the lines do not meet, then calls them parallel. Two errors, zero marks, on lines that actually cross.

Learn by heart

• $\overrightarrow{AB} = B - A$ , end minus start, and $\|c\mathbf{v}\| = |c|\,\|\mathbf{v}\|$ , never negative.
• Parallelogram on $\mathbf{u}$ , $\mathbf{v}$ : diagonals $\mathbf{u} + \mathbf{v}$ and $\mathbf{u} - \mathbf{v}$ . $ABCD$ is a parallelogram when $\overrightarrow{AB} = \overrightarrow{DC}$ .
• $AP : PB = m : n$ gives $P = \frac{nA + mB}{m + n}$ : the BIGGER weight goes on the endpoint $P$ is near.
• Centroid $\frac{1}{3}(A + B + C)$ , two thirds of the way from each vertex along its median.
• Symmetric form: coordinate MINUS number, coefficient $1$ , and a zero component becomes its own equation.
• Two lines: DIRECTION test first, then the POINT test or the system, with one parameter per line and the THIRD equation checked.
• No common point in space: parallel OR skew. Only the directions decide.

Frequently asked questions

How do I know if two lines in 3D intersect, are parallel or are skew?

Compare the direction vectors first. If one is a multiple of the other with the same factor on every component, the lines are parallel or identical, and testing one point of one line on the other decides. If not, give each line its own parameter, solve two of the three coordinate equations and check the third: if it holds the lines meet, if it fails they are skew.

Why can't I use the same parameter t for both lines?

Because the parameter only says where you are along one line. Using the same letter for both asks whether the two lines reach a common point at the same parameter value, which is a much stronger condition. Lines that really cross usually fail it, so the method declares them disjoint. Always rename the second parameter before writing the equations.

How do I write the symmetric equations of a line when a direction component is zero?

You never divide by zero. If the direction is two, zero, three through the point one, minus one, zero, the y coordinate never changes, so you write y equals minus one as a separate equation, and the symmetric form only links x and z: x minus one over two equals z over three. The parametric equations are always the safe route in this case.

What is the formula for the point dividing a segment in a given ratio?

If P divides AB with AP to PB equal to m to n, then P equals A plus m over m plus n times the vector AB, which is the same as n times A plus m times B, all over m plus n. The larger weight goes on the endpoint that P is closer to. For a ratio of two to one, P is two thirds of the way from A, and the midpoint is the special case of one to one.

Practise it

Corrected exercises: Geometric vectors and lines in space, MATH 133 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Determinants, adjugate and Cramer's rule Next sheet Dot product, projections and planes

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Revision sheet: geometric vectors and lines in space (MATH 133)

The essentials

A vector is a move, a point is a place

A line is one point plus one direction, and neither is unique

The rules in table form

Two lines in space: what the two tests say

The mistakes that cost marks

1. The same parameter letter for both lines

2. Solving two coordinate equations and stopping

3. Proportional directions read as the same line

4. Calling two lines that never meet parallel

5. Dividing by a zero component in the symmetric form

6. Reading the point and direction off a symmetric form that is not in standard shape

7. The weight on the wrong end of a ratio

8. The order of the letters in a parallelogram

Which method to choose

Two lines are given: which test, in which order

How the answer is expected to be written

Deciding the position of two lines

Writing a vector proof

Check before you hand in

The typical problem, taken apart

Two lines written with the same letter

Learn by heart

Frequently asked questions

How do I know if two lines in 3D intersect, are parallel or are skew?

Why can't I use the same parameter t for both lines?

How do I write the symmetric equations of a line when a direction component is zero?

What is the formula for the point dividing a segment in a given ratio?

Corrected exercises: Geometric vectors and lines in space, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?