Quantities carrying both a magnitude and a direction, and the algebra they obey. Defines a vector as a directed segment, establishes that position is irrelevant, and records it as a pair of components by subtracting the tail from the head. Adds vectors tip to tail and componentwise, showing the two agree, applies the resultant to a navigation problem via the Law of Cosines, and closes with the properties of vector addition and of scalar multiplication.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.8 Vectors, pp. 1012-1020
Objectives
Five outcomes, and the first is a conceptual one that the rest depend on.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1012-1020 — the pages these objectives are drawn from
Warm-up
Most quantities so far have been single numbers with units attached.
Discussion prompt
A plane flies at 175 miles per hour and a 35 mile per hour wind blows. Is the plane's speed over the ground 210, 140, or something else?
Hint: Does the answer depend on anything not yet given?
Answer:
It depends entirely on which way the wind is blowing. A tailwind gives 210 and a headwind gives 140, and a crosswind gives something in between and also pushes the plane sideways.
So the speeds alone are insufficient. What is needed is a quantity carrying both a size and a direction, and there is no way to package that as a single number.
Such a quantity is a vector, and this lesson builds the algebra for combining them. The particular problem above is worked out later in this deck, and the answer is 184 miles per hour.
Concept
A vector is a directed line segment: an arrow with a length and a direction. Two arrows with the same length and direction are the same vector, wherever they are drawn, so a vector records a displacement rather than a position.
component form — For a vector with initial point at one pair of coordinates and terminal point at another, the component form is the ordered pair obtained by subtracting the initial coordinates from the terminal ones. It records how far across and how far up the vector moves.
\[ \vec{v} = \overrightarrow{PQ} = \langle x_1 - x_0, \; y_1 - y_0 \rangle \]
The angle brackets are deliberate: they mark a vector rather than a point. The same two numbers mean different things in the two notations.
Figure (svg): The same vector drawn three times at different places in the plane, each with the same length and direction, showing that position does not matter
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1012-1013
Section
Section 1
Concept
A vector is defined by two characteristics: how long it is and which way it points. Any two directed segments agreeing on both are the same vector, however far apart they are drawn.
This is a deliberate abstraction. A displacement of three east and four north is the same displacement wherever it is performed, and vectors are built to make that literally true.
Figure (svg): The same vector drawn three times at different places in the plane, each with the same length and direction, showing that position does not matter
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1012-1013
Picture it
Subtract the tail from the head, coordinate by coordinate.
Figure (svg): A vector drawn from an initial point to a terminal point with its horizontal and vertical components marked as the differences of the coordinates
Two vectors are equal exactly when their components agree, which turns a geometric question into an arithmetic one.
Worked example
The book's introductory vector.
\[ \text{Find the component form of } \overrightarrow{PQ} \text{ with } P(1,2) \text{ and } Q(4,6). \]
Identify tail and head
Why: P is the initial point and Q the terminal one.
Subtract the x-coordinates
Why: Head minus tail.
\[ 4 - 1 = 3 \]
Subtract the y-coordinates
Why: Same order.
\[ 6 - 2 = 4 \]
Write in angle brackets
Why: The vector notation.
\[ < 3, 4 > \]
Figure (svg): The solution to Worked example finding the component form shown as a ladder of expressions, one row per legal move
\[ \vec{v} = \langle 3, 4\rangle \]
Verify: check by redrawing elsewhere
Why: Starting at negative 2 comma 3 and moving 3 across and 4 up lands at 1 comma 7. That arrow has the same length and direction as the original, so it is the same vector — which is exactly what the component form was designed to express.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1012-1013
Faded example
Find the component form of the vector from (5, -1) to (2, 4).
Fill in the blanks
\langle 2 - 5, \; 4 - (-1)\rangle = \langle -3, \; 5\rangle
Why: The x-component is negative because the motion is to the left, and the y-component is positive because the motion is upward. Reading the signs off the picture before computing is the fastest check on the subtraction order.
Worked example
The reverse operation, and a reminder that the tail is free.
\[ \text{Draw } \langle 3, 4\rangle \text{ with initial point } (-2, 3). \]
Read the components
Why: Three across and four up.
\[ \text{over } 3,\text{ up } 4 \]
Add to the x-coordinate
Why: Negative 2 plus 3.
\[ x = 1 \]
Add to the y-coordinate
Why: Three plus 4.
\[ y = 7 \]
Draw the arrow
Why: From the given tail to that head.
\[ \text{to } (1, 7) \]
Figure (svg): The solution to Worked example reconstructing a vector shown as a ladder of expressions, one row per legal move
\[ \text{tail } (-2,3), \quad \text{head } (1, 7) \]
Verify: confirm it is the same vector
Why: Subtracting tail from head gives 1 minus negative 2, which is 3, and 7 minus 3, which is 4. The components are unchanged, so this is the same vector as before, just drawn somewhere else.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1013-1013
Trap
\[ P(1,2), \; Q(4,6) \;\Longrightarrow\; \overrightarrow{PQ} = \langle 1 - 4, \; 2 - 6\rangle = \langle -3, -4\rangle \]
Subtract the terminal point from the initial one
Why: Subtraction has to go one way or the other, and this order is as plausible as the other.
But that gives the vector pointing from Q back to P, which has the opposite direction. The two vectors have the same length and are not equal.
\[ \overrightarrow{PQ} = \langle 4 - 1, \; 6 - 2\rangle = \langle 3, 4\rangle \]
Head minus tail, always
Why: The arrow's name lists the tail first, and the arithmetic subtracts it.
A sanity check that never fails: the components should have the signs of the motion. Going right and up gives two positive components, and any answer with negatives has the order backwards.
Prediction
Two arrows are drawn in different parts of the plane with the same length and the same direction.
Predict first
Are they the same vector?
Correct: Yes, always.
Why: A vector is defined by magnitude and direction alone, so any two directed segments agreeing on both are the same vector regardless of where they are drawn. This is exactly what makes the component form well-defined: subtracting tail from head gives the same pair whichever drawing is used. It is also what distinguishes a vector from a point, whose location is its entire content.
Sorting
A vector needs both a magnitude and a direction.
Sort into buckets
Sort each quantity.
Socratic
The definition deliberately ignores where a vector is drawn.
Discussion prompt
What is gained by that, and what would be lost if position were kept?
Hint: How many different situations does one vector then describe?
Answer:
What is gained is that one vector describes every occurrence of the same displacement. A wind of 35 miles per hour from the north-west is the same vector whether it is blowing over one airport or another, so a single calculation serves both.
If position were kept, every arrow would be a different object and no two situations could share a calculation. The algebra would have nothing to generalise over.
The cost is that a vector cannot say where anything is, only how far and in what direction something moved. Positions are points and displacements are vectors, and keeping the two straight is the main conceptual discipline of the topic — a vector added to a point gives a point, and a point subtracted from a point gives a vector.
Section
Section 2
Concept
To add geometrically, place the tail of the second at the head of the first and draw from the first tail to the second head. To add algebraically, add the corresponding components. The two agree by construction.
resultant — The sum of two or more vectors, understood as the net result of performing each displacement in turn.
\[ \langle v_1, v_2\rangle + \langle w_1, w_2\rangle = \langle v_1 + w_1, \; v_2 + w_2\rangle \]
Subtraction is defined as adding the additive inverse, and works componentwise too. Its picture is the other diagonal of the same parallelogram, pointing from the head of the second vector to the head of the first.
Figure (svg): Two vectors added tip to tail, with the resultant drawn from the tail of the first to the head of the second, and the same sum shown as a parallelogram diagonal
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1013-1017
Picture it
Each row is one operation described twice.
Figure (svg): Two columns pairing each geometric description of a vector operation with its componentwise formula
The right column is not a discovery. Each formula was defined to reproduce the picture beside it, and checking that it does is the content of the theorems.
Worked example
Example 11.8.2. One vector is given in components and one by two points.
\[ \text{With } \vec{v} = \langle 3,4\rangle \text{ and } \vec{w} = \overrightarrow{PQ}, \; P(-3,7), \; Q(-2,5), \text{ find } \vec{v} + \vec{w}. \]
Convert w to components
Why: Head minus tail.
\[ < 1, -2 > \]
Add the first components
Why: Three plus 1.
\[ 4 \]
Add the second components
Why: Four plus negative 2.
\[ 2 \]
Write the sum
Why: In angle brackets.
\[ < 4, 2 > \]
Figure (svg): The solution to Worked example a sum, both ways shown as a ladder of expressions, one row per legal move
\[ \vec{v} + \vec{w} = \langle 4, 2\rangle \]
Verify: check geometrically
Why: Drawing v from the origin puts its head at 3 comma 4. Drawing w from there moves one right and two down, landing at 4 comma 2. The arrow from the origin to that point is the sum, with components 4 and 2 — the two methods agree.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1014-1015
Faded example
Add the vectors with components 7 and -3, and -2 and 8.
Fill in the blanks
\langle 7, -3\rangle + \langle -2, 8\rangle = \langle 5, \; 5\rangle
Why: Seven plus negative 2 is 5, and negative 3 plus 8 is 5. The two components are added completely independently of one another, which is the whole content of componentwise addition.
Worked example
Defined as adding the inverse, and computed componentwise.
\[ \text{With } \vec{v} = \langle 3,4\rangle \text{ and } \vec{w} = \langle 5,1\rangle, \text{ find } \vec{v} - \vec{w}. \]
Write the definition
Why: Adding the additive inverse.
\[ v + (-w) \]
Negate both components of w
Why: The inverse flips both signs.
\[ < -5, -1 > \]
Add componentwise
Why: Three minus 5, and 4 minus 1.
\[ < -2, 3 > \]
Interpret it
Why: From the head of w to the head of v.
Figure (svg): The solution to Worked example a difference shown as a ladder of expressions, one row per legal move
\[ \vec{v} - \vec{w} = \langle -2, 3\rangle \]
Verify: check the interpretation
Why: Adding w to the answer should give v back: 5 minus 2 is 3 and 1 plus 3 is 4, which is v. So walking along w and then along the difference lands at the head of v, which is exactly what the arrow from the head of w to the head of v does.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1017-1017
Error analysis
A student adds two vectors of lengths 5 and 12.
Annotate
On: \( |\vec{v}| = 5, \; |\vec{w}| = 12 \;\Longrightarrow\; |\vec{v} + \vec{w}| = 17 \)
This is the same fact as the missing sum rule for the modulus in Lesson 11.7a, and the same triangle inequality bounds it. The length of a sum lies between the difference and the sum of the lengths, and where in that range depends on the directions.
Prediction
Two vectors have lengths 6 and 10.
Predict first
What is the range of possible lengths for their sum?
Correct: From 4 to 16.
Why: Pointing the same way gives 16 and opposite ways gives 4, and every value between is achieved by some angle. This is exactly the triangle inequality bound met in Lesson 11.3, and it is the same computation: the Law of Cosines with the included angle running from 0 to 180 degrees.
Matching
The parallelogram has two diagonals and four sides.
Match the pairs
Why: The two diagonals of the parallelogram are the sum and the difference, and which is which is decided by where the arrow starts. The additive inverse reverses direction without changing length, and adding it to the original gives the zero vector — a displacement with no length and no direction.
Edge cases
The zero vector is included so that addition has an identity.
Discussion prompt
What is its direction, and why is including it worth the awkwardness?
Hint: What direction does a displacement of zero have?
Answer:
Its direction is undefined, which sits awkwardly with the claim that a vector is a magnitude and a direction. A displacement of nothing does not point anywhere.
It is included because without it addition has no identity and no inverses, and the whole algebraic structure collapses. Every property of vector addition would need an exception clause.
The pattern is one seen twice already this chapter: the pole had no polar angle and zero had no principal argument, and now the zero vector has no direction. In each case the degenerate object is admitted for the algebra's sake and its missing property is simply declared undefined. That is the standard trade, and recognising it as the same trade each time is worth more than memorising three exceptions.
Section
Section 3
Concept
Two velocities added tip to tail produce a triangle whose third side is the resultant. Its length is the true speed and its direction the true bearing, and both come from the Law of Cosines.
The vector language does not replace the triangle trigonometry; it organises the problem so that the triangle can be found. Everything after that is Lesson 11.3.
Figure (svg): A plane's velocity and a wind velocity drawn tip to tail, with the resultant giving the true speed and bearing, and the triangle they form labelled for the Law of Cosines
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1013-1014
Picture it
The plane's velocity and the wind's, and what actually happens.
Figure (svg): A plane's velocity and a wind velocity drawn tip to tail, with the resultant giving the true speed and bearing, and the triangle they form labelled for the Law of Cosines
The plane is faster over the ground than through the air here, and pushed east of its heading. Both effects fall out of the one triangle.
Worked example
Example 11.8.1, the first half.
\[ \text{A plane flies } 175 \text{ mph at } N40^\circ E; \text{ a } 35 \text{ mph wind blows at } S60^\circ E. \text{ Find the true speed.} \]
Draw both velocities from the bearings
Why: Both measured from north.
Find the angle between them
Why: From the geometry of the bearings.
\[ 100 ^\circ \]
Apply the Law of Cosines
Why: Two sides and the included angle.
\[ c ^{2} = 31850 - 12250 \cos 100 \]
Evaluate
Why: The cosine is negative, so the term adds.
\[ c = 184 \text{mph} \]
Figure (svg): The solution to Worked example the true speed shown as a ladder of expressions, one row per legal move
\[ c = \sqrt{31850 - 12250\cos(100^\circ)} \approx 184 \text{ mph} \]
Verify: check against the bounds
Why: The true speed must lie between 140 and 210, the difference and the sum of the two speeds. One hundred eighty-four is inside that range and closer to the top, which is right for an angle a little over a right angle where the wind is largely helping.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1014-1014
Prediction
A plane heads due north at 200 mph with a 30 mph wind blowing due east.
Predict first
What is the true speed?
Correct: About 202 mph.
Why: The two velocities are at right angles, so the resultant is the hypotenuse of a right triangle with legs 200 and 30. Its length is the root of 40000 plus 900, about 202.2. A crosswind barely changes the speed but does change the direction, pushing the plane east of north — which is why a pure crosswind costs almost nothing in speed and a great deal in heading.
Worked example
Example 11.8.1, the second half. A second application of the same law.
\[ \text{Find the angle } \alpha \text{ between the plane's heading and its true course.} \]
Identify the angle wanted
Why: Between the 175 side and the resultant.
\[ \text{opposite the } 35\text{ side} \]
Write the cosine form
Why: With the side of length 35 opposite.
\[ \cos a = \frac{c ^{2} + 29400}{350 c} \]
Evaluate
Why: Using the computed c.
\[ \alpha = 11 ^\circ \]
Add to the given bearing
Why: The wind pushes east.
\[ N 51 ^\circ E \]
Figure (svg): The solution to Worked example the true bearing shown as a ladder of expressions, one row per legal move
\[ \alpha \approx 11^\circ \;\Longrightarrow\; \text{true bearing } N51^\circ E \]
Verify: sanity-check the direction of the shift
Why: The wind blows towards the south-east, which has an easterly component, so it should push the plane east of its heading. The bearing moved from N40E to N51E, which is eastward. The sign of the effect matches the physics.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1014-1014
Trap
\[ N40^\circ E \text{ and } S60^\circ E \;\Longrightarrow\; \text{angle between} = 60^\circ - 40^\circ = 20^\circ \]
Subtract the two bearing numbers
Why: Both are measured from a north-south line, so the difference looks like the angle between them.
But one is measured from north and the other from south, and they turn in different senses. The actual angle between the two directions is 100 degrees.
Draw the two directions first. Bearings are a notation for describing a direction, not angles that can be subtracted.
Here the plane's heading is 40 degrees east of north and the wind's is 60 degrees east of south, which are 40 plus 60 degrees apart from the north-south line's two ends — giving 180 minus 100, and the interior angle of 100 degrees.
The check that catches this every time: the angle you found must look like the angle in your drawing. Twenty degrees would put the wind nearly parallel to the plane, which no reading of the bearings supports.
Faded example
Two forces of 40 and 60 newtons act with 120 degrees between them.
Fill in the blanks
R^2 = 1600 + 3600 - 2(40)(60)\cos(120^\circ) = 5200 + 2400 \;\Longrightarrow\; R \approx 87.2 \text___
Why: The cosine of 120 degrees is negative one half, so subtracting 4800 times it adds 2400. That gives 7600 under the root, and the resultant is about 87.2 newtons. Note it exceeds each individual force but falls well short of their sum of 100, which is what an obtuse angle between them should produce.
Sorting
The vector diagram gives a triangle; the usual rules then apply.
Sort into buckets
Sort each situation.
Real world
A river flows at 3 miles per hour and a swimmer can manage 2 miles per hour in still water. The swimmer wants to reach the point directly opposite.
Discussion prompt
Explain what vector question is being asked, and whether it can be answered.
Hint: What must the resultant point along?
Answer:
The swimmer's velocity relative to the water and the current's velocity add to give the velocity over the ground, and the requirement is that this resultant point straight across, with no downstream component.
That means the swimmer's own velocity must have an upstream component exactly cancelling the 3 mile per hour current. But the swimmer's whole speed is only 2, so no direction gives an upstream component of 3 and the requirement cannot be met.
So the answer is that it is impossible — the swimmer will be carried downstream whatever heading is chosen. The best available is to minimise the drift, which is a different question with a different answer, and it is a genuinely useful distinction: the vector algebra says clearly which goals are achievable before any effort is spent pursuing them.
With a swimmer faster than the current the problem does have a solution, and it is found by requiring the upstream component to equal the current — one equation in the heading angle. Whether a vector problem has a solution is itself a question the components answer.
Section
Section 4
Concept
Vector addition is commutative and associative, has the zero vector as an identity, and gives every vector a unique additive inverse. Each property is inherited from the corresponding property of real numbers, one component at a time.
The proofs are all the same shape: write both vectors in components, apply the corresponding real-number property to each component separately, and reassemble.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1015-1017
Picture it
The two routes round the parallelogram end in the same place.
Figure (svg): Two vectors added tip to tail, with the resultant drawn from the tail of the first to the head of the second, and the same sum shown as a parallelogram diagonal
This is the geometric content of the algebraic proof, and it is worth having both: the picture explains why one would expect it, and the components establish that it is so.
Worked example
The proof pattern that all four properties follow.
\[ \text{Prove } \vec{v} + \vec{w} = \vec{w} + \vec{v}. \]
Write both in components
Why: Two ordered pairs.
\[ < v 1, v 2 >\text{ and } < w 1, w 2 > \]
Apply the definition of addition
Why: Componentwise.
\[ < v 1 + w 1, v 2 + w 2 > \]
Swap within each component
Why: By commutativity of real addition.
\[ < w 1 + v 1, w 2 + v 2 > \]
Reassemble
Why: That is the other sum.
\[ = w + v \]
Figure (svg): The solution to Worked example proving commutativity shown as a ladder of expressions, one row per legal move
\[ \langle v_1+w_1, v_2+w_2\rangle = \langle w_1+v_1, w_2+v_2\rangle \]
Verify: check the picture agrees
Why: Geometrically, walking v then w and walking w then v both trace two sides of the same parallelogram to the same opposite corner. The algebra and the picture say the same thing, which is the standard for a definition being the right one.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1016-1016
Sorting
Points and vectors combine in specific ways.
Sort into buckets
Sort each expression.
Worked example
Existence, uniqueness, and the formula.
\[ \text{Find the vector } \vec{w} \text{ with } \vec{v} + \vec{w} = \vec{0}. \]
Write the requirement componentwise
Why: The sum's components are both zero.
\[ v 1 + w 1 = 0 \]
Solve each component
Why: One equation per component.
\[ w 1 = -v 1 \]
Do the same for the second
Why: Same reasoning.
\[ w 2 = -v 2 \]
Note uniqueness
Why: Each equation has one solution.
Figure (svg): The solution to Worked example the additive inverse shown as a ladder of expressions, one row per legal move
\[ -\vec{v} = \langle -v_1, -v_2\rangle \]
Verify: check the geometry
Why: The negated vector has the same length, since negating both components does not change how far the displacement goes, and the opposite direction. Walking v then negative v returns to the start, which is the zero displacement — exactly what the algebra says.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1016-1016
Error analysis
A student adds a position to a displacement without distinguishing them.
Annotate
On: \( (3, 5) + \langle 2, -1\rangle = \langle 5, 4\rangle \)
Points and vectors have the same components and different meanings. A point plus a vector is a point, a point minus a point is a vector, and a point plus a point is nothing at all — a distinction the brackets are there to keep.
Prediction
You add a vector to its own additive inverse.
Predict first
What is the result?
Correct: The zero vector.
Why: The inverse has the same length and the opposite direction, so walking one and then the other returns to the starting point — a net displacement of nothing. Componentwise, each component adds to its own negative and gives zero. This is the defining property of an additive inverse and it is why the zero vector had to be admitted in the first place.
Faded example
Find the additive inverse of the vector with components -6 and 2.
Fill in the blanks
-\langle -6, 2\rangle = \langle 6, \; -2\rangle
Why: Both components change sign. Geometrically the arrow keeps its length and points the opposite way, and adding the two gives the zero vector, as the definition requires.
Explain it to yourself
Every property of vector addition is proved the same way.
Discussion prompt
Describe that common proof pattern, and say what it relies on.
Hint: What happens to each component separately?
Answer:
Write everything in components, apply the definition of vector addition to reduce the statement to a statement about the two components separately, invoke the corresponding real-number property on each, and reassemble.
It relies entirely on the fact that vector addition was defined componentwise. Each component is a completely independent copy of real addition, so every property real addition has is inherited automatically.
Worth noticing what that means: vectors do not have interesting addition of their own. The addition is two copies of familiar addition running in parallel, and everything new about vectors comes from the interaction between the components — which is the length, the direction, and eventually the dot product two lessons from now.
Section
Section 5
Concept
Multiplying a vector by a real number multiplies both components by it. Geometrically the vector is scaled by the size of the number, and reversed as well if the number is negative.
scalar — A real number, in a context where vectors are also present. The word distinguishes it from a vector and names its role: it scales.
\[ k\langle v_1, v_2\rangle = \langle kv_1, kv_2\rangle \]
The properties are the expected ones: scalar multiplication is associative with real multiplication, has 1 as an identity, and distributes over both kinds of addition.
Figure (svg): A vector shown alongside several scalar multiples of itself, illustrating stretching, shrinking, and reversal
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1017-1018
Picture it
Same line, different lengths, two directions.
Figure (svg): A vector shown alongside several scalar multiples of itself, illustrating stretching, shrinking, and reversal
Every scalar multiple of a vector lies on the same line through the tail, which is why the scalar multiples of a single non-zero vector describe a line.
Worked example
The same componentwise pattern as before.
\[ \text{Prove } (kr)\vec{v} = k(r\vec{v}) \text{ for scalars } k \text{ and } r. \]
Apply the definition on the left
Why: Multiply both components by the product.
\[ < (k r) v 1, (k r) v 2 > \]
Regroup within each component
Why: By associativity of real multiplication.
\[ < k(r v 1), k(r v 2) > \]
Factor k out of the pair
Why: By the definition, backwards.
\[ k < r v 1, r v 2 > \]
Recognise the inner vector
Why: It is r times v.
\[ = k(r v) \]
Figure (svg): The solution to Worked example proving associativity shown as a ladder of expressions, one row per legal move
\[ (kr)\vec{v} = k(r\vec{v}) \]
Verify: check with numbers
Why: Take k equal to 2, r equal to 3, and v with components 1 and 4. The left gives 6 times the vector, namely 6 and 24. The right gives 2 times the vector 3 and 12, which is also 6 and 24. They agree.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1018-1018
Faded example
Compute negative 3 times the vector with components 2 and -5.
Fill in the blanks
-3\langle 2, -5\rangle = \langle -6, \; 15\rangle
Why: Both components are multiplied by negative 3, which changes the sign of each and triples the length. The resulting vector points in the opposite direction from the original and is three times as long.
Worked example
A product is the zero vector only in the obvious ways.
\[ \text{Show } k\vec{v} = \vec{0} \text{ forces } k = 0 \text{ or } \vec{v} = \vec{0}. \]
Write the condition componentwise
Why: Both components vanish.
\[ k v 1 = 0\text{ and } k v 2 = 0 \]
Suppose k is not zero
Why: Then divide.
\[ v 1 = 0\text{ and } v 2 = 0 \]
Conclude in that case
Why: The vector is the zero vector.
\[ v = 0 \]
Note the other case
Why: If k is zero the product is zero regardless.
Figure (svg): The solution to Worked example the zero product property shown as a ladder of expressions, one row per legal move
\[ k\vec{v} = \vec{0} \;\Longleftrightarrow\; k = 0 \text{ or } \vec{v} = \vec{0} \]
Verify: check why this needs proving
Why: The statement is not automatic — there are algebraic systems in which a product of two non-zero things can be zero. Here it holds because it holds for real numbers in each component, which is again the same inheritance pattern.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1018-1018
Trap
\[ \langle 3, 4\rangle \cdot \langle 2, 5\rangle = \langle 6, 20\rangle \]
Multiply the vectors component by component, as addition does
Why: Addition worked that way, so multiplication should too.
But no such operation is defined. Scalar multiplication takes a number times a vector, and the expression above multiplies two vectors, which this lesson gives no meaning to.
Only a scalar may multiply a vector, and the result is a vector.
There is a product of two vectors, but it is not componentwise multiplication and it does not produce a vector. It is the dot product, defined two lessons from now, and it produces a number.
Worth being firm about: an operation exists only if it has been defined. The componentwise product looks natural and is almost never useful, which is why no one defines it.
Prediction
You multiply a vector by the scalar negative one half.
Predict first
What happens to it?
Correct: Half as long, opposite direction.
Why: The size of the scalar sets the scaling, and one half shortens the vector. The sign sets the direction, and a negative scalar reverses it. The two effects are independent, which is why any scalar multiple can be described by naming a factor and whether or not the direction flipped.
Sorting
Some combinations of vectors and scalars are defined and some are not.
Sort into buckets
Sort each expression.
Counterexample
A student proposes: multiplying a vector by a scalar always makes it longer.
Discussion prompt
Give two counterexamples of different kinds, and state the correct rule.
Hint: What scalars are there besides those greater than 1?
Answer:
Multiplying by one half halves the length, so the vector gets shorter. Multiplying by zero gives the zero vector, of no length at all.
Even multiplying by negative 2 does not simply lengthen: it lengthens and reverses, so the resulting vector points the other way and is not a longer version of the original in any useful sense.
The correct rule is that the length is multiplied by the absolute value of the scalar, and the direction is preserved for a positive scalar and reversed for a negative one. Two separate effects, decided by the size and the sign — which is precisely how multiplication by a complex number worked in the previous lesson, with the argument playing the role the sign plays here.
Comparison
Fill the blanks from memory. Each row is one thing described twice.
Comparison matrix
| Operation | Geometrically | In components |
|---|---|---|
| the vector itself | a directed segment | head minus tail, coordinate by coordinate |
| addition | place tip to tail; the resultant closes the path | add the corresponding entries |
| subtraction | from the head of w to the head of v | subtract the corresponding entries |
| scalar multiple | scale, reversing if the scalar is negative | multiply both entries |
| the zero vector | no displacement; direction undefined | both entries zero |
Every componentwise formula was defined to match the geometry beside it, and the theorems are the verification that it does. Neither column is the definition on its own.
Pattern
Five moves, covering both the algebra and the applications.
Keep points and vectors distinct: round brackets for a position, angle brackets for a displacement.
Check
Component form.
Check your understanding
What is the component form of the vector from (7, -2) to (3, 5)?
Answer: B
Why: Head minus tail gives 3 minus 7, which is negative 4, and 5 minus negative 2, which is 7. The motion is to the left and upward, so the signs should be negative then positive — which they are.
Check
Combining operations.
Check your understanding
With v having components 2 and -1 and w having components -3 and 4, what is 2v minus w?
Answer: A
Why: Doubling v gives components 4 and negative 2. Subtracting w means subtracting negative 3 and 4, giving 4 plus 3 and negative 2 minus 4, namely 7 and negative 6.
Check
Resultants.
Check your understanding
Two forces of 5 and 12 newtons act at right angles. What is the magnitude of the resultant?
Answer: C
Why: At right angles the two vectors form the legs of a right triangle and the resultant is the hypotenuse, so its length is the root of 25 plus 144, which is 13. The Law of Cosines gives the same answer, since the cosine of 90 degrees is zero.
Real world
A cargo container hangs from two cables attached to the same point on it, each running up to a different anchor. The container weighs 800 kilograms-force, straight down. An engineer must check that neither cable exceeds its rated tension.
Discussion prompt
Explain what vector condition the situation imposes, and how the cable tensions are found.
Hint: What must the three forces sum to?
Answer:
The container is not moving, so the three forces on it — the weight and the two cable tensions — must sum to the zero vector. That single condition is the whole physics of the problem.
Written out, it says the two tensions add to a vector of 800 pointing straight up. Drawing them tip to tail gives a triangle with one known side and two known directions, which is AAS — so the Law of Sines finishes it, giving both tensions.
The engineering point is that the tensions can each be larger than the weight. As the cables approach horizontal, the angle between them widens and the tensions grow without bound, which is why a load is never hung from two nearly horizontal cables. That fact is invisible in the weight alone and immediate in the vector triangle.
This is the general shape of statics: every equilibrium problem is the statement that a sum of vectors is zero, and every such statement is a closed polygon whose sides are the forces. The trigonometry of this chapter is what turns that polygon into numbers.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Two vectors both have length 10. What can you say about the length of their sum?
Correct: It is between 0 and 20.
\[ 0 \le |\vec{v} + \vec{w}| \le |\vec{v}| + |\vec{w}| \]
Why: Pointing the same way gives 20 and pointing opposite ways gives 0, with everything between achievable. It is 10 root 2 only in the special case of a right angle. And it is not undetermined — the range is completely pinned down by the two lengths, even though the exact value needs the angle as well.
Explain it
They have been told a vector has a magnitude and a direction, and are confused that the same vector can be drawn in different places without being a different vector.
Discussion prompt
In no more than five sentences, explain what a vector is actually recording.
Hint: What does an arrow tell you to do, as opposed to where it is?
Answer:
An arrow records an instruction: go this far in this direction. The instruction go three east and four north is the same instruction wherever you happen to be standing when you follow it.
So the arrow's position is not part of the information. Only its length and its direction are, and those are what the component form records — three across and four up, with no mention of where it started.
Contrast a point, whose entire content is where it is. A point says where; a vector says how far and which way, and confusing them is what makes the topic feel slippery at first.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Component form is fixed by head minus tail, checked against the signs of the motion. Addition is fixed by componentwise arithmetic and the tip-to-tail picture. Resultants are fixed by drawing the directions first rather than subtracting bearing numbers. Scalar multiplication is fixed by the two independent effects, size from the magnitude and direction from the sign. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw the same vector three times in different places and write beside them why all three are the same vector, along with the head-minus-tail rule for the component form. Underneath, draw a parallelogram from two vectors with both diagonals marked, labelling one as the sum and one as the difference and noting which way each arrow points. In the middle of the page write the four properties of vector addition and the six properties of scalar multiplication, marking beside each that it is inherited componentwise from the real numbers. In the bottom left, draw the plane and wind problem as a vector diagram, mark the 100 degree angle, and write the Law of Cosines computation that gives the true speed. In the bottom right, draw one vector together with two, one half and negative two times it. Finally, circle the one object on the page whose direction is undefined.
The circled object is the zero vector. It is admitted for the algebra's sake despite failing the definition of a vector, which is the same trade made for the pole in polar coordinates and for the argument of zero — worth recognising as one pattern rather than three exceptions.
Recap
Five things, and the first is the one that makes the rest coherent.
| If the question says | Your first move is |
|---|---|
| Find the component form | Head minus tail, coordinate by coordinate |
| Add these vectors | Add the corresponding components |
| Find the resultant velocity | Draw both, tip to tail, and find the triangle |
| A bearing is given | Draw the direction; never subtract bearing numbers |
| Multiply by a negative scalar | Scale by its size and reverse the direction |
Vectors can now be added, subtracted and scaled, but nothing yet computes how long one is or which way it points from its components. The next lesson supplies both, along with the unit vectors that let any vector be written as a combination of two standard ones.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1012-1020 — everything on these slides traces back here
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