How long a vector is and which way it points, extracted from its components. Defines the magnitude as the distance from tail to head and the direction as the unit vector pointing the same way, establishing that every vector is its magnitude times its direction. Uses that to resolve a vector into components from a stated size and bearing, introduces normalising and the principal unit vectors, and closes with a static equilibrium problem solved by setting a vector sum to zero.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.8 Vectors, pp. 1020-1027
Objectives
Five outcomes, and the third is what makes vectors usable in applications.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1020-1027 — the pages these objectives are drawn from
Warm-up
Vectors can now be added, subtracted and scaled. One obvious thing is still missing.
Discussion prompt
The vector with components 3 and 4 was called a displacement of three across and four up. How far does it actually move something?
Hint: Draw the arrow and drop a perpendicular.
Answer:
Five units. The horizontal and vertical displacements are the legs of a right triangle and the vector is the hypotenuse, so the Pythagorean Theorem gives the length.
That length is the magnitude of the vector, and nothing in the previous lesson computed it — the algebra there worked entirely with components and never asked how long anything was.
With the magnitude available, a vector can be split into how big and which way, which is what every application actually needs. That splitting is this lesson.
Concept
Plot the vector in standard position and convert its head to polar coordinates. The distance is the magnitude and the angle gives the direction, packaged as a unit vector; the original vector is the product of the two.
Definition 11.8 — The magnitude of a vector, written with double bars, is the square root of the sum of the squares of its components. For a non-zero vector, its direction is the vector of the cosine and sine of any polar angle of its head, and is written with a hat.
\[ \vec{v} = \|\vec{v}\|\,\hat{v}, \qquad \|\vec{v}\| = \sqrt{v_1^2 + v_2^2}, \quad \hat{v} = \langle \cos\theta, \sin\theta\rangle \]
This is Definition 11.2 from the complex number lesson wearing different notation. The magnitude plays the part of the modulus and the direction the part of the argument.
Figure (svg): A vector in standard position with its magnitude marked as the length of the arrow and its direction marked as the angle from the positive x-axis
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1020-1020
Section
Section 1
Concept
The magnitude is the distance from tail to head, computed by the Pythagorean Theorem from the components. It is non-negative, vanishes only for the zero vector, and interacts predictably with scalar multiplication.
Theorem 11.20 — The magnitude is non-negative and zero only for the zero vector; scaling a vector by k multiplies its magnitude by the absolute value of k; and every non-zero vector equals its magnitude times its direction.
\[ \|k\vec{v}\| = |k|\,\|\vec{v}\| \]
The proof of the scaling rule turns on the square root of k squared being the absolute value of k rather than k itself, which is the one place a sign could go wrong.
Figure (svg): Two columns separating the quantities in this lesson that are numbers from those that are vectors
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1020-1021
Picture it
Every expression is one or the other, and the double bars mark the boundary.
Figure (svg): Two columns separating the quantities in this lesson that are numbers from those that are vectors
Reading an expression aloud and naming its type before evaluating catches most errors in this lesson, particularly the ones where the bars are in an unexpected place.
Worked example
Example 11.8.4, part 3, parts b and c. The bars change everything.
\[ \text{With } \vec{v} = \langle 3,4\rangle \text{ and } \vec{w} = \langle 1,-2\rangle, \text{ find } \|\vec{v}\| - 2\|\vec{w}\| \text{ and } \|\vec{v} - 2\vec{w}\|. \]
Compute both magnitudes
Why: Nine plus 16, and 1 plus 4.
\[ 5\text{ and } \sqrt{5} \]
Evaluate the first expression
Why: Two numbers combined.
\[ 5 - 2 \sqrt{5} \]
For the second, do the vector arithmetic first
Why: Inside the bars.
\[ < 1, 8 > \]
Then take the magnitude
Why: One plus 64.
\[ \sqrt{65} \]
Figure (svg): The solution to Worked example magnitudes and combinations shown as a ladder of expressions, one row per legal move
\[ \|\vec{v}\| - 2\|\vec{w}\| = 5 - 2\sqrt{5}, \qquad \|\vec{v} - 2\vec{w}\| = \sqrt{65} \]
Verify: note how different they are
Why: The first is about 0.53 and the second about 8.06 — the same symbols in a different order give answers that differ by a factor of fifteen. The bars are not decoration, and where they sit determines the whole computation.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1022-1022
Faded example
Find the magnitude of the vector with components -5 and 12.
Fill in the blanks
\|\vec169\| = \sqrt13 = \sqrt___} = ___
Why: Both components are squared, so the negative sign disappears. The result 13 is a length and cannot be negative, which is a check worth making on every magnitude computation.
Worked example
Where the absolute value appears.
\[ \text{Prove } \|k\vec{v}\| = |k|\,\|\vec{v}\|. \]
Apply scalar multiplication
Why: Both components scaled.
\[ < k v 1, k v 2 > \]
Write the magnitude
Why: Sum of squares under a root.
\[ \sqrt{k ^{2} v 1 ^{2} + k ^{2} v 2 ^{2}} \]
Factor out k squared
Why: It is a common factor.
\[ \sqrt{k ^{2}} \sqrt{v 1 ^{2} + v 2 ^{2}} \]
Simplify the first root
Why: The root of a square is the absolute value.
\[ | k | \times | | v | | \]
Figure (svg): The solution to Worked example proving the scaling rule shown as a ladder of expressions, one row per legal move
\[ \|k\vec{v}\| = \sqrt{k^2}\,\sqrt{v_1^2+v_2^2} = |k|\,\|\vec{v}\| \]
Verify: test a negative scalar
Why: Take k equal to negative 3 and v with components 1 and 0. The scaled vector has components negative 3 and 0, with magnitude 3. The rule gives the absolute value of negative 3 times 1, which is 3. Writing k instead of its absolute value would have given negative 3, which no length can be.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1021-1021
Trap
\[ \|\vec{v} + \vec{w}\| = \|\vec{v}\| + \|\vec{w}\| \]
Distribute the bars over the addition
Why: Scalar multiplication distributed cleanly, so addition should too.
But the magnitude of the sum of two unit vectors pointing opposite ways is zero, while the sum of the magnitudes is 2. The rule fails.
\[ \|\vec{v} + \vec{w}\| \le \|\vec{v}\| + \|\vec{w}\| \]
Only an inequality holds
Why: The triangle inequality, again.
This is the third appearance of the same fact: no sum rule for the modulus, no sum rule for the magnitude, and the triangle inequality bounding both. They are the same statement in three notations, and recognising that is worth more than memorising three separate cautions.
Sorting
The double bars always produce a number.
Sort into buckets
Sort each expression.
Prediction
A vector is multiplied by the scalar negative 4.
Predict first
What happens to its magnitude?
Correct: It is multiplied by 4.
Why: The rule involves the absolute value of the scalar, so the magnitude is multiplied by 4 rather than negative 4. A magnitude is a length and can never be negative, which is precisely why the absolute value appears. The negative sign does something real, but to the direction rather than to the length.
Socratic
The magnitude uses a different symbol from the absolute value and the modulus.
Discussion prompt
Are these three different ideas, and if not, why the different notations?
Hint: What does each of them measure?
Answer:
They are one idea in three settings: the distance from the object to zero. For a real number that is the absolute value, for a complex number the modulus, and for a vector the magnitude.
The notations differ because the objects differ, and the symbol records what kind of thing is inside. Single bars around a vector would suggest it were a number, which is exactly the confusion the lesson keeps warning against.
The deeper reason to keep them distinct is that the objects support different operations. All three have a product rule of some kind, none has a sum rule, and all three obey the triangle inequality — but a vector has no multiplication yet, so its version of the product rule is the one about scalars.
Section
Section 2
Concept
The direction of a non-zero vector is the unit vector pointing the same way, obtained by dividing by the magnitude. Its head lies on the unit circle, and its components are the cosine and sine of the vector's angle.
unit vector — A vector of magnitude 1. The unit vector in the direction of a given non-zero vector is obtained by normalising it — multiplying by the reciprocal of its magnitude.
\[ \hat{v} = \frac{1}{\|\vec{v}\|}\vec{v} \]
The direction is well-defined even though the angle is not: different polar angles for the same head are coterminal, and coterminal angles have the same cosine and sine.
Figure (svg): A vector being normalised, with its head pulled back along the same line onto the unit circle to give the unit vector in its direction
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1020-1024
Picture it
Pull the head back along the same ray until it reaches the unit circle.
Figure (svg): A vector being normalised, with its head pulled back along the same line onto the unit circle to give the unit vector in its direction
For a vector shorter than 1 the head is pushed outward instead, but the description is the same: multiply by the reciprocal of the length.
Worked example
Example 11.8.4, part 3a.
\[ \text{Find } \hat{v} \text{ for } \vec{v} = \langle 3, 4\rangle. \]
Compute the magnitude
Why: Nine plus 16, then the root.
\[ 5 \]
Multiply by its reciprocal
Why: One fifth of the vector.
\[ (\frac{1}{5}) < 3, 4 > \]
Scale both components
Why: Componentwise.
\[ < \frac{3}{5}, \frac{4}{5} > \]
Check the length
Why: Nine plus 16 over 25.
\[ 1 \]
Figure (svg): The solution to Worked example normalising a vector shown as a ladder of expressions, one row per legal move
\[ \hat{v} = \left\langle \tfrac{3}{5}, \tfrac{4}{5} \right\rangle \]
Verify: confirm the magnitude
Why: Nine twenty-fifths plus sixteen twenty-fifths is 1, whose root is 1. The result really is a unit vector, and this check is worth doing every time since it catches an arithmetic slip immediately.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1022-1022
Faded example
Normalise the vector with components -6 and 8.
Fill in the blanks
\|\vec10\| = 4 \;\Longrightarrow\; \hat___ = \left\langle -\tfrac______}, \tfrac______}\right\rangle = \left\langle -\tfrac______, \tfrac___}___\right\rangle
Why: The magnitude is the root of 36 plus 64, namely 10. Dividing both components by 10 and reducing gives negative three fifths and four fifths, whose squares sum to 1 as required.
Worked example
Example 11.8.4, part 3d. The answer is known in advance.
\[ \text{Find } \|\hat{w}\| \text{ for } \vec{w} = \langle 1, -2\rangle. \]
Compute the magnitude of w
Why: One plus 4.
\[ \sqrt{5} \]
Normalise
Why: Divide both components.
\[ < 1 / \sqrt{5}, -2 / \sqrt{5} > \]
Take the magnitude of that
Why: One fifth plus four fifths.
\[ \sqrt{1} \]
Read the answer
Why: The root of 1.
\[ 1 \]
Figure (svg): The solution to Worked example the magnitude of a direction shown as a ladder of expressions, one row per legal move
\[ \|\hat{w}\| = 1 \]
Verify: note that no computation was needed
Why: A unit vector has magnitude 1 by definition, so the answer was determined before any arithmetic. The computation is a confirmation rather than a discovery, and recognising which questions are like that saves time.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1022-1022
Error analysis
A student tries to shorten a vector to length 1.
Annotate
On: \( \vec{v} = \langle 3, 4\rangle, \; \|\vec{v}\| = 5 \;\Longrightarrow\; \hat{v} = \langle 3 - 4, \; 4 - 4\rangle \)
Direction is preserved by positive scalar multiplication and by nothing else. Any operation that touches the components unequally is a rotation as well as a scaling, and normalising must not rotate.
Prediction
You normalise a vector of magnitude 0.4.
Predict first
What happens to it?
Correct: It gets longer.
Why: Normalising multiplies by one over the magnitude, and one over 0.4 is 2.5, which is greater than 1. So a vector shorter than a unit vector is stretched out to reach the unit circle. The word normalising suggests shrinking because vectors longer than 1 are the common case, but the operation goes whichever way is needed.
Matching
Some of these are determined without any computation.
Match the pairs
Why: The first and last follow from the definition and the scaling rule, needing no components at all. The zero vector has magnitude zero, correctly and unproblematically, but no direction, since no angle is distinguished — the same exception as the pole and the argument of zero.
Edge cases
The direction is defined only for a non-zero vector.
Discussion prompt
Why can no convention rescue the zero vector's direction, when a convention did rescue its magnitude?
Hint: How many angles reach the origin?
Answer:
The magnitude is rescued because there is exactly one non-negative distance to the origin, namely zero. The requirement that a magnitude be non-negative picks it out with no ambiguity.
The direction cannot be rescued because every angle reaches the origin equally. There is no property distinguishing one from another, so any choice would be arbitrary rather than canonical.
The general principle is worth naming: a convention can pick a representative from a set, but only if some property singles one out. For the magnitude, non-negativity does; for the direction of zero, nothing does. That is why one is defined and the other is left undefined, in every one of the three settings where this has now come up.
Section
Section 3
Concept
Given a vector's magnitude and its direction as an angle, its components are the magnitude times the cosine and sine of that angle. The work is in converting the stated angle into a standard-position one.
\[ \vec{v} = \|\vec{v}\|\langle \cos\theta, \sin\theta\rangle \]
This is the operation that lets a physical description enter the algebra. A stated speed and bearing becomes a pair of numbers, and from there everything is componentwise arithmetic.
Figure (svg): A vector of known magnitude and known angle being resolved into its horizontal and vertical components
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1021-1023
Picture it
Sixty degrees from the negative x-axis in the second quadrant is 120 degrees in standard position.
Figure (svg): A vector of known magnitude and known angle being resolved into its horizontal and vertical components
That conversion is the entire difficulty. Once the standard-position angle is known, two multiplications finish the job.
Worked example
Example 11.8.4, part 1. The angle is given relative to the wrong axis.
\[ \text{Find } \vec{v} \text{ with } \|\vec{v}\| = 5, \text{ in Quadrant II, at } 60^\circ \text{ to the negative } x\text{-axis.} \]
Convert to a standard-position angle
Why: One hundred eighty minus 60.
\[ 120 ^\circ \]
Write the direction
Why: Cosine and sine of that angle.
\[ < -\frac{1}{2}, \sqrt{3} / 2 > \]
Multiply by the magnitude
Why: Five times each component.
\[ < -\frac{5}{2}, 5 \sqrt{3} / 2 > \]
Check the quadrant
Why: Negative then positive.
Figure (svg): The solution to Worked example from magnitude and angle shown as a ladder of expressions, one row per legal move
\[ \vec{v} = \left\langle -\tfrac{5}{2}, \; \tfrac{5\sqrt{3}}{2}\right\rangle \]
Verify: check the magnitude
Why: Twenty-five quarters plus seventy-five quarters is 100 quarters, namely 25, whose root is 5. The magnitude is right and the signs put the vector in the second quadrant, as required.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1021-1022
Faded example
Resolve a vector of magnitude 8 at 150 degrees in standard position.
Fill in the blanks
\vec-sqrt3/2 = 8\langle \cos 150^\circ, \sin 150^\circ\rangle = 8\left\langle -4 sqrt 3, \; \tfrac______\right\rangle = \langle ___, \; 4\rangle
Why: One hundred fifty degrees has reference angle 30 in the second quadrant, so the cosine is negative root three over two and the sine is positive one half. Multiplying by 8 gives negative four root three and 4, which is in the second quadrant as expected.
Worked example
Example 11.8.5. The same problem as before, with no triangle at all.
\[ \text{Redo the plane and wind problem by resolving both velocities.} \]
Convert both bearings to standard-position angles
Why: North 40 east is 50 degrees; south 60 east is negative 30.
\[ 50\text{ and } -30 \]
Resolve each velocity
Why: Magnitude times cosine and sine.
Add componentwise
Why: The resultant.
Take the magnitude and the angle
Why: For the true speed and true bearing.
\[ 184 \text{mph}, N 51 E \]
Figure (svg): The solution to Worked example the plane problem, componentwise shown as a ladder of expressions, one row per legal move
\[ \|\vec{v}+\vec{w}\| \approx 184, \qquad \theta \approx 39^\circ \;\Longrightarrow\; N51^\circ E \]
Verify: compare with the earlier method
Why: The triangle method in the previous lesson gave the same 184 miles per hour and the same N51E. The component method needed no diagram-derived angle, which makes it more reliable, but it did need both bearings converted to standard position — the difficulty moved rather than vanished.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1022-1023
Trap
\[ N40^\circ E \;\Longrightarrow\; \vec{v} = 175\langle \cos 40^\circ, \sin 40^\circ\rangle \]
Read the number in the bearing as the angle
Why: It is the only number given, so it looks like the angle wanted.
But a bearing is measured clockwise from north, and a standard-position angle is measured counter-clockwise from east. The correct angle here is 50 degrees, not 40.
\[ N40^\circ E \;\Longrightarrow\; \theta = 90^\circ - 40^\circ = 50^\circ \]
Convert the bearing before resolving
Why: Subtract from 90 for a bearing east of north.
The conversion depends on which quadrant the bearing names, so the reliable route is to draw it. South 60 east becomes negative 30 degrees, which no single subtraction rule would have produced.
Sorting
Standard position is counter-clockwise from the positive x-axis.
Sort into buckets
Sort each bearing by its standard-position angle.
Prediction
You resolve a vector whose standard-position angle is in the third quadrant.
Predict first
What signs do its components have?
Correct: Both negative.
Why: In the third quadrant both the cosine and the sine are negative, and the magnitude is positive, so both components come out negative. Checking the signs against the quadrant is the fastest way to catch an angle-conversion error, since a wrong quadrant almost always shows up as a wrong sign.
Explain it to yourself
The plane problem was solved twice, once with a triangle and once with components.
Discussion prompt
Compare the two methods and say what each demands of you.
Hint: Where does the geometry enter in each?
Answer:
The triangle method needs the angle between the two vectors, which must be worked out from a drawing, and then applies the Law of Cosines twice. The geometry is concentrated in one hard step and the rest is a formula.
The component method needs each bearing converted separately into a standard-position angle, and then everything is componentwise arithmetic. The geometry is spread across two easier steps.
Neither avoids the geometry, and both give 184 miles per hour. The component method scales better: adding a third velocity means one more resolution and one more addition, while the triangle method would need a whole new diagram. When more than two vectors are involved, components win, which is why physics is done that way.
Section
Section 4
Concept
The unit vectors along the two axes are named, and every vector is the sum of a multiple of each. It is the component form written as a sum rather than as a pair.
Theorem 11.21 — With the first principal unit vector having components 1 and 0 and the second having components 0 and 1, every vector equals its first component times the first plus its second component times the second.
\[ \vec{v} = v_1\hat{\imath} + v_2\hat{\jmath} \]
Writing a vector as a combination of two chosen vectors is a habit that becomes central in later courses. This is the simplest instance, with the most convenient possible choice of pair.
Figure (svg): A vector decomposed into a horizontal multiple of the unit vector i and a vertical multiple of the unit vector j
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1024-1025
Picture it
Four steps east and three north, written as a sum.
Figure (svg): A vector decomposed into a horizontal multiple of the unit vector i and a vertical multiple of the unit vector j
The two multiples are exactly the components, so nothing has been computed — only rewritten in a form where each piece names its own direction.
Worked example
Immediate from the two definitions.
\[ \text{Show } v_1\hat{\imath} + v_2\hat{\jmath} = \langle v_1, v_2\rangle. \]
Write each principal vector
Why: By definition.
\[ < 1, 0 >\text{ and } < 0, 1 > \]
Apply scalar multiplication
Why: Each scaled by its component.
\[ < v 1, 0 >\text{ and } < 0, v 2 > \]
Add componentwise
Why: Each has a zero where the other does not.
\[ < v 1, v 2 > \]
Recognise the result
Why: That is the vector.
\[ = v \]
Figure (svg): The solution to Worked example proving the decomposition shown as a ladder of expressions, one row per legal move
\[ v_1\langle 1,0\rangle + v_2\langle 0,1\rangle = \langle v_1, v_2\rangle \]
Verify: note that nothing was assumed
Why: The proof uses only the two definitions from the previous lesson, so the theorem holds for every vector without exception, including the zero vector, whose decomposition has both coefficients zero.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1024-1025
Faded example
Write the vector with components -6 and 0 in terms of the principal unit vectors.
Fill in the blanks
\langle -6, 0\rangle = -6\hat0 + ___\hat___ = -6\hat___
Why: The second coefficient is zero, so that term contributes nothing and is usually dropped. The vector points along the negative x-axis, which is what a purely horizontal decomposition should give.
Worked example
Both directions, which is all there is to it.
\[ \text{Write } \langle -2, 7\rangle \text{ in } \hat{\imath}, \hat{\jmath} \text{ form, and } 5\hat{\imath} - 3\hat{\jmath} \text{ in component form.} \]
Read the components of the first
Why: Negative 2 and 7.
Attach the unit vectors
Why: First to i, second to j.
\[ -2 i + 7 j \]
Read the coefficients of the second
Why: Five and negative 3.
Write as a pair
Why: In angle brackets.
\[ < 5, -3 > \]
Figure (svg): The solution to Worked example converting between the two forms shown as a ladder of expressions, one row per legal move
\[ \langle -2, 7\rangle = -2\hat{\imath} + 7\hat{\jmath}, \qquad 5\hat{\imath} - 3\hat{\jmath} = \langle 5, -3\rangle \]
Verify: confirm nothing changed
Why: The magnitudes agree, as they must: the root of 4 plus 49 either way. Converting between the forms is pure notation and cannot change any property of the vector.
Trap
\[ 3\hat{\imath} + 4\hat{\jmath} = 7\hat{\imath}\hat{\jmath} \]
Combine the two terms as though they were like terms
Why: They look like an algebraic expression in two symbols.
But the two unit vectors point in perpendicular directions and cannot be combined at all. There is no such object as their product, and adding them gives a vector rather than a number times a symbol.
\[ 3\hat{\imath} + 4\hat{\jmath} = \langle 3, 4\rangle, \quad \text{of magnitude } 5 \]
Leave the two terms separate
Why: They live in independent directions.
The two coefficients are coordinates, not like terms. The expression is fully simplified as it stands, in the same way that the point 3 comma 4 cannot be simplified to a single number.
Prediction
You add the two principal unit vectors.
Predict first
What is the magnitude of the result?
Correct: root 2.
Why: The sum has components 1 and 1, and its magnitude is the root of 1 plus 1. It is not 2, because the two unit vectors point in perpendicular directions rather than the same one — which is the missing sum rule for magnitudes appearing yet again. The resulting vector points at 45 degrees, halfway between the two axes.
Sorting
The same vector can be written several ways.
Sort into buckets
Sort each expression by whether it equals the vector with components 2 and -5.
Counterexample
A student proposes: any two vectors can serve as building blocks the way i and j do.
Discussion prompt
Give a pair that cannot, and say what property i and j have that makes them work.
Hint: What if the two vectors point along the same line?
Answer:
Take the vectors with components 1 and 2, and 2 and 4. The second is twice the first, so every combination of them lies along the same line through the origin — and no combination reaches, say, the vector with components 1 and 0.
The property needed is that the two are not scalar multiples of one another, so that they point in genuinely different directions. Any such pair works, though the arithmetic is messier.
What makes i and j the standard choice is that they are perpendicular and of unit length, so reading off the coefficients requires no computation at all — the coefficients are just the components. Any pair pointing in different directions can serve; this pair is chosen because it makes the coefficients free, and the next lesson's dot product is what makes that precise.
Section
Section 5
Concept
An object at rest has all the forces on it summing to the zero vector. Resolving each force into components turns that single vector equation into a system of two ordinary equations.
\[ \vec{w} + \vec{T}_1 + \vec{T}_2 = \vec{0} \]
The unknowns are the magnitudes of the tensions, and their directions are given. That is the usual shape of a statics problem: directions known, sizes to be found.
Figure (svg): A hanging weight supported by two braces at different angles, with the three force vectors drawn from a common point summing to zero
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1025-1027
Picture it
Three forces at one point, summing to nothing.
Figure (svg): A hanging weight supported by two braces at different angles, with the three force vectors drawn from a common point summing to zero
The steeper support carries the larger tension, which is the answer to the practical question of which brace to make stronger.
Worked example
Example 11.8.6. Resolving all three forces.
\[ \text{A } 50 \text{ lb speaker hangs from braces at } 60^\circ \text{ and } 30^\circ \text{ to the ceiling. Set up the system.} \]
Resolve the weight
Why: Straight down, magnitude 50.
\[ < 0, -50 > \]
Resolve the first tension
Why: At 60 degrees in standard position.
\[ T 1 < \cos 60, \sin 60 > \]
Resolve the second
Why: At 30 degrees from the negative x-axis, so 150.
\[ T 2 < \cos 150, \sin 150 > \]
Set each component sum to zero
Why: Two equations.
Figure (svg): The solution to Worked example setting up the equations shown as a ladder of expressions, one row per legal move
\[ \tfrac{T_1}{2} - \tfrac{\sqrt{3}T_2}{2} = 0, \qquad \tfrac{\sqrt{3}T_1}{2} + \tfrac{T_2}{2} = 50 \]
Verify: check the angle for the second brace
Why: The second brace makes 30 degrees with the ceiling on the other side, so measured from the positive x-axis it is at 180 minus 30, namely 150 degrees. Using 30 there would put both braces on the same side and the speaker could not hang at all.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1025-1026
Prediction
A weight hangs from two cables at equal angles to the ceiling, and the angles are made shallower.
Predict first
What happens to the tensions?
Correct: They increase.
Why: Only the vertical component of each tension opposes the weight, and that component is the tension times the sine of the angle with the horizontal. As the angle shrinks the sine shrinks, so a larger tension is needed to supply the same vertical force. In the limit of horizontal cables the required tension grows without bound, which is why no load is ever hung that way.
Worked example
Two equations, two unknown tensions.
\[ \text{Solve the system for } T_1 \text{ and } T_2. \]
Solve the first equation
Why: It says the horizontal forces balance.
\[ T 1 = \sqrt{3} T 2 \]
Substitute into the second
Why: Multiplied through by 2.
\[ 3 T 2 + T 2 = 100 \]
Solve for the second tension
Why: Four T2 is 100.
\[ T 2 = 25 \]
Back-substitute
Why: Root three times 25.
\[ T 1 = 25 \sqrt{3} \]
Figure (svg): The solution to Worked example solving the system shown as a ladder of expressions, one row per legal move
\[ \|\vec{T}_1\| = 25\sqrt{3} \approx 43.3 \text{ lb}, \qquad \|\vec{T}_2\| = 25 \text{ lb} \]
Verify: check the vertical balance
Why: The vertical components are 43.3 times the sine of 60, about 37.5, and 25 times the sine of 150, namely 12.5. Those sum to 50, matching the weight. And the horizontal components, 21.65 and negative 21.65, cancel. Both equations are satisfied.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1026-1027
Trap
\[ \text{two braces, } 50 \text{ lb} \;\Longrightarrow\; 25 \text{ lb each} \]
Divide the weight between the supports
Why: Two supports, one weight, so half each.
But the braces are at different angles, so they contribute different amounts vertically. The steeper one carries about 43.3 pounds and the shallower one 25.
Resolve and solve. Only the vertical components sum to the weight, and each vertical component is the tension times the sine of its angle.
Note too that the tensions sum to about 68.3 pounds, which exceeds the weight. That is not a paradox: the horizontal components cancel and only the vertical parts oppose gravity.
The practical consequence is real. Shallower braces carry larger tensions, which is why a load hung from nearly horizontal cables can snap them, and why the guy wires on a mast are anchored well out from its base.
Faded example
Two cables at 45 degrees on each side support a 100 pound weight.
Fill in the blanks
2T\sin 45^\circ = 100 \;\Longrightarrow\; T\sqrt100 = 50 sqrt 2 \;\Longrightarrow\; T = ___
Why: By symmetry both tensions are equal, and each contributes T times the sine of 45 degrees vertically. Two of those must equal 100, giving T root two equal to 100 and so T equal to fifty root two, about 70.7 pounds — noticeably more than half the weight.
Sorting
In a typical statics problem some things are given and some are sought.
Sort into buckets
Sort each quantity in the speaker problem.
Real world
A tightrope walker of 70 kilograms-force stands at the middle of a rope strung between two poles. The rope sags very slightly, making an angle of about 3 degrees with the horizontal on each side.
Discussion prompt
Estimate the tension in the rope, and explain what the answer implies about how ropes are rigged.
Hint: What fraction of each tension is vertical?
Answer:
The two vertical components share the weight, so each is 35, and each equals the tension times the sine of 3 degrees, which is about 0.052. So the tension is about 35 divided by 0.052, or roughly 670 kilograms-force in each half of the rope.
That is nearly ten times the walker's weight, from a rope that looks almost straight. The shallower the sag, the worse it gets, and there is no angle at which the rope is genuinely horizontal under load.
The implication is direct: a rope must be rated far above the load it carries, and rigging is designed to allow some sag rather than to eliminate it. The same arithmetic explains why a car pulled sideways from the middle of a taut tow rope can be moved by a person, and why that trick is dangerous — the forces involved are far larger than they look.
All of it is one vector equation read componentwise. The whole of statics is the statement that forces sum to zero, and the trigonometry of this chapter is what turns that statement into numbers.
Comparison
Fill the blanks from memory. Note how many rows echo the complex number lesson.
Comparison matrix
| Quantity | Notation | What it is |
|---|---|---|
| magnitude | double bars | the length, a non-negative number |
| direction | a hat | the unit vector pointing the same way |
| the vector itself | an arrow, or angle brackets | magnitude times direction |
| normalising | multiply by one over the magnitude | keeps the direction, sets the length to 1 |
| the principal unit vectors | i-hat and j-hat | the unit vectors along the two axes |
The magnitude and direction are the modulus and argument under new names, with one difference: the direction is packaged as a vector rather than an angle, so it needs no principal-value convention.
Pattern
Five moves, covering both directions between the two descriptions.
Check every resolved vector by its quadrant: the signs of the two components must match where the arrow points.
Check
Magnitudes.
Check your understanding
With v having components 3 and 4 and w having components 1 and -2, what is the magnitude of v minus 2w?
Answer: A
Why: The arithmetic inside the bars comes first: v minus 2w has components 3 minus 2 and 4 plus 4, namely 1 and 8. Its magnitude is the root of 1 plus 64, which is the root of 65, about 8.06.
Check
Resolving.
Check your understanding
A vector of magnitude 10 makes a 30 degree angle with the negative x-axis in the third quadrant. What are its components?
Answer: B
Why: Thirty degrees below the negative x-axis is 180 plus 30, namely 210 degrees in standard position. The cosine there is negative root three over two and the sine is negative one half, giving negative five root three and negative 5 — both negative, as the third quadrant requires.
Check
Unit vectors.
Check your understanding
What is the unit vector in the direction of the vector with components 8 and -6?
Answer: A
Why: The magnitude is the root of 64 plus 36, which is 10, so dividing both components by 10 gives four fifths and negative three fifths. Their squares sum to 1, confirming a unit vector.
Real world
A quadcopter hovers in a steady horizontal wind. Its four rotors together produce a single thrust vector, which the flight controller can point in any direction by tilting the aircraft. The wind pushes horizontally with a force of 8 newtons, and the drone weighs 25 newtons.
Discussion prompt
Explain what the controller must arrange, and find the required thrust and tilt.
Hint: What must the three forces sum to for a hover?
Answer:
For the drone to hover the three forces — weight, wind, and thrust — must sum to the zero vector. So the thrust must exactly cancel the other two combined, meaning it points opposite to their sum.
The weight and wind are perpendicular, with components 0 and negative 25, and negative 8 and 0 if the wind blows in the negative x-direction. Their sum has components negative 8 and negative 25, so the thrust must have components 8 and 25 — magnitude the root of 64 plus 625, about 26.3 newtons, tilted about 17.7 degrees from vertical.
Two consequences fall straight out. The drone must tilt into the wind rather than staying level, which is why a hovering drone visibly leans on a windy day. And it must produce more thrust than its own weight, which is why battery life falls in wind even though the aircraft is not going anywhere.
The controller is solving this vector equation many times a second, from measured accelerations. Resolving a force into components and requiring the sum to vanish is the entire content, and it is the same computation as the hanging speaker with one force replaced by a measurement.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Two vectors each have magnitude 6. What is the magnitude of their sum?
Correct: Between 0 and 12.
\[ 0 \le \|\vec{v} + \vec{w}\| \le \|\vec{v}\| + \|\vec{w}\| \]
Why: There is no rule giving the magnitude of a sum from the magnitudes alone. Pointing the same way gives 12, opposite ways gives 0, at right angles gives 6 root 2, and every value between 0 and 12 occurs for some angle. This is the same missing sum rule met for the modulus in Lesson 11.7a and bounded by the same triangle inequality.
Explain it
They have computed the magnitude of a sum by adding the two magnitudes and cannot see what is wrong with it.
Discussion prompt
In no more than five sentences, show them with a picture rather than a formula.
Hint: Ask them to walk it.
Answer:
Ask them to walk 3 metres east and then 4 metres north. They have walked 7 metres of path, but they are only 5 metres from where they started — and the vector sum measures the second thing, not the first.
The magnitudes add only when the two vectors point the same way, which is the one case where the path and the displacement coincide. Any other angle makes the displacement shorter than the path.
The extreme case makes it obvious: walk 3 metres east and then 3 metres west, and you are back where you began. The magnitudes add to 6 and the sum has magnitude 0.
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: Magnitude expressions are fixed by doing the vector arithmetic inside the bars first and naming each piece a scalar or a vector. Resolving is fixed by converting the stated angle into standard position before anything else, and checking the signs against the quadrant. Normalising is fixed by multiplying by the reciprocal of the magnitude and verifying the result has length 1. Equilibrium is fixed by resolving every force and setting each component sum to zero. 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 a vector in standard position with its magnitude, its angle, and both components labelled, and write beside it the equation saying a vector equals its magnitude times its direction. Underneath, write the three properties of the magnitude, marking which one needs an absolute value and why. In the middle of the page draw a vector together with its normalised version on the unit circle, and write the normalising formula; beside it, draw the two principal unit vectors and decompose one vector along them. In the bottom left, resolve a vector of magnitude 5 lying in the second quadrant at 60 degrees to the negative x-axis, showing the angle conversion. In the bottom right, draw the hanging speaker with its three force vectors, write the vector equation, and write the two scalar equations it becomes. Finally, circle the step in the bottom-left problem where most marks are lost.
The circled step is the angle conversion. Sixty degrees to the negative x-axis is 120 degrees in standard position, and every stated angle has to be converted before the cosine and sine mean anything — a bearing especially, since those are measured from a different axis in the opposite sense.
Recap
Five things, and the second and third are the ones that connect the algebra to the world.
| If the question says | Your first move is |
|---|---|
| Find the magnitude | Square both components, add, take the root |
| Find the direction | Divide the vector by its magnitude |
| Given a magnitude and a bearing | Convert to a standard-position angle first |
| Bars appear around an expression | Do the vector arithmetic inside them first |
| The object is at rest | Set the sum of the forces to the zero vector |
Vectors can now be measured, pointed and combined, but there is still no way to multiply two of them or to ask what angle lies between them. The next lesson supplies both at once with the dot product, which takes two vectors and returns a number.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.8 Vectors §11.8, pp. 1020-1027 — everything on these slides traces back here
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