Module 1: Mathematical Foundations · Practice · assumes Foundations 5 and 6
Five worked examples on the two vector products — projection, power, torque in three dimensions, and angular momentum.
What this set is for
The earlier worked examples treated the dot product only as a way of computing work, and the cross product only in a plane. Both do more than that. The dot product resolves a vector along any direction you name, not only along the axes; the cross product handles geometries where the answer points somewhere no coordinate axis does.
Same three-part structure throughout: The Physics before any symbol, The Mathematics in full, The Contemplation on what the working revealed.
On this page
1. Splitting a force along a direction that is not an axis Foundation 5
Problem. A constant force F = (6î + 8ĵ) N acts on a bead that is constrained to slide along a guide track running in the direction û = 0.8î + 0.6ĵ. Find the component of the force along the track, and split F into a part along the track and a part across it.
The physics
The bead can only move along the track, so only the part of the force lying along the track can accelerate it. The rest pushes the bead sideways into the guide, which pushes back — that part does no work and never appears in the equation of motion. Foundation 4 did this with a tilted axis on an inclined plane; here the direction is given as a vector rather than an angle, and the dot product handles it without any trigonometry.
Left: only the black part accelerates the bead. Right: the position vector changes, the perpendicular distance does not.
The mathematics
First check that û really is a unit vector, since the whole method depends on it: √(0.8² + 0.6²) = √1 = 1. Good.
The component along the track is the scalar projection:
F∥ = F · û = (6)(0.8) + (8)(0.6) = 4.8 + 4.8 = 9.6 N
Sensible: |F| = √(36 + 64) = 10 N, so 9.6 N is most of it, and the track is nearly aligned with the force.
As a vector, point that magnitude back along û:
F∥ = 9.6(0.8î + 0.6ĵ) = 7.68î + 5.76ĵ N
The part across the track is whatever is left over:
F⊥ = F − F∥ = −1.68î + 2.24ĵ N
Check. The two parts must be perpendicular, so their dot product must vanish: (7.68)(−1.68) + (5.76)(2.24) = −12.9024 + 12.9024 = 0. And their magnitudes should recombine: √(9.6² + 2.8²) = √(92.16 + 7.84) = 10 N, the original force.
The contemplation
Nothing here required the track to be one of the coordinate axes, and that is the point worth taking. Foundation 4 taught resolution as a process for splitting a vector along x and y; this is the same operation freed from that restriction, and the physical situation almost never obliges by lining up with the page. Notice too that the perpendicular part was found by subtraction rather than by another projection. Once you know one piece and the whole, the remainder is fixed — no second calculation is needed, and none can disagree with the first.
2. A force that never does any work Foundation 5
Problem. A particle has velocity v(t) = (3tî − 2ĵ) m/s while the net force on it is F(t) = (4î + 6tĵ) N. Find the instantaneous power, and say what happens to the particle’s speed.
The physics
Both vectors are changing in time, and neither is obviously related to the other. Power is the rate at which the force feeds energy into the particle, so a natural expectation is that it varies as the two vectors swing about. Whether it does is the question — and by the work–energy theorem, whatever the power does determines what the kinetic energy, and hence the speed, does.
The mathematics
P(t) = F · v = (4)(3t) + (6t)(−2)
= 12t − 12t = 0 for every t
Not zero at some particular instant — zero identically. Since a dot product vanishes only when the vectors are perpendicular or one of them is zero, and neither is zero here, the force is perpendicular to the velocity at every moment of the motion.
The consequences follow immediately. The work done over any interval is ∫P dt = 0, so by the work–energy theorem the kinetic energy never changes, so the speed is constant. Check it directly: |v| = √(9t² + 4), which grows with time.
That contradiction is worth resolving rather than glossing. The speed is not constant, so something is wrong — and it is the premise. A force perpendicular to velocity at all times is consistent with these two expressions only if they are not in fact the velocity and net force of a single particle obeying Newton’s second law. Check: F = mdv/dt would require dv/dt = 3î, a constant, whereas F has a component growing as 6t. The two given expressions are inconsistent.
So the honest answer to the problem as posed: the power is zero at every instant, and that fact reveals the data are not physically consistent. A genuine case of perpendicular force and velocity is uniform circular motion — v = Rωθ̂ with F along −r̂ — where the speed really does stay constant.
The contemplation
The dot product did its job perfectly and the physics still did not hold together — because the numbers in the question were invented rather than measured. This is worth meeting once. A calculation cannot tell you whether its inputs describe something real; it can only tell you what follows if they do. The check that exposed it was a second, independent route to the same quantity: the work–energy theorem said the speed was constant, and direct computation said it was not. Whenever two correct methods disagree, the fault lies upstream of both.
3. The angle between two vectors in space Foundation 5
Problem. Two structural struts meet at a joint, running along A = 2î + 3ĵ + 6k̂ and B = 4î − 4ĵ + 2k̂. Find the angle between them.
The physics
Two directions in three dimensions, and the angle between them is a real geometric quantity that no amount of drawing on paper will give you reliably. This is where the dot product earns its keep: two definitions of the same product, one geometric and one from components, and setting them equal converts six numbers into an angle.
The mathematics
A · B = (2)(4) + (3)(−4) + (6)(2) = 8 − 12 + 12 = 8
A = √(4 + 9 + 36) = √49 = 7
B = √(16 + 16 + 4) = √36 = 6
cos θ = 8 / (7 × 6) = 8/42 = 4/21 ≈ 0.190
θ = cos−1(0.190) ≈ 79.0°
Close to perpendicular but not quite. Note that the dot product came out positive, which already told us the angle was acute before any inverse cosine was taken — a useful sanity check that costs nothing.
The contemplation
Compare this with Foundation 3, where finding an angle from coordinates required an inverse tangent and then a quadrant correction, because the arctangent could not tell opposite directions apart. Nothing of that kind is needed here. The inverse cosine returns a value between 0° and 180°, and that is exactly the range an angle between two vectors can occupy — there is no second candidate to rule out. The formula is not merely convenient; it is unambiguous by construction, and it works in as many dimensions as you like.
4. Torque when nothing lies along an axis Foundation 6
Problem. A mechanic applies a force F = (10î + 20ĵ − 5k̂) N at a point whose position relative to the bolt is r = (0.3î + 0.4ĵ) m. Find the torque about the bolt, its magnitude, and the axis about which the bolt tends to turn.
The physics
Every earlier torque example had everything lying in a plane, so the answer came out along one axis and could almost be guessed. Here the force has a component out of the plane of the spanner, and there is no reason to expect the resulting axis to line up with anything. Expect all three components of the answer to be non-zero, and expect that the axis of rotation is genuinely tilted in space.
The mathematics
Set out the determinant, with r in the middle row and F below — the order matters, and reversing the rows would reverse the answer:
| î | ĵ | k̂ |
| 0.3 | 0.4 | 0 |
| 10 | 20 | −5 |
τx = (0.4)(−5) − (0)(20) = −2.0
τy = −[(0.3)(−5) − (0)(10)] = −(−1.5) = +1.5
τz = (0.3)(20) − (0.4)(10) = 6.0 − 4.0 = +2.0
τ = −2.0î + 1.5ĵ + 2.0k̂ N m
Note the minus sign in front of the ĵ term. It comes from the alternating signs of determinant expansion and is the single most common error in this calculation.
Check before going further. The torque must be perpendicular to both inputs:
τ · r = (−2.0)(0.3) + (1.5)(0.4) + (2.0)(0) = −0.6 + 0.6 = 0 ✔
τ · F = (−2.0)(10) + (1.5)(20) + (2.0)(−5) = −20 + 30 − 10 = 0 ✔
Magnitude: |τ| = √(4.0 + 2.25 + 4.0) = √10.25 ≈ 3.20 N m.
The axis is the direction of τ itself: τ/|τ| ≈ −0.62î + 0.47ĵ + 0.62k̂. The bolt tends to turn about a line tilted well away from every coordinate axis — which is what a force applied out of the plane of the spanner does.
The contemplation
The two dot-product checks cost about fifteen seconds and would have caught any sign error in the determinant, including the notorious middle term. They work because the cross product’s defining property — perpendicularity to both inputs — is independently testable using the other product. Notice what that means: the dot and cross products are not two separate tools kept in different drawers. Each can verify the other, and a student who computes a cross product without checking it this way is declining a free guarantee.
5. Angular momentum of a particle going straight Foundation 6
Problem. A particle of mass 2 kg moves at a constant 5 m/s in the +x direction along the line y = 3 m. Find its angular momentum about the origin when it is at (4, 3, 0) m, and show that this does not change as the particle continues along the line.
The physics
Angular momentum is usually met in the context of spinning bodies, so a particle travelling in a perfectly straight line seems an odd thing to compute it for. But L = r × p asks only for a position and a momentum, and both exist here. The position vector is certainly changing as the particle moves. Whether the cross product changes with it is the question — and since no torque acts on the particle, we should expect not.
The mathematics
The momentum is p = mv = 2(5î) = 10î kg m/s. At the given moment r = 4î + 3ĵ, so:
L = (4î + 3ĵ) × (10î)
= 40(î × î) + 30(ĵ × î) = 0 + 30(−k̂)
= −30 k̂ kg m²/s
Now let the particle move. At any later time its position is r = xî + 3ĵ, with x growing steadily. Repeating the calculation:
L = (xî + 3ĵ) × (10î) = 10x(î × î) + 30(ĵ × î)
= 0 − 30k̂ = −30 k̂, whatever x is
The x-dependence multiplies î × î, which is zero, so it disappears entirely. What survives is the product of the momentum and the perpendicular distance from the origin to the line of motion — here 10 × 3 = 30 — and that distance never changes.
The contemplation
A particle moving in a straight line has angular momentum, and it is constant. Both halves of that sentence sound wrong on first hearing, and both follow directly from the definition — which is a good reason to trust definitions over intuitions built from spinning wheels. What the cross product actually measured was the perpendicular distance from the origin to the line the particle travels along, and that is a property of the line rather than of the particle’s progress along it. Move the origin, though, and the answer changes completely, exactly as torque did. Angular momentum, like torque, is a relation between a body and a point you chose.
The two products, side by side
Across these five examples the dot product measured how much of one vector lay along another — along a track, along a velocity, along another strut. The cross product measured what was left perpendicular, and pointed at the axis about which that leftover acts.
And each verified the other. Example 1 confirmed a decomposition by showing a dot product vanished; example 4 confirmed a cross product the same way. That is worth building into habit: a cross product costs three lines to compute and thirty seconds to check, and the check catches the error the calculation is most likely to contain.
Assumes: Foundations 5 and 6 · Type: Worked examples · Reading: 22 min · Difficulty: Intermediate