University Physics I · Mathematical & Physical Foundations · 1.6
Dot & Cross Products
Vectors multiply in two ways. The dot product asks how much of one vector lies along the other. The cross product asks how strongly one swings the other around.
Build the model
Connect the measurement to the mechanism.
Every mechanical quantity built from two vectors picks one of two products. The dot product returns a scalar, A·B = AB cos θ — biggest when the vectors point the same way, zero when they are perpendicular — and it carries work, W = F·d, and power, P = F·v, because only the part of a force running along the motion changes speed. The cross product returns a vector of magnitude AB sin θ — zero for parallel vectors, biggest at 90° — lying perpendicular to both, with its sense fixed by the right-hand rule.
It carries torque, τ = r×F, and angular momentum, L = r×p, because only the part running across the lever arm turns anything. In components the same two rules read A·B = AₓBₓ + AyBy + AzBz and (A×B)ₓ = AyBz − AzBy, cycled through x→y→z.
- Simple definition
- The dot product of two vectors is a scalar measuring how much they line up. The cross product is a vector whose magnitude measures how much they do not, perpendicular to both, with the right-hand rule choosing which way it points.
- Example
- One 40 N pull, one 30° angle, two jobs: over a 5.0 m drag it does work, W = 40 × 5.0 × cos 30° ≈ 173 J; on a 0.50 m spanner it turns a bolt, |τ| = 0.50 × 40 × sin 30° = 10 N m.
Positive when they lean the same way, zero at 90°, negative beyond it.
scalar out · θ is the angle between them, drawn tail to tail
One product recovers the angle, the projection of A onto B, and the length of A.
B̂ = B/B is dimensionless, so A·B̂ keeps the units of A
Zero for parallel vectors, largest at 90° — the opposite of the dot product.
equal to the area of the parallelogram that A and B span
Order matters: A×B = −B×A, and A×A = 0.
planar case: (A×B)z = AₓBy − AyBₓ, one signed number
Expanding both products term by term produces the component formulas above.
reverse the order and the sign flips: ĵ×î = −k̂
Power P = F·v and angular momentum L = r×p follow the same two patterns.
W in joules (J) · torque in newton metres (N m), never joules
Two questions, two products
Multiplying two numbers is one operation. Multiplying two vectors is not, because a pair of directions raises two separate questions. How much do these vectors agree — how much of one runs along the other? How much do they disagree — how much of one runs across the other? The dot product answers the first and hands back a scalar. The cross product answers the second and hands back a vector. Neither is a shortcut for the other, and choosing the wrong one is choosing the wrong question.
The dot product measures alignment
Draw the vectors tail to tail and read the angle θ between them. Then A·B = AB cos θ. The factor B cos θ is the part of B lying along A, so the dot product is the length of A times how much of B goes with it. The sign carries physics: positive below 90°, exactly zero at 90°, negative beyond it. Order does not matter, A·B = B·A, and a vector dotted with itself gives A·A = A², which squares a magnitude without any trigonometry. That makes the perpendicularity test the cheapest check in vector algebra: if A·B = 0 and neither vector is zero, the two are at right angles.
The cross product measures turning
The cross product keeps what the dot product discards. Its magnitude is |A×B| = AB sin θ, which is A times the part of B perpendicular to A: zero for parallel vectors, largest at 90°. Geometrically that number is the area of the parallelogram the two vectors span. The result is itself a vector, and perpendicular to both is not yet an answer — two opposite directions qualify, so the right-hand rule picks one. Point the fingers of your right hand along A, curl them towards B, and your thumb gives the direction of A×B; curling the other way reverses it, which is why A×B = −B×A and A×A = 0. The construction needs three dimensions to have somewhere to point. In a plane problem only the component perpendicular to the plane survives, and it is quoted as a signed number.
Components do the arithmetic
Both products are distributive, so expanding the î, ĵ, k̂ terms one at a time produces the component formulas. Take A = (3.0, −2.0, 1.0) and B = (4.0, 5.0, −2.0). Their dot product is 12 − 10 − 2 = 0, so the two are perpendicular — a fact no sketch would have settled. The x-component of the cross product is (−2)(−2) − (1)(5) = −1, and cycling the labels gives A×B = (−1, 10, 23), whose magnitude is √630 ≈ 25.1. That is a check, not a coincidence: at 90° sin θ = 1, so |A×B| must equal AB = √14 × √45 = √630 exactly. Components are faster; the geometric definitions are what make the answer mean something.
Work is a dot product
A constant force F acting through a straight displacement d does work W = F·d = Fd cos θ, measured in joules. The cosine is not decoration. A force perpendicular to the motion does no work at all: the normal force on a sliding block and the tension in a string whirling a ball in a circle both change the speed by exactly nothing. The perpendicular part is not wasted, only irrelevant to speed — the upward part of a slanted pull lightens a crate on the floor and cuts the friction, but it never enters W. When the force or the path varies, the same product moves inside an integral, W = ∫F·dr, and its rate is power, P = F·v.
Torque is a cross product
Torque about a point is τ = r×F, where r runs from the pivot to the point where the force acts. Its magnitude rF sin θ reads two ways: rF⊥, the part of the force running across the arm, or r⊥F, the perpendicular distance from the pivot to the force's line of action. Either way the along-the-arm part does nothing, so pushing a door straight towards its hinge or pulling a spanner along its own handle turns nothing at all. Pull with 80 N at 60° to a 0.25 m spanner and |τ| = 0.25 × 80 × sin 60° ≈ 17.3 N m. The direction of τ lies along the rotation axis, and the right-hand rule fixes its sign, which turns clockwise and anticlockwise into ordinary vector arithmetic. Angular momentum L = r×p and the magnetic force qv×B are assembled the same way.
Change one variable at a time
Make the relationship visible.
Set θ to 90°: the dot bar collapses to zero while the cross bar stands tallest. Push θ past 90° and the dot bar drops below the line; pull θ below 0° and the cross bar drops instead.
ALONG ARM F cos θ43.3 N
ACROSS ARM F sin θ25.0 N
DOT r·F12.99 N m
CROSS (r×F)z7.50 N m
Live interpretationALONG ARM F cos θ: 43.3 N. ACROSS ARM F sin θ: 25.0 N. DOT r·F: 12.99 N m. CROSS (r×F)z: 7.50 N m
Catch the common trap
Explain before calculating.
A spanner handle runs 0.30 m from the bolt to your hand, and you pull with 50 N at 25° to the handle. What is the magnitude of the torque about the bolt?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
01 · EasyA rope drags a crate 8.0 m across a level floor with a steady 120 N held at 35° above the horizontal. Find the work the rope does, and say what the vertical part of the pull contributes to it.
- Work is a dot product, W = F·d = Fd cos θ, where θ is the angle between the force and the displacement with both drawn from the same point — here 35°.
- W = (120 N)(8.0 m)(cos 35°) = 960 × 0.8192 = 786 J.
- The vertical part of the pull is F sin θ = 120 × sin 35° = 68.8 N ≈ 69 N. It runs perpendicular to the displacement, so it contributes exactly nothing to W — it only presses less of the crate's weight onto the floor.
AnswerW ≈ 7.9 × 10² J. The 69 N vertical component does no work.
02 · MediumA bolt sits at the origin. A spanner applies F = (6.0, −3.0, 0) N at the point r = (0, 0.40, 0) m from it. Find r·F, the z-component of r×F, and the angle between r and F.
- Dot product from components: r·F = rₓFₓ + ryFy + rzFz = (0)(6.0) + (0.40)(−3.0) + (0)(0) = −1.2 N m. It is negative, so F leans back past 90° from the arm.
- Both vectors lie in the xy-plane, so only the z-component of the cross product survives: (r×F)z = rₓFy − ryFₓ = (0)(−3.0) − (0.40)(6.0) = −2.4 N m. Negative means the torque points into the page — a clockwise turn.
- Magnitudes: |r| = 0.40 m and |F| = √(6.0² + (−3.0)²) = √45 = 6.708 N, so |r||F| = 2.683 N m.
- cos θ = (r·F)/(|r||F|) = −1.2/2.683 = −0.4472, giving θ = 116.6°, i.e. 117° to the nearest degree. Check it against the other product: |r||F| sin 116.6° = 2.683 × 0.8944 = 2.40 N m, matching |(r×F)z|.
Answerr·F = −1.2 N m; (r×F)z = −2.4 N m, a torque of 2.4 N m into the page (clockwise); θ ≈ 117°.
03 · HardA force F = (−1.0, 4.0, 2.0) N acts at the point r = (2.0, 1.0, −3.0) m from a pivot. Find r·F, the torque τ = r×F and its magnitude, and the angle between r and F — then check the two products against each other.
- Dot: r·F = (2.0)(−1.0) + (1.0)(4.0) + (−3.0)(2.0) = −2.0 + 4.0 − 6.0 = −4.0 N m. The unit is newton metres, but this is a scalar, not the torque — the sign only says F leans past 90° from the arm.
- Cross, one component at a time: τₓ = ryFz − rzFy = (1.0)(2.0) − (−3.0)(4.0) = 14 N m; cycling x→y→z, τy = rzFₓ − rₓFz = (−3.0)(−1.0) − (2.0)(2.0) = −1 N m; τz = rₓFy − ryFₓ = (2.0)(4.0) − (1.0)(−1.0) = 9 N m.
- |τ| = √(14² + 1² + 9²) = √278 = 16.67 N m. The vector magnitudes are |r| = √(4.0 + 1.0 + 9.0) = √14 = 3.742 m and |F| = √(1.0 + 16.0 + 4.0) = √21 = 4.583 N, so |r||F| = 17.15 N m.
- Angle from the dot product: cos θ = (r·F)/(|r||F|) = −4.0/17.15 = −0.2333, so θ = 103.5°, i.e. 103° to the nearest degree.
- Check: (r·F)² + |τ|² = 16 + 278 = 294, and (|r||F|)² = 14 × 21 = 294. The two products divide |r||F| between them — cos²θ + sin²θ = 1, written in vectors.
Answerr·F = −4.0 N m; τ = (14, −1, 9) N m with |τ| = 16.7 N m; θ ≈ 103°.