Physics I course sequence · Interactive lesson
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.
UPI-01.06 · Lesson 6 of 9 in this unit. Position is not a completion score.Two vectors, two questions
The dot product gives a scalar. The cross product gives a vector perpendicular to both inputs.
Dot product · alignment
A · B = AB cos θ
How much of A lies along B?
Cross product · perpendicular contribution
|A × B| = AB sin θ
How large is the spanned parallelogram, and which way is its normal?
The full physical model
Two common ways of combining vectors occur throughout mechanics. 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.
Definition and a two-job example
The dot product of two vectors is a scalar measuring how much they line up. The cross product is a vector whose magnitude measures their perpendicular contribution, perpendicular to both, with the right-hand rule choosing which way it points. 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.
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 of one runs perpendicular to 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.
Dot product
A scalar that measures alignment.
A · B = AₓBₓ + AᵧBᵧ + A𝓏B𝓏
- Positive for an acute angle.
- Zero for perpendicular nonzero vectors.
- Negative for an obtuse angle.
- Commutative: A · B = B · A.
Signed projection
B̂ = B / |B|
A · B̂ = A cos θ
B̂ has magnitude 1 and no physical unit. The scalar projection can be negative; it is undefined if B = 0.
A · A = |A|². Recovering an angle requires A ≠ 0 and B ≠ 0.
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.
Cross product
A vector normal to the input plane.
|A × B| = AB sin θ
- θ is the included angle, 0–180°.
- Magnitude is the parallelogram area.
- A × B = −B × A.
- Parallel or antiparallel inputs give zero.
Right-hand direction
Point your right-hand fingers along A, curl toward B, and follow your thumb. A zero cross product has no direction.
τ = r × F
|τ| = rF sin θ
r runs from the pivot to the force’s application point. Torque is perpendicular to r and F; use N·m, not J.
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.
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 is perpendicular to the plane of r and F, with its sense fixed by the right-hand rule, 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.
Component & unit-vector rules
Use a right-handed orthonormal basis.
(A × B)ᵧ = A𝓏Bₓ − AₓB𝓏
(A × B)𝓏 = AₓBᵧ − AᵧBₓ
î · î = ĵ · ĵ = k̂ · k̂ = 1
î · ĵ = ĵ · k̂ = k̂ · î = 0
î × ĵ = k̂
ĵ × k̂ = î
k̂ × î = ĵ
Reverse the order to flip the sign. A basis vector crossed with itself gives zero.
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.
Dot vs cross in physics
Choose the product by the quantity you need—not just by the presence of an angle.
| Feature | Dot product | Cross product |
|---|---|---|
| Result | Scalar | Vector |
| Reverse the order | Unchanged | Sign reverses |
| Geometric meaning | Signed alignment / projection | Oriented area / perpendicular contribution |
| Zero for nonzero inputs | Perpendicular | Parallel or antiparallel |
| Common uses | Work F · d; power F · v | Torque r × F; angular momentum r × p |
| Units | Product of the input units | Product of the input units |
Work & power
For a constant force, W = F · d. A force along the motion does positive work; a force against it does negative work. In general, W = ∫F · dr and P = F · v.
Torque & lever arm
|τ| = rF⊥ = r⊥F. A force along the lever arm produces no torque about the pivot. The scalar r · F is not work: r is a position from the pivot, not a displacement.
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.
Interactive vector product lab
Adjust the geometry or signed components, then connect the calculation to work and torque.
Dimensionless lengths, 0–10. θ is the included angle, 0–180°.
Product readouts
- A · B · scalar
- 6.000
- |A × B| · nonnegative magnitude
- 10.392
- A × B · vector
- (0.000, 0.000, 10.392)
- Included angle · 0–180°
- 60.000°
- Signed projection of A onto B
- 1.500
- Vector projection of A onto B
- (0.750, 1.299, 0.000)
(A · B)² + |A × B|² = 144.000; A²B² = 144.000.
Readouts are rounded; calculations use unrounded inputs. Values within floating-point cancellation error are treated as zero. This lab does not change saved course progress.
Explore the original force and lever-arm lab
This comparison uses the same zero line and bar scale. The cross bar is a signed z-component, not a magnitude; r · F is not work.
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
Right-hand-rule explorer
Swap the order and watch the normal reverse.
Use your right hand
- Point your fingers along the first vector.
- Curl through the smaller angle toward the second.
- Your thumb gives the cross-product direction.
î × ĵ = ?
The diagram shows the direction. Cover it first if you want to predict.
Catch the common trap
Cosine measures along; sine measures across. Check the physical meaning and the units.
Choosing sine or cosine by habit is a common error. Work is a dot product; torque is a cross product. A zero dot product implies a right angle only when both vectors are nonzero.
Explain before calculating
The trap is choosing the trig function by habit — cos for anything at an angle. Ask instead which part of the vector does the job. If only the part running along the other vector counts, as when a slanted rope drags a sledge, you want cos and a dot product. If only the part running across it counts, as when you push a door perpendicular to the line from its hinge, you want sin and a cross product. And A·B = 0 is not a failed calculation: for two nonzero vectors, it means their directions are at right angles. If either vector is zero, their angle is undefined.
0 / 5 checked · 0 correct · this visit only
These answers are not saved course completion. Refreshing clears this practice.
Worked examples
A rope doing work, a planar spanner, and a three-dimensional torque.
01 · Easy · Work by a rope
A 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.
Show worked solution →
- 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.
W ≈ 7.9 × 10² J. The 69 N vertical component does no work.
02 · Medium · Spanner and bolt
A 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.
Show worked solution →
- 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|.
r·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 · Hard · 3D torque & product identity
A 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.
Show worked solution →
- 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 ≈ 103.49° from unrounded values, 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.
r·F = −4.0 N m; τ = (14, −1, 9) N m with |τ| = 16.7 N m; θ ≈ 103°.
Summary & applications
The dot product returns a signed scalar; the cross product returns an oriented normal vector. Use components for the calculation and geometry to check the sign and direction.
(A · B)² + |A × B|² = |A|²|B|²
Both products are useful in mechanics: work and power use alignment, while torque and angular momentum use the cross product. A magnetic force qv × B is perpendicular to both velocity and magnetic field.
Full formula reference and conditions
For angle recovery, both vectors must be nonzero. For projection onto B, only B must be nonzero. |A × B| ≥ 0; in a planar problem the z-component may be negative. W = F · d assumes a constant force.
Dot product · geometry and components
A·B = AB cos θ = AₓBₓ + AyBy + AzBz
Positive when they lean the same way, zero at 90°, negative beyond it.
scalar out · θ is the angle between them, drawn tail to tailAngle, projection, and length
cos θ = (A·B)/(AB); A·B̂ = A cos θ; A·A = A²
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 ACross product · magnitude and direction
|A×B| = AB sin θ; A×B ⊥ both, sense from the right-hand rule
Zero for parallel vectors, largest at 90° ; the dot product is zero at 90°.
equal to the area of the parallelogram that A and B spanCross product · components
(A×B)ₓ = AyBz − AzBy; cycle x→y→z→x for the other two
Order matters: A×B = −B×A, and A×A = 0.
planar case: (A×B)z = AₓBy − AyBₓ, one signed numberUnit vector rules
î·î = 1; î·ĵ = 0; î×ĵ = k̂; ĵ×k̂ = î; k̂×î = ĵ; î×î = 0
Expanding both products term by term produces the component formulas above.
reverse the order and the sign flips: ĵ×î = −k̂Where the products go: work and torque
W = F·d = Fd cos θ; τ = r×F; |τ| = rF sin θ = rF⊥ = r⊥F
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