A door turns most easily when you push near its handle, perpendicular to its surface. Push toward the hinge and the same force may do very little turning. The cross product records both the size of that perpendicular effect and its direction.
We will connect its geometry to a component calculation, then use torque and magnetic force to show why the order of the vectors and the sign of a charge matter. The examples state their coordinate frames so that every sign can be checked.
What you will learn to do
- Distinguish dot and cross products by the question they answer.
- Find cross-product directions in a right-handed frame.
- Calculate components and verify perpendicularity.
- Use a perpendicular lever arm for torque.
- Apply the charge sign separately in a magnetic force.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: The cross product: perpendicular effects and direction Learning goals: - Distinguish dot and cross products by the question they answer. - Find cross-product directions in a right-handed frame. - Calculate components and verify perpendicularity. - Use a perpendicular lever arm for torque. - Apply the charge sign separately in a magnetic force. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BEFORE YOU READ
Bring back one idea
Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.
Why does pushing directly along a line through the hinge give zero torque about that hinge?
Compare your reasoning
The perpendicular lever arm is zero. Equivalently, position and force are parallel, so the sine factor in the torque magnitude vanishes.
Revisit the explanation →01 / The cross product: perpendicular effects and direction
A different question from the dot product
A dot product measures how much one vector lies along another and produces a scalar. Some physical effects depend instead on the component perpendicular to a direction. Turning a door provides a useful example: a push directed along the line from the hinge to the handle produces no turning effect about the hinge, whereas a sideways push can.
The cross product combines this perpendicular dependence with an oriented normal direction. For two vectors a and b, its magnitude is ab sin θ, where θ is the angle between them from 0 to π. It is zero for parallel or antiparallel vectors and greatest for perpendicular vectors of fixed magnitudes.
The result is a vector perpendicular to the plane of the two inputs. Its direction carries information about orientation. Reversing the order of the inputs reverses that direction, even though the magnitude is unchanged. This is why the order in a torque or magnetic-force formula is part of the physics.
A zero cross product has no unique direction. It can result from parallel inputs or from one input being the zero vector.
WORKED EXAMPLE
Separate the angle from the vector lengths
Two displacement vectors have magnitudes 3 m and 4 m, with angle 30° between them. Find the magnitude of their cross product.
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STOP & THINK
Two nonzero vectors have zero cross product. What must be true?
- They have equal magnitudes
- They are parallel or antiparallel
- They are perpendicular
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: The cross product: perpendicular effects and direction Focus: A different question from the dot product Learning goals: - Distinguish dot and cross products by the question they answer. - Find cross-product directions in a right-handed frame. - Calculate components and verify perpendicularity. - Use a perpendicular lever arm for torque. - Apply the charge sign separately in a magnetic force. Equations in this section (LaTeX): |\mathbf a\times\mathbf b|=ab\sin\theta,\qquad \mathbf b\times\mathbf a=-\mathbf a\times\mathbf b Meaning: A zero cross product has no unique direction. It can result from parallel inputs or from one input being the zero vector. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
02 / The cross product: perpendicular effects and direction
Build the direction in a right-handed frame
Choose a right-handed coordinate system. With x to the right and y upward on a page, +z points out of the page. Curl the fingers of your right hand from the first vector toward the second through the smaller angle; your thumb gives the cross-product direction. The order must be read from the written product, not chosen after guessing the result.
For unit basis vectors, x crossed with y gives z, y crossed with z gives x, and z crossed with x gives y. These are cyclic permutations. Reversing any pair introduces a minus sign. Crossing a basis vector with itself gives zero.
For a flat diagram it is useful to state both the vector direction and its rotational meaning. A position vector to the right crossed with an upward force points out of the page. The corresponding turning tendency is anticlockwise when the page is viewed from the +z side. Viewed from the opposite side, the same physical rotation looks clockwise.
| Product | Result | Reverse order |
|---|---|---|
| x̂ × ŷ | ẑ | ŷ × x̂ = −ẑ |
| ŷ × ẑ | x̂ | ẑ × ŷ = −x̂ |
| ẑ × x̂ | ŷ | x̂ × ẑ = −ŷ |
| x̂ × x̂ | 0 | The same direction has no perpendicular part |
STOP & THINK
In a right-handed frame, what is the direction of x̂ × (−ẑ)?
- +y
- −y
- +z
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: The cross product: perpendicular effects and direction Focus: Build the direction in a right-handed frame Learning goals: - Distinguish dot and cross products by the question they answer. - Find cross-product directions in a right-handed frame. - Calculate components and verify perpendicularity. - Use a perpendicular lever arm for torque. - Apply the charge sign separately in a magnetic force. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
03 / The cross product: perpendicular effects and direction
Calculate each component without losing its sign
A component method is more dependable than trying to picture a complicated three-dimensional angle. Expand each vector in x, y and z components, distribute the cross product across sums, and use the basis products. Terms with identical basis directions vanish.
The resulting x component uses only the y and z components of the inputs. The y component uses z and x; the z component uses x and y. Each has a forward product minus its reverse. A determinant is a compact memory aid for this pattern, but expanding its middle term carelessly is a common sign error.
Check the result by taking its dot product with each input. Both should be zero, up to numerical rounding, because a cross product is perpendicular to both original vectors. This catches many component errors independently of memorising the formula.
All components must be expressed in the same right-handed orthonormal frame.
WORKED EXAMPLE
Compute a product and verify both perpendicularities
Let a = (2, −1, 3) and b = (1, 4, −2), expressed in a right-handed orthonormal basis. Compute a × b. Treat these vectors as dimensionless for this exercise.
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STOP & THINK
For a = (3, 2, 0) and b = (−1, 4, 0), what is the z component of a × b?
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: The cross product: perpendicular effects and direction
Focus: Calculate each component without losing its sign
Learning goals:
- Distinguish dot and cross products by the question they answer.
- Find cross-product directions in a right-handed frame.
- Calculate components and verify perpendicularity.
- Use a perpendicular lever arm for torque.
- Apply the charge sign separately in a magnetic force.
Equations in this section (LaTeX):
\mathbf a\times\mathbf b=(a_yb_z-a_zb_y)\hat{\mathbf x}+(a_zb_x-a_xb_z)\hat{\mathbf y}+(a_xb_y-a_yb_x)\hat{\mathbf z}
Meaning: All components must be expressed in the same right-handed orthonormal frame.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.04 / The cross product: perpendicular effects and direction
Torque uses a position measured from a stated point
Torque about an origin O is r × F, where r runs from O to the point where the force acts. Its magnitude is force times the perpendicular distance from O to the force’s line of action. That perpendicular distance is the lever arm. It is not necessarily the full length of r.
Moving a force’s application point along the same line of action leaves its torque about O unchanged for the usual rigid-body force description: the added displacement is parallel to F and has zero cross product with it. Moving the origin generally changes the torque. Always identify the point or axis about which you are calculating.
Torque has units N·m. Energy has the same dimensions, but it is a different quantity: work uses a dot product with displacement, while torque uses a cross product with a position vector. Do not infer that a large torque necessarily performs work at an instant; the body may be held still.
For a fixed axis, the relevant turning component is the projection of this torque onto that axis.
WORKED EXAMPLE
A force with both radial and turning components
Relative to an origin, a force is applied at r = (0.40, 0, 0) m. The force is F = (6, 8, 0) N. Find its torque about the origin.
MEMBERSThe approach, step-by-step working and check are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: The cross product: perpendicular effects and direction Focus: Torque uses a position measured from a stated point Learning goals: - Distinguish dot and cross products by the question they answer. - Find cross-product directions in a right-handed frame. - Calculate components and verify perpendicularity. - Use a perpendicular lever arm for torque. - Apply the charge sign separately in a magnetic force. Equations in this section (LaTeX): \boldsymbol\tau_O=\mathbf r\times\mathbf F,\qquad |\boldsymbol\tau_O|=F d_\perp Meaning: For a fixed axis, the relevant turning component is the projection of this torque onto that axis. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
05 / The cross product: perpendicular effects and direction
Separate the geometric cross product from an extra sign
The magnetic force on a moving point charge is q(v × B). First calculate v × B in the chosen frame. Then multiply by the charge. A negative charge reverses the cross-product direction; it does not change which two vectors were crossed.
Only the velocity component perpendicular to B contributes. A particle moving exactly along the magnetic field experiences no magnetic force from that field even if both its speed and the field are large. When the magnetic force acts alone, it is perpendicular to velocity, so its instantaneous power F·v is zero. It can turn the velocity without changing the speed.
These statements concern the magnetic part of the force. An electric field can add a force along the velocity and change the kinetic energy. Specifying “magnetic field” does not imply that all other forces are absent unless the model says so.
WORKED EXAMPLE
Reverse the charge only after finding the geometric direction
A charge q = −2 μC moves at 300 m/s along +x through B = 0.50 T along +z. Find the magnetic force.
MEMBERSThe approach, step-by-step working and check are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: The cross product: perpendicular effects and direction Focus: Separate the geometric cross product from an extra sign Learning goals: - Distinguish dot and cross products by the question they answer. - Find cross-product directions in a right-handed frame. - Calculate components and verify perpendicularity. - Use a perpendicular lever arm for torque. - Apply the charge sign separately in a magnetic force. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
06 / The cross product: perpendicular effects and direction
Use geometry as an independent error detector
Before calculating components, predict which components of the answer must vanish. If both input vectors lie in a plane, their cross product must be normal to that plane. If one input reverses, the answer reverses. If one input doubles while its direction stays fixed, the answer doubles.
After calculating, check units, both perpendicularities and a simple limiting case. A small angle should give a small product; a right angle should give the product of magnitudes. The identity |a × b|² = a²b² − (a·b)² connects the two vector products and provides another arithmetic check.
Be careful with more than two vectors. Cross products distribute over addition but are not generally associative: a × (b × c) need not equal (a × b) × c. Keep parentheses and use components or a derived identity when a nested product appears. A valid two-vector shortcut should not be extended by visual pattern alone.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: The cross product: perpendicular effects and direction Focus: Use geometry as an independent error detector Learning goals: - Distinguish dot and cross products by the question they answer. - Find cross-product directions in a right-handed frame. - Calculate components and verify perpendicularity. - Use a perpendicular lever arm for torque. - Apply the charge sign separately in a magnetic force. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BRING THE IDEAS TOGETHER
What should stay with you
- A cross product is perpendicular to both inputs and has magnitude ab sin θ.
- Reversing the input order reverses the result.
- Component calculations require one common right-handed orthonormal frame.
- Torque depends on the stated origin and force line of action.
- A negative scalar reverses the geometric product direction.
RETRIEVE, THEN EXPLAIN
Try it without the examples
Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.
SELF CHECK 1
A 12 N force acts perpendicular to a 0.25 m position vector from a pivot. What is the torque magnitude about that pivot?
Answer in N·m.
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SELF CHECK 2
A charged particle moves exactly parallel to a uniform magnetic field. What is its magnetic force?
- Zero
- qvB in the field direction
- qvB opposite the field direction
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SELF CHECK 3
Can the torque of a force change when you change the reference origin?
- Yes, because the position vector changes
- No, torque is always origin-independent
- Only if the force magnitude changes
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SELF CHECK 4
If both vectors have units of velocity, what units does their cross product have?
- m/s
- m²/s²
- m/s²
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APPLY THE CHAPTER
Take the ideas into a problem
Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.
Which component turns the arm?Intermediate · Hints and guided reasoningCOME BACK TO THE IDEA
What can you explain later?
Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.
Next study session
What changes when you reverse the order of two vectors in a cross product?
Compare your reasoning
The direction reverses while the magnitude stays the same: a × b = −b × a. This matters when translating a physical law with a specified order.
Revisit the explanation →A few days later: change the situation
A negative charge has the same velocity as a positive charge in the same magnetic field. How do you obtain its force direction?
Compare your reasoning
First find v × B from the right-hand rule, then reverse it for negative charge. Changing the charge sign is different from changing the velocity direction.
Revisit the explanation →Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.
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