Understanding Cross Product: Vector Product and Angle Separation Explained

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SUMMARY

The discussion focuses on understanding the vector product, specifically the expression x × (x × y) and the calculation of the cross product for two vectors with a given angle. It is established that the result of x × (x × y) is orthogonal to x, contrary to the initial assumption that it would be parallel to y. Additionally, the correct formula for the magnitude of the cross product is clarified as |a × b| = |a||b|sin(θ), emphasizing that direction must also be considered alongside magnitude.

PREREQUISITES
  • Understanding of vector operations, particularly cross products
  • Familiarity with vector notation and components in three-dimensional space
  • Knowledge of trigonometric functions, specifically sine
  • Ability to visualize geometric relationships between vectors
NEXT STEPS
  • Explore the geometric interpretation of the cross product in three-dimensional space
  • Learn about the properties of vector products, including distributive and anti-commutative properties
  • Study specific examples of cross products with varying angles and vector orientations
  • Investigate applications of cross products in physics, such as torque and angular momentum
USEFUL FOR

Students studying physics or mathematics, educators teaching vector calculus, and professionals in engineering or computer graphics who require a solid understanding of vector operations.

theBEAST
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I have two problems:


Homework Statement


In general, what can be said about the vector product x×(x×y)

The Attempt at a Solution


I thought the result of this would be parallel to y. However the answer suggests it is orthogonal to x. Can anyone explain how I could approach this question? I tried to visualize it in my head but it was very difficult.


Homework Statement


Given two vectors of length 2 and 3 separated by an angle of 30 degrees, what is the cross product of the two vectors?

The Attempt at a Solution


I know that the cross product = (length of vector a)*(length of vector b)*sin(theta)
This gives us 3.0, however the answer key suggests there is not enough information to answer the question.
 
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theBEAST said:
I have two problems:

Homework Statement


In general, what can be said about the vector product x×(x×y)

The Attempt at a Solution


I thought the result of this would be parallel to y. However the answer suggests it is orthogonal to x. Can anyone explain how I could approach this question? I tried to visualize it in my head but it was very difficult.
You seem to be making an assumption about the relative orientations of ##\vec{x}## and ##\vec{y}##. Try look at a few specific cases and see if it clears up any misconceptions you have. Concentrate on the direction of the cross product. Don't worry about the magnitudes for now. Draw pictures!

First case, let ##\vec{x} = (1, 0, 0)## and ##\vec{y} = (0, 1, 0)##. That is the two vectors lie along the x-axis and y-axis respectively. What is ##\vec{x}\times\vec{y}##? And when you cross that result again with ##\vec{x}##, what do you get?

Second case, let ##\vec{x} = (1, 0, 0)## and ##\vec{y} = (1, 1, 0)##. This time, the second vector still lies in the xy-plane, but it's no longer aligned to the y-axis. Again, what is ##\vec{x}\times\vec{y}##? And when you cross that result again with ##\vec{x}##, what do you get? What effect did changing ##\vec{y}## have on the direction of the final answer?

Homework Statement


Given two vectors of length 2 and 3 separated by an angle of 30 degrees, what is the cross product of the two vectors?

The Attempt at a Solution


I know that the cross product = (length of vector a)*(length of vector b)*sin(theta)
This gives us 3.0, however the answer key suggests there is not enough information to answer the question.
This isn't quite right. The cross product gives you a vector, but quantity on the righthand side is a number. The two sides of your equation can't be equal. What you mean is
$$|\vec{a}\times\vec{b}| = |a||b|\sin\theta.$$ You found the magnitude of the cross product, but that's only half the answer. You still have to give its direction.
 
vela said:
You seem to be making an assumption about the relative orientations of ##\vec{x}## and ##\vec{y}##. Try look at a few specific cases and see if it clears up any misconceptions you have. Concentrate on the direction of the cross product. Don't worry about the magnitudes for now. Draw pictures!

First case, let ##\vec{x} = (1, 0, 0)## and ##\vec{y} = (0, 1, 0)##. That is the two vectors lie along the x-axis and y-axis respectively. What is ##\vec{x}\times\vec{y}##? And when you cross that result again with ##\vec{x}##, what do you get?

Second case, let ##\vec{x} = (1, 0, 0)## and ##\vec{y} = (1, 1, 0)##. This time, the second vector still lies in the xy-plane, but it's no longer aligned to the y-axis. Again, what is ##\vec{x}\times\vec{y}##? And when you cross that result again with ##\vec{x}##, what do you get? What effect did changing ##\vec{y}## have on the direction of the final answer?



This isn't quite right. The cross product gives you a vector, but quantity on the righthand side is a number. The two sides of your equation can't be equal. What you mean is
$$|\vec{a}\times\vec{b}| = |a||b|\sin\theta.$$ You found the magnitude of the cross product, but that's only half the answer. You still have to give its direction.

Wow, thanks for clearing everything up! Makes sense now.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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