Vectors - Geometric interpretation of cross product

  • In this lecture, we will explore the geometric interpretation of the cross product.
  • The cross product of two vectors gives us a vector that is orthogonal to the two original vectors.
  • It is denoted by the symbol × or ⨯.

Geometric Interpretation

  • The magnitude of the cross product of two vectors A and B is given by:
    • |A × B| = |A| |B| sinθ
    • θ is the angle between the two vectors.
  • The direction of the cross product is determined by the right-hand rule:
    • If you extend your right hand’s fingers in the direction of vector A and then curl them towards vector B, your thumb points in the direction of the cross product.

Properties of Cross Product

  1. The cross product is not commutative:
    • A × B = - (B × A)
  1. The length of the cross product is:
    • |A × B| = |A| |B| sinθ
  1. The direction of the cross product is orthogonal to both A and B.
  1. If A and B are parallel, the cross product is zero:
    • A × B = 0
  1. If A and B are perpendicular, the magnitude of the cross product is:
    • |A × B| = |A| |B|

Example 1

  • Let A = i + 2j and B = 3i + k be two vectors.
  • Find the cross product of A and B.
  • Solution:
    • (i × 3i) + (i × k) + (2j × 3i) + (2j × k)
    • = 0 + (-k) + (6k) + 0
    • = 5k
  • Therefore, A × B = 5k.

Example 2

  • Given two vectors A = 2i - 3j + 4k and B = 5i + 2j - k, find the angle between them.
  • Solution:
    • |A × B| = |A| |B| sinθ
    • |A × B| = sqrt(29) sqrt(30) sinθ
    • |A × B| = 5 sqrt(2) sqrt(3) sinθ
  • Since A × B = 0 (cross product is zero for parallel vectors), sinθ = 0.
  • Therefore, θ = 0 degrees.

Example 3

  • Let A = i + j + k and B = 2i - j - 2k be two vectors.
  • Find the vector that is orthogonal to both A and B.
  • Solution:
    • A × B = (i × (2i - j - 2k)) + (j × (2i - j - 2k)) + (k × (2i - j - 2k))
    • = (-5i - 4j + 3k)
  • Therefore, the vector orthogonal to both A and B is -5i - 4j + 3k.

Note: Slide numbering is not allowed, so the slides will be labeled with headings only. Also, expressions and equations may not be accurately represented in this text format.

Slide 11: Properties of Cross Product (continued)

  • The cross product is distributive:
    • A × (B + C) = (A × B) + (A × C)
  • The cross product satisfies the scalar triple product rule:
    • A · (B × C) = B · (C × A) = C · (A × B)
    • This rule is useful in solving problems involving vectors.

Slide 12: Geometric Interpretation (continued)

  • The cross product of two vectors can also be interpreted as the area of the parallelogram formed by those vectors.
  • The magnitude of the cross product is equal to the area of the parallelogram.

Slide 13: Example 4

  • Given two vectors A = 2i - 3j + k and B = 4i - 2j + 3k, find the area of the parallelogram formed by these vectors.

  • Solution:

    • A × B = (-3i - 5j - 2k)
    • |A × B| = sqrt((-3)^2 + (-5)^2 + (-2)^2) = sqrt(38)

  • Therefore, the area of the parallelogram formed by A and B is sqrt(38) square units.

Slide 14: Example 5

  • Find the position vector of the point where the line defined by A = 2i - 3j + 4k + λ(i - 2j + k) intersects the plane defined by B = 3i - j + 2k + μ(2i + j - k) + v(i + j - 3k).

  • Solution:

    • The normal vector to the plane is the cross product of the two direction vectors of the plane, B and C.
    • N = B × C = (9i + 5j + 9k).

  • The point where the line intersects the plane can be found using the formula:

    • P = A + λ(N dot (D - A))/ (N dot (D - A)), where D is a point on the line.

Slide 15: Example 5 (continued)

  • Applying the formula, we get:
    • P = (2i - 3j + 4k) + λ((9i + 5j + 9k) dot (D - (2i - 3j + 4k)))/ ((9i + 5j + 9k) dot (D - (2i - 3j + 4k))).
  • Now, substitute the values of A, B, C, and D to find the position vector of the point of intersection.

Slide 16: Application - Torque

  • In physics, the cross product of a force vector and a displacement vector gives us the torque or moment of the force.
  • Torque is a measure of the rotational effect of a force on an object.
  • Torque = r × F
    • r is the displacement vector from the point of rotation to the point of application of the force.
    • F is the force vector.

Slide 17: Application - Torque (continued)

  • The magnitude of the torque is given by:
    • |Torque| = |r| |F| sinθ
    • θ is the angle between the displacement and force vectors.
  • The direction of the torque is given by the right-hand rule:
    • If you extend your right hand’s fingers in the direction of the displacement vector and then curl them towards the force vector, your thumb points in the direction of the torque.

Slide 18: Summary of Cross Product

  • The cross product of two vectors has a geometric interpretation as a vector orthogonal to both vectors.
  • It can also be interpreted as the area of the parallelogram formed by the two vectors.
  • The cross product has various properties, such as distributivity and scalar triple product rule.
  • It is useful in solving problems involving torque and finding points of intersection between lines and planes.

Slide 19: Summary of Geometric Interpretation

  • The magnitude of the cross product is equal to the area of the parallelogram formed by the two vectors.
  • The direction of the cross product is orthogonal to both vectors.
  • The right-hand rule can be used to determine the direction of the cross product.

Slide 20: Conclusion

  • The geometric interpretation of the cross product is essential to understand its properties and applications.
  • The cross product provides a useful tool in various fields like physics and engineering.
  • Practice solving problems involving cross product to enhance your understanding and problem-solving skills. Vectors - Geometric interpretation of cross product

Slide 21:

  • The cross product can also be used to find the angle between two vectors.
  • The dot product of the two vectors can be used in combination with the cross product to find the angle.
  • The formula is given by: |A × B| = |A| |B| sinθ
  • Solving for θ gives: θ = arcsin(|A × B| / (|A| |B|))

Slide 22:

  • Another important property of the cross product is its use in finding the projection of a vector onto a plane.
  • The projection of vector A onto a plane with normal vector N is given by: A_proj = A - (A · N / |N|²)N
  • The cross product is used in this formula to find the perpendicular component of A.

Slide 23:

  • The cross product is also used in finding the volume of a parallelepiped.
  • The volume of a parallelepiped formed by three vectors A, B, and C is given by: V = |A · (B × C)|
  • This formula is derived using the scalar triple product rule.

Slide 24:

  • Example 6:
    • Let A = i - j and B = j - k be two vectors.
    • Find the angle between A and B.

Slide 25:

  • Example 7:
    • Given two vectors A = i + 2j + k and B = 3i + 2j - k, find the projection of A onto B.

Slide 26:

  • Example 8:
    • Let A = i + j + k and B = 2i - j + 3k be two vectors.
    • Find the volume of the parallelepiped formed by A, B, and C = 2i + 3j - 4k.

Slide 27:

  • Example 9:
    • Find the torque produced by a force vector F = 3i + 5j + 2k acting at a point r = 2i - 3j + 4k with respect to the origin.

Slide 28:

  • Example 10:
    • Let A = 2i - 3j + 4k and B = i + j + 2k be two vectors.
    • Determine if these vectors are parallel, orthogonal, or neither.

Slide 29:

  • Example 11:
    • Given three vectors A = i + j, B = i + 2j, and C = 3i - 2j, find the angle between vector A and the plane formed by vectors B and C.

Slide 30:

  • Conclusion:
    • The geometric interpretation of the cross product is valuable for understanding its applications in various fields.
    • Practice solving problems involving the cross product to enhance your problem-solving skills.
    • Review the formulas and properties discussed in this lecture to reinforce your understanding.