Vectors - Problems - Proof using STP

  • Problem: Given two vectors A and B, find their sum A + B.
  • Solution: To find the sum of two vectors, add their corresponding components together. For example, if A = &

##10216;3, 4&

##10217; and B = &

##10216;-2, 1&

##10217;, then A + B = &

##10216;3 + (-2), 4 + 1&

##10217; = &

##10216;1, 5&

##10217;.

  • Problem: Given two vectors A = &

##10216;2, -3, 1&

##10217; and B = &

##10216;5, 2, -4&

##10217;, find their difference A - B.

  • Solution: To find the difference between two vectors, subtract their corresponding components. In this case, A - B = &

##10216;2 - 5, -3 - 2, 1 - (-4)&

##10217; = &

##10216;-3, -5, 5&

##10217;.

  • Problem: Given a vector A = &

##10216;3, -2, 4&

##10217;, find the magnitude of A.

  • Solution: The magnitude of a vector is calculated using the formula |A| = √(Ax2 + Ay2 + Az2). In this case, |A| = √(32 + (-2)2 + 42) = √(9 + 4 + 16) = √29.

  • Problem: Given two vectors A = &

##10216;1, -2, 3&

##10217; and B = &

##10216;4, 2, -1&

##10217;, find their dot product A · B.

  • Solution: The dot product of two vectors is calculated using the formula A · B = AxBx + AyBy + AzBz. In this case, A · B = (1)(4) + (-2)(2) + (3)(-1) = 4 - 4 - 3 = -3.

  • Problem: Given a vector A = &

##10216;2, -1, 3&

##10217;, find its unit vector u.

  • Solution: A unit vector is a vector with magnitude equal to 1. To find the unit vector u, divide the vector A by its magnitude. In this case, |A| = √(22 + (-1)2 + 32) = √(4 + 1 + 9) = √14. Therefore, u = &

##10216;2/√14, -1/√14, 3/√14&

##10217;.

  • Problem: Given two vectors A = &

##10216;2, -1, 3&

##10217; and B = &

##10216;-3, 2, -4&

##10217;, find their cross product A × B.

  • Solution: The cross product of two vectors is calculated using the formula A × B = &

##10216;AyBz - AzBy, AzBx - AxBz, AxBy - AyBx&

##10217;. In this case, A × B = &

##10216;(-1)(-4) - (3)(2), (3)(-3) - (2)(-4), (2)(2) - (-1)(-3)&

##10217; = &

##10216;8, -3, 1&

##10217;.

  • Problem: Given a vector A = &

##10216;2, -5, 1&

##10217;, find a vector parallel to A with magnitude 10.

  • Solution: To find a vector parallel to A, we can multiply A by a scalar. Let’s call this scalar k. Since we want the magnitude of the parallel vector to be 10, we can set up the equation |k&

##10216;2, -5, 1&

##10217;| = 10. Using the magnitude formula, we get √(k2(22 + (-5)2 + 12)) = 10. Simplifying, we have √(30k2) = 10. Squaring both sides, we get 30k2 = 100. Solving for k, we find k = ±√(10/3). Therefore, a vector parallel to A with magnitude 10 can be either ±(√(10/3))&

##10216;2, -5, 1&

##10217;.

  • Problem: Given the position vectors r1 = &

##10216;3, 2, -5&

##10217; and r2 = &

##10216;-4, 1, 6&

##10217;, find the vector r joining these two points.

  • Solution: The vector r joining two points can be found by subtracting their position vectors. In this case, r = r2 - r1 = &

##10216;-4 - 3, 1 - 2, 6 - (-5)&

##10217; = &

##10216;-7, -1, 11&

##10217;.

  • Proof using STP: To prove that the vectors A, B, and C are coplanar, we can use the scalar triple product (STP). If STP(A, B, C) = 0, then the three vectors are coplanar.
  • STP Formula: STP(A, B, C) = A · (B × C).
  • Proof: If STP(A, B, C) = 0, then A · (B × C) = 0.
    • Expand the right side of the equation: A · (B × C) = A · &

##10216;ByCz - BzCy, BzCx - BxCz, BxCy - ByCx&

##10217;.

  • Substitute the vector values and simplify.
  • If the result is 0, then the vectors A, B, and C are coplanar.

Dot Product - Properties

  • The dot product is commutative: A · B = B · A
  • The dot product is distributive over vector addition: A · (B + C) = A · B + A · C
  • The dot product is distributive over scalar multiplication: k(A · B) = (kA) · B = A · (kB)

Dot Product - Geometrical Interpretation

  • The dot product of two vectors can be positive, negative, or zero:
    • If A and B are in the same direction, the dot product is positive.
    • If A and B are in opposite directions, the dot product is negative.
    • If A and B are orthogonal (perpendicular), the dot product is zero.
  • The dot product measures the projection of one vector onto another.

Dot Product - Example

  • Given A = &

##10216;3, 4&

##10217; and B = &

##10216;2, -1&

##10217;, find their dot product A · B.

  • Using the dot product formula, A · B = (3)(2) + (4)(-1) = 6 - 4 = 2.

Cross Product - Properties

  • The cross product is anti-commutative: A × B = -B × A
  • The cross product is distributive over vector addition: A × (B + C) = A × B + A × C
  • The cross product is not associative: A × (B × C) ≠ (A × B) × C

Cross Product - Geometrical Interpretation

  • The cross product of two vectors is perpendicular to both vectors.
  • 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 follows the right-hand rule.

Cross Product - Example

  • Given A = &

##10216;2, 3, 1&

##10217; and B = &

##10216;4, -2, 5&

##10217;, find their cross product A × B.

  • Using the cross product formula, A × B = &

##10216;(3)(5) - (1)(-2), (1)(4) - (2)(5), (2)(-2) - (3)(4)&

##10217; = &

##10216;17, 6, -14&

##10217;.

Scalar Triple Product (STP) - Properties

  • STP(A, B, C) = -STP(B, A, C) = -STP(A, C, B) = -STP(C, B, A)
  • STP(A, B, C) is invariant under cyclic permutation:
    • STP(A, B, C) = STP(B, C, A) = STP(C, A, B)
  • STP(A, B, C) = 0 when A, B, and C are coplanar.

Scalar Triple Product (STP) - Example

  • Given A = &

##10216;2, -1, 3&

##10217;, B = &

##10216;3, 4, -2&

##10217;, and C = &

##10216;1, -5, 2&

##10217;, find their scalar triple product STP(A, B, C).

  • Using the STP formula, STP(A, B, C) = A · (B × C) = A · &

##10216;ByCz - BzCy, BzCx - BxCz, BxCy - ByCx&

##10217;.

  • Substitute the vector values and simplify.

Projection of a Vector

  • The projection of a vector v onto a vector u is given by the formula proju v = (v · u) / (∥u∥2) u.

Projection of a Vector - Example

  • Given u = &

##10216;3, 2&

##10217; and v = &

##10216;4, 1&

##10217;, find the projection of v onto u.

  • Using the projection formula, proju v = (v · u) / (∥u∥2) u = ((4)(3) + (1)(2)) / ((3)(3) + (2)(2)) &

##10216;3, 2&

##10217;.

Vector Equation of a Line

  • The vector equation of a line passing through a point A and parallel to a vector u is given by r = A + tu, where t is a scalar parameter.
  • This equation represents all the points on the line.

Vector Equation of a Line - Example

  • Given the point A = &

##10216;1, 2, -3&

##10217; and the vector u = &

##10216;2, -1, 3&

##10217;, find the vector equation of the line passing through A and parallel to u.

  • Using the vector equation formula, r = A + tu, the equation of the line is r = &

##10216;1 + 2t, 2 - t, -3 + 3t&

##10217;.

Scalar Projection of a Vector

  • The scalar projection of a vector v onto a vector u is given by the formula scal-proju v = (∥v∥ cos θ) / ∥u∥, where θ is the angle between v and u.

Scalar Projection of a Vector - Example

  • Given u = &

##10216;3, 2&

##10217; and v = &

##10216;7, 1&

##10217;, find the scalar projection of v onto u.

  • Using the scalar projection formula, scal-proju v = (∥v∥ cos θ) / ∥u∥ = ((√(72 + 12)) cos θ) / (√(32 + 22)).

Parametric Equations of a Line

  • The parametric equations of a line passing through a point A and parallel to vectors u and v are given by:
    • x = x0 + tu1 + sv1
    • y = y0 + tu2 + sv2
    • z = z0 + tu3 + sv3
  • Here, A = &

##10216;x0, y0, z0&

##10217; is the point and u = <span