Given two vectors, a 3i+4)-5k and b-2j+ 6k a) find the magnitude of each vector. b) find the angle of each vector to the +x axis. c) find a b d) find the included angle between a and b. e) find a -b. (leave in component form.)

Explanation:

One of the easiest ways to work with vectors is to use their components

a) To find the magnitude let's use the Pythagorean theorem

    A² = Ax² + Ay² + Az²

    A = √ 3² + 4² + (-5)² = √ 50

    A = 7.07

    B = √ Bx² + By² + Bz²

     B = √ (-2)² + 0 + 6² = √ 40

     B = 6.32

b) to find these angles the most practical use the concept of cosine directors with the formulas

Vector A

      cos α = X / A

       cos β = y / A

       cos γ = z / A

       cos α = 3 / 7.07

      α = cos⁻¹ 0.424

      α = 64.9º

      cos β = 4 / 7.07

      β = cos⁻¹ 0.5658

      β = 55.5º

      Cos γ = -5 / 7.07

      γ=Cos⁻¹ (-0.7079)

      γ= 135º

Vector B

    cos α = X / B

    cos β = y / B

    cos γ = z / B

    cos α = 0

    α = 90º

    cos β = -2 / 6.32

    β = 108.4º

    cos γ = 6 / 6.32

    γ = 18.3º

c) find A + B

     R = A + B = (3 + 0) i ^ + (4-2) j ^ + (-5 +6) k ^

     R = 3 i ^ + 2j ^ + 1 k ^

d) to find the angles we use the scalar product

     cos θ = A.B / | A | | B |

     A.B = 0 i ^ -8 j ^ -30 k ^

     Cos θ = [R (8 2 + 30 2)] / 7.07 6.32

     Cos θ = 31.05 / 44.68

     θ = 46º

e) find A-B

   R = A-B = (3-0) i ^ + (4- (2)) j ^ + (-5 - 6) k ^

   R = 3 i ^ + 6j ^ -11 k ^

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