Question

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.)

Answer 1

Answer:

a) 6.32

b) 69.4° ; 55.5°;  135°

c) 3i + 2j + k

d) 46°

e)3i + 6j - 11k

Explanation:

a) The magnitude is determined by the Pythagoras' theorem:

$C^{2} = Ax^{2} + By^{2} + Cz^{2} \\ = \sqrt{3^{2}+ 4^{2}+ (-5)^{2} } \\ = \sqrt{50} \\ = 7.07$

Similarly, the magnitude of B = 6.32

b) the cosine rule is used:

$cos\alpha = \frac{x}{a}$

       = $\frac{3}{7.07}$

$\alpha = cos^{-1}(0.424)\\ = 64.9$

similarly, for the second angle:

$\beta = cos^{-} (\frac{4}{7.07})\\ \\ = 55.5$

The third angle:

$\gamma = cos^{-1}(\frac{-5}{7})\\ = 135$

c) the vector will be 3i + 2j + k

d)the angle will be:

$\theta = cos^{-}(\frac{31.05}{44.68)}$

  = 46°

e) AB = 3i + 6j - 11k

Answer 2

Answer:

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 ^