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

a) 6.32

b) 69.4° ; 55.5°; 135°

c) 3i + 2j + k

d) 46°

e)3i + 6j - 11k

Step-by-step 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) } \\ = √(50) \\ = 7.07`

Similarly, the magnitude of B = 6.32

b) the cosine rule is used:

`cos\alpha = (x)/(a)`

=
`(3)/(7.07)`

`\alpha = cos^(-1)(0.424)\\ = 64.9`

similarly, for the second angle:

`\beta = cos^(-) ((4)/(7.07))\\ \\ = 55.5`

The third angle:

`\gamma = cos^(-1)((-5)/(7))\\ = 135`

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

d)the angle will be:

`\theta = cos^(-)((31.05)/(44.68))`

= 46°

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

Answer 2

Step-by-step 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 ^