Respuesta :
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 ^
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:
[tex]C^{2} = Ax^{2} + By^{2} + Cz^{2} \\ = \sqrt{3^{2}+ 4^{2}+ (-5)^{2} } \\ = \sqrt{50} \\ = 7.07[/tex]
Similarly, the magnitude of B = 6.32
b) the cosine rule is used:
[tex]cos\alpha = \frac{x}{a}[/tex]
= [tex]\frac{3}{7.07}[/tex]
[tex]\alpha = cos^{-1}(0.424)\\ = 64.9[/tex]
similarly, for the second angle:
[tex]\beta = cos^{-} (\frac{4}{7.07})\\ \\ = 55.5[/tex]
The third angle:
[tex]\gamma = cos^{-1}(\frac{-5}{7})\\ = 135[/tex]
c) the vector will be 3i + 2j + k
d)the angle will be:
[tex]\theta = cos^{-}(\frac{31.05}{44.68)}[/tex]
= 46°
e) AB = 3i + 6j - 11k
