Answer:
The magnitude = 10.30 m
The direction of the vector proceeds at angle of 119.05°
Explanation:
Given that:
A vector [tex]\bar A[/tex] has component [tex]A_x[/tex] = -5 m and [tex]A_y[/tex] = 9 m
The magnitude of vector [tex]\bar A[/tex] can be represented as:
[tex]\bar A[/tex] = [tex]\sqrt{A_x^2 + A_y^2}[/tex]
[tex]\bar A[/tex] = [tex]\sqrt{(-5)^2 + (9)^2}[/tex]
[tex]\bar A[/tex] = [tex]\sqrt{25 + 81}[/tex]
[tex]\bar A[/tex] = [tex]\sqrt{106}[/tex]
[tex]\bar A[/tex] = 10.30 m
If we make [tex]\bar A[/tex] an angle [tex]\theta[/tex] with y- axis:
Then; tan [tex]\theta[/tex] = [tex]\frac{A_x}{A_y}[/tex]
tan [tex]\theta[/tex] = [tex]\frac{5}{9}[/tex]
tan [tex]\theta[/tex] = 0.555
[tex]\theta[/tex] = tan⁻¹ (0.555)
[tex]\theta[/tex] = 29.05°
Angle with positive x-axis = 90 + [tex]\theta[/tex]
= 90° + 29.05°
= 119.05°
