The vector magnitude is 10.29 lb and the vector direction is 150.9⁰.
Vector:
As per the given question:
[tex]Ay = 5.00 \ lb\\\\Ax = 9.00 \ lb\\\\[/tex]
To find:
magnitude and direction=?
Solution:
Vector representation of the given value:
[tex]\to (9i + 5j)[/tex]
Calculating the vector magnitude:
[tex]\to \sqrt{(9)^2 +(5)^2}\\\\ \to \sqrt{81+25}\\\\ \to \sqrt{106} \\\\\to 10.29 \ lb[/tex]
Calculating the vector direction:
When [tex]90 < \theta < 180[/tex]
Let's solve the value for [tex]\theta\\\\[/tex]
[tex]\tan \theta = \frac{y}{x}\\\\\tan \theta = \frac{5}{9}\\\\\tan \theta =0.555\\\\\theta = \tan^{-1}(0.555555556) \\\\\theta= 29.1^{\circ}\\\\[/tex]
[tex]\sin \theta[/tex] lies in the second quadrant,
[tex]\to \theta = 29.1^{\circ} = 180 -29.1 = 150.9^{\circ}\\\\[/tex]
Find out more information about the angle here:
brainly.com/question/24797859
