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