what are the magnitude and direction of u+v+w ?

The magnitude and the direction of the resultant are approximately 57.871 and 198.676°.
Vectors are elements with two given characteristics: Magnitude and direction. A resultant is derived from the sum of vectors, the magnitude is the norm of a vector and the direction is the orientation of a vector. The resultant and its characteristics are described below:
[tex]\vec r = \left(\sum\limits_{i=1}^{n}x_{i}, \sum\limits_{i=1}^{n}y_{i}\right)[/tex] (1)
[tex]\|\vec r\| = \sqrt{\left(\sum\limits_{i=1}^{n} x_{i}\right)^{2}+\left(\sum\limits_{i=1}^{n} y_{i}\right)^{2}}[/tex] (2)
[tex]\theta = \tan^{-1}\frac{\sum\limits_{i=1}^{n}y_{i}}{\sum\limits_{i=1}^{n}x_{i}}[/tex] (3)
Where [tex]\theta[/tex] is resultant angle, measured in degrees.
If we know that [tex]\|\vec u\| = 90[/tex], [tex]\theta_{u} = 40^{\circ}[/tex], [tex]\|\vec v\| = 60[/tex], [tex]\theta_{v} = 20^{\circ}[/tex], [tex]\|\vec w\| = 40[/tex] and [tex]\theta_{w} = 30^{\circ}[/tex], then the resultant, its magnitude and its direction:
[tex]\vec r = (-90\cdot \cos 40^{\circ}-60\cdot \sin 20^{\circ}+40\cdot \cos 30^{\circ}, 40\cdot \sin 30^{\circ}+60\cdot \cos 20^{\circ}-90\cdot \sin 40^{\circ})[/tex]
[tex]\vec r = (-54.824, 18.531)[/tex]
[tex]\|\vec r\| = \sqrt{(-54.824)^{2}+18.531^{2}}[/tex]
[tex]\|\vec r\| \approx 57.871[/tex]
[tex]\theta = \tan^{-1} \left(\frac{18.531}{-54.824}\right)[/tex]
[tex]\theta \approx 198.676^{\circ}[/tex]
The magnitude and the direction of the resultant are approximately 57.871 and 198.676°. [tex]\blacksquare[/tex]
To learn more on vectors, we kindly invite to check this verified question: https://brainly.com/question/13322477