Question

How do you find the magnitude and direction angle of the vector `v=-5i+4j`?

Answer 1

`|vec v|=sqrt(41)*unit`
`theta = pi/2+tan^-1(5/4)`
`~~2.46685 rad~~141.34^@~~141^@ 20' 24.69''`

Explanation:

`|vec v|=sqrt(x^2+y^2)=sqrt((-5)^2+4^2)=sqrt(25+16)=sqrt(41)*unit`

Now it is clear that the head of the vector lies in the second quadrant, and we can use a little trigonometry to find out the angle.

`theta = 90^@+tan^-1(5/4)`