Vector A has a magnitude 2.4m and direction 130 degrees relative to the positive x-axis. Vector B has magnitude 3.0m and direction 250 degrees relative to the positive x-axis.

Question

1) The resultant R=A+B has magnitude(in m) ?

I know the R =2.75m but I am not sure how to get this. I know you have to use sqrt of rx^2 + ry^2 but I am not sure how to get Ry.

Answer 1

You can simply compute the squared norm of R as follows. Writing the squared norm of a vector X as X^2 (which is also the inner product of X with itself, so this notation makes sense):

R^2 = (A + B)^2 =

A^2 + B^2 + 2 A dot B.

A dot B = |A| |B| cos(angle between A and B)

The angle is 250° - 130° = 120°

Cos(120°) = Cos(90° + 30°) = -Sin(30°) = -1/2

So:

|R| = sqrt[A^2 + B^2 - |A||B|] =

sqrt[7.56] = 2.74

Answer 2

why is sin negative?

Answer 3

You can use the identity:

Sin(x)= Cos(90° - x)

So, if you have Cos(90° + x), you can replace x by -x in the above identity:

Cos(90° + x) = Sin(-x) = -Sin(x)

Note that Sin is an odd function so it changes sign when the sign of x is changed:

Sin(-x) = -Sin(x)

Cos is an even function:

Cos(-x) = Cos(x)