# Math Help - Magnitude and direction of a vector

1. ## Magnitude and direction of a vector

Hi,

I didn't know exactly where to put this post, but I see in the posts here that there are a lot of things similar to what I am learning.

The problem I have is:
The resultant of P and Q (both vectors_ is a force F (vector). F is 80N [N70degreesE] and P is 25N[E]. Find the magnitude and direction of Q.

My answer is: 53.5 N [N59.1degreesW].

Is this correct? I have solutions from my teacher; her process work is the same as mine but her values are different, so I just want to know if my answer above is correct. Thanks!

2. Originally Posted by bhuang
Hi,

I didn't know exactly where to put this post, but I see in the posts here that there are a lot of things similar to what I am learning.

The problem I have is:
The resultant of P and Q (both vectors_ is a force F (vector). F is 80N [N70degreesE] and P is 25N[E]. Find the magnitude and direction of Q.

My answer is: 53.5 N [N59.1degreesW].

Is this correct? I have solutions from my teacher; her process work is the same as mine but her values are different, so I just want to know if my answer above is correct. Thanks!
There are two ways to do this- trigonometrically and by components.

Trigonometrically: draw a picture and you see a triangle. One side, corresponding to F, has length 80 (I assume these are force vectors and the "N" is just "Newtons") making an angle of 70 degrees with the vertical (north). Another is P which has length 25 and is directly "east", at 90 degrees to the vertical. That means that the angle between F and P is 90- 70= 30 degrees. Q is the third side of that triangle and we can use the
"cosine law" to find that length. The "cosine law" says that if we label the sides of any triangle and a, b, and c and use A, B, and C for the angles opposite sides a, b, and c, respectively, then $c^2= a^2+ b^2- 2ab cos(C)$. Here, that says $|Q|^2= 80^2+ 25^2- 2(80)(25)cos(30)= 6400+ 625- 4000(.8660)= 3561$ so that the magnitude of Q is [tex]\sqrt{3561}= 59.7, not quite what you got. Now that you know the lengths of all three sides of the triangle, you can use the sine law, $\frac{sin(A)}{a}= \frac{sin(B}}{b}= \frac{sin(C)}{c}$, to determine the angle Q makes $\frac{sin(30)}{59.7}= \frac{sin(C}}{80}$, where C is the angle at the junction of P and Q. $sin(C)= 80\frac{sin 30}{59.7}= 0.6700$ so C= 42 degrees East of North.

By components: resolve each vector into x and y (East and North) components. F has length 80 at 70 degrees to the vertical so its x component is 80 sin(70)= 75.17 and its y component is 80 cos(70)= 27.36.

P has length 25 and points directly east so its x component is 25 and its y component is 0.

P+ Q= F so Q= F- P= (75.17, 27.36)- (25, 0)= (50.17, 27.36).

That is, Q has x component 50.17 and y component 27.36. Its magnitude is [tex]\sqrt{(50.17)^2+ (27.36)^2}= 57.1, not quite what we got before, probably because of round off error. The angle is given by [tex]tan^{-1}(27.36}{50.17}= 28.6 degrees, again slightly different from what we got before, again probably due to round off error.