MHB What is the method for solving vector problems with component form and addition?

[Coder74]

Hi everyone, I am taking my finals next week and I am really excited. But I forgot how to solve problems like these I know I'm supposed to separate them and solve for x and y seperately but its still not ringing any bells.

Vector L is 303m long in a 205* direction. Vector M is 5552 m long in a 105* direction.
Find the magnitude of their vector sumx=303 + 5552

y= 205+105

^ Am I on the right track so far?
Thanks so much for the help!

 

[MarkFL]

Here are two ways I know of to find the magnitude of the vector sum. The first is by vector addition, so let's write:

$$\vec{L}=303\left\langle \cos\left(205^{\circ}\right),\sin\left(205^{\circ}\right) \right\rangle$$

$$\vec{M}=5552\left\langle \cos\left(105^{\circ}\right),\sin\left(105^{\circ}\right) \right\rangle$$

And so the vector sum $\vec{S}$ is:

$$\vec{S}=\left\langle 303\cos\left(205^{\circ}\right)+5552\cos\left(105^{\circ}\right),303\sin\left(205^{\circ}\right)+5552\sin\left(105^{\circ}\right) \right\rangle$$

And thus, the magnitude of the sum (in m) is:

$$\left|\vec{S}\right|=\sqrt{\left(303\cos\left(205^{\circ}\right)+5552\cos\left(105^{\circ}\right)\right)^2+\left(303\sin\left(205^{\circ}\right)+5552\sin\left(105^{\circ}\right)\right)^2}\approx5507.47415994496$$

The other approach is to draw a diagram and then use the Law of Cosines like this:

$$\left|\vec{S}\right|=\sqrt{303^2+5552^2-2\cdot303\cdot5552\cos\left(80^{\circ}\right)}\approx5507.47415994496$$

 

[Coder74]

Thanks, Mark!

If I were to put this on a graph(triangle) how would I know which number goes on the corresponding side? How did you figure out that we needed to use cos to solve the answer?
I really appreciate the help! :D

 

[MarkFL]

Thanks, Mark!

If I were to put this on a graph(triangle) how would I know which number goes on the corresponding side?

Each vector has a magnitude and a direction...you have to match those up. :D

How did you figure out that we needed to use cos to solve the answer?
I really appreciate the help! :D

When you know two sides of a triangle and the angle they subtend, then the Law of Cosines will give you the length of the unknown side. :D

 

[Coder74]

Thank you Mark,

I took my finals today and as it turns out since there were new students coming in for the late semester, they took off most of the material from the beginning of the year.. So this wasn't on the test.. But I'm still confused. Earlier when you were solving like this

And so the vector sum S⃗ S→ is:

=⟨303cos(205∘)+5552cos(105∘),303sin(205∘)+5552sin(105∘)⟩S→=⟨303cos⁡(205∘)+5552cos⁡(105∘),303sin⁡(205∘)+5552sin⁡(105∘)⟩

And thus, the magnitude of the sum (in m) is:

S =(303cos(205∘)+5552cos(105∘))2+(303sin(205∘)+5552sin(105∘))2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√≈5507.47415994496|S→|=(303cos⁡(205∘)+5552cos⁡(105∘))2+(303sin⁡(205∘)+5552sin⁡(105∘))2≈5507.47415994496I think I understand why you used cos, but I thought only 1 of the trig functions were applicable per equation. Why are you using sin and how come there are two of the same values in the same line? (I apologize if my questions seem repetitive.. I really want to grasp this to perfection so I could know this for the future and help anyone else having trouble with it) I really appreciate you taking the time to help me. It means a lot

 

[MarkFL]

If we know the magnitude $M$ of a vector $\vec{v}$, and the direction $\theta$ in which it points, then we may express the vector in component form as:

$$\vec{v}=M\left\langle \cos(\theta),\sin(\theta) \right\rangle$$

And when adding vectors, we can sum the components to get the resultant vector:

$$\vec{v_1}+\vec{v_2}=M_1\left\langle \cos(\theta_1),\sin(\theta_1) \right\rangle+M_2\left\langle \cos(\theta_2),\sin(\theta_2) \right\rangle=\left\langle M_1\cos(\theta_1)+M_2\cos(\theta_2),M_1\sin(\theta_1)+M_2\sin(\theta_2) \right\rangle$$