Solve+Vector+Word+Problems+Using+Both+Methods

Solve Vector Word Problems Using Both Methods
Intro : I choose this topic, because it was a topic I struggle with at first, but after I practiced it I began to understand it more fully. Due to the fact that I struggled to understand this topic, I feel that my explaination for another person to learn this will be more step by step than something I understood right away, even if I may not be trying to do this on purpose. Hopefully, my explaination will help those who read this understand vector word problems more clearly, therefore, succeeding in this section in mathematics.

Description: To fully understand this topic one must use their knowledge of vectors, and then apply them to the real world through a word problem.

To first understand how to do these word problems, we will look at an example from a vectors test, number 3. The problem stated:

"A ship sails for 20 miles on a bearing of 325 degrees then turns and sails on a bearing of 250 degrees for 7 more mile. What is the distance and bearing of the ship from its starting point?"

Sometimes drawing a picture can be a simple, yet effective way to understand a problem. Likewise, it is the case with vector problems. It defiantly helps to a certain "degree". First, you must draw a dotted line, then a line coming 325 degrees off of this line, since the ship is sailing at a bearing of 325 degrees. Label this line 20 miles, since the ship is on this course for 20 miles. After this, draw a line 250 degrees off of the line labeled 20 miles, and label this line 7 miles. This is, because the ship turns to a bearing of 250 degrees for 7 more miles. The picture is shown below:



This picture can be the key to unlocking the answer to this problem. Our first method to solving this problem is the component method. The equation for the component method is the following:

(Acos(a))+(Bcos(b))=X

Asin(a))+(Bsin(b))=Y

x^2+y^2=C

This equation is justified through the following picture:



As one can see, in this hypothetical example, the component method allows one to use the properties of right triangles and trig functions we already know, to find the distance of both X and Y. Therefore, the Pythagorean Theorem allows us to find the length of line C, which is the distance in this problem.

The same concept applies to this vectors word problem.

In this case, the following is true for the variables in this problem:

A=20 miles B=7 miles a=325 degrees b=250 degrees

Therefore, all you need to do is plug in the variables for this problem:

(20cos(325))+(7cos(250))= 13.98889988 (20sin(125))+(7sin(250))= -18.04937707

-18.04937707^2+13.98889988^2=521.4693326

square root of 521.4693326= 22.83570302 miles, the distance of the ship from its starting point.

As one can see, all that is need, once you grasp understanding of the equations, is to plug in the variables and, from there, solve the equation.

To find the bearing the ship is from its starting point is quite simple. For our first method, we will use the inverse of tangent. All one needs to do is plug in the variables X and Y, derived from the previous equation, and do the following:

arctan(Y/X)=bearing

This is, because we can apply opposite/adjacent to the right triangle we created from the previous formula.

In this case, the following would be plugged in for the variables to get the bearing the ship was from its starting point:

arctan(-18.04937707/13.98889988)=-52.22306103

Since this is a negative answer, you can get a positive angle by subtracting 52.22306103 from 360:

360-52.22306103=307.776939 degrees

Next, for the purpose of the real-world situation in the problem, convert 307.776939 degrees into a bearing. To do this, one must only cross multiply:

77/100=x/60 100x=4620 x=46.2

Therefore, the bearing the ship is from its original starting point is 307 degrees 46'.

Next, we'll continue with another way to solve this problem, which I hope will "point" you in the "right direction". For this method, one can simply use the Law of Cosines, which, as you may know is the following:

A^2+B^2-2(A)(B)cos(c)=C^2

For this problem, all one needs to do is plug in the variables for the Law of Cosines equation. In this case, it is:

A=20 B=7 c=250 degrees

Therefore, the equation is:

20^2+7^2-2(20)(7)cos(250)=544.7656401

the square root of 544.7656401 = 22.83520245 miles

Obviously, this is the same distance we got for the component method. Thus, both ways are equally effective when used correctly.

Next, another way to find the bearing of the boat from its original point is to use the Law of Sines, which, as you may know is:

sina/A=sinb/B=sinc/C

To use this method, one merely plugs in the variables again:

sin(120)/22.83570245=sinx/27 (27=20+7)

17.32050808=22.83571245sinx

x=-52.22306183 degrees

From there, one only needs to do what we did in the pervious method once we got to -52.22306183 degrees.

All in all, once fully explained, it is quite simple to do vector word problems and solve them both ways.

Now, to test your knowledge, here is another vector word problem:

A plane travels 70 miles at a bearing of 85 degrees. Then, it changes course and travels 50 miles at a bearing of 60 degrees. What is the distance and degrees of the plane from its starting point?

Once again, you should try to draw a picture, which should look something like the following:



Next, you can use the component method. In this case, the equation would be:

(70cos(85))+(50cos(60))=31.10090199 (70sin(85))+(50sin60))=113.0348991

31.10090199^2+113.0348991^2=13,744.15451

the square root of 13,744.15451= 117.2354661 miles away from its original point.

Next the bearing would be done as such:

arctan(113.0348991/31.10090199)= 74.62461819 degrees from its starting point.

Furthermore, the other method using the Law of Cosines would look like this:

70^2+50^2-2(70)(50)cos(60)=13,744.15451 the square root of 13,744.15451=117.2354661 miles away from its original point.

Using the Law of Sines, it would look like this:

sin60/117.2354661=sinx/130

x=74.62461819 degrees

Here is video explaination of this concept:

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Another example problem you can try on your own is:

Batman is riding in the batmobile and traveled 30 miles at a 40 degree bearing, then he turns and travels 10 miles at a 97 degrees bearing. What is the distance and degree of batman from his starting point?