Vector Exercises: Solve for a & b

[Electro]

Hello Everyone,
In school I had a test of Vector exercises. The problem is that I have to submit them online and I need the exact value for the questions. Anyway I don't want you to solve them for me. I just need some help where I am stuck.

The first exercise: A boy runs 3.3 m North, 7.1 m NorthEast and 15.6 m West. a)Determine the length of the displacement vector tha goes from the starting point to his final position. b)Determine the direction of the displacement vector.

I was confused by this exercise because there isn't any angle given. So I assumed that "NorthEast" means that he moves 7.1 m with an angle 45 Deg.
Then part a) is easy. For part b) I found the angle formed by the North line and the Resultant vector. Am I right here?

The next exercise is a little fuzzy I think: Let aA + bB + C=0, where
A=(17,-99), B=(-99,28) and C=(4.3,78), where Ax=17 units, Ay=-99 units etc. What is the value of a and b? (All the Upper case letters A,B and C have the vector arrow above).
What I did is: I opened aA + bB + C=0 using unit vectors i, j and k (k=0) and the numbers given above.
Still I have a and b and it's an equation with 2 unknowns.
The final expression I concluded is: (a*17i-b*99i+4.3i)+(b*28j-a*99j+78j)=0...?
What I think now, is forming two simultaneous equations.
1) 17a-99b= -4.3
2) 28b-99a= -78
Solving them, I think is the answer for a and b.
Since the i units can't be added to the j units I formed two equations just neglecting i and j (the brackets give 0). So for i group it will give 0 and for j group too.

Please I need an answer whether I am right or not.

Thank you
Electro

 

[jcsd]

Hello Everyone,
In school I had a test of Vector exercises. The problem is that I have to submit them online and I need the exact value for the questions. Anyway I don't want you to solve them for me. I just need some help where I am stuck.

The first exercise: A boy runs 3.3 m North, 7.1 m NorthEast and 15.6 m West. a)Determine the length of the displacement vector tha goes from the starting point to his final position. b)Determine the direction of the displacement vector.

I was confused by this exercise because there isn't any angle given. So I assumed that "NorthEast" means that he moves 7.1 m with an angle 45 Deg.
Then part a) is easy. For part b) I found the angle formed by the North line and the Resultant vector. Am I right here?

Yes north-east means 45 degrees, but you should find the angle formed with the East-West axis.

The next exercise is a little fuzzy I think: Let aA + bB + C=0, where
A=(17,-99), B=(-99,28) and C=(4.3,78), where Ax=17 units, Ay=-99 units etc. What is the value of a and b? (All the Upper case letters A,B and C have the vector arrow above).
What I did is: I opened aA + bB + C=0 using unit vectors i, j and k (k=0) and the numbers given above.
Still I have a and b and it's an equation with 2 unknowns.
The final expression I concluded is: (a*17i-b*99i+4.3i)+(b*28j-a*99j+78j)=0...?
What I think now, is forming two simultaneous equations.
1) 17a-99b= -4.3
2) 28b-99a= -78
Solving them, I think is the answer for a and b.
Since the i units can't be added to the j units I formed two equations just neglecting i and j (the brackets give 0). So for i group it will give 0 and for j group too.

Please I need an answer whether I am right or not.

Thank you
Electro

You have: (17a - 99b + 4.3)i + (28b - 99a + 78)j = 0 = 0i + 0j

which is correct.

 

[Electro]

Thanks a lot Jcdsd,
I still have a question which I couldn't understand from the book and the professor. Why should we take the angle of the displacement from the East West direction and not from the North line. So I found that the angle is about 53 degrees with the North, but taking it with respect to WE line, it would be 53+90=143 degrees. I attached a simple picture to demonstrate what I think. But an advice will be really helpful.

ThankYou

 

[jcsd]

It's just because the East line is analagous to the postive x-axis, which by convention we usually measure the angles of our vectors from. Of course as long as you are clear in what you are doing it doesn't really matter which angle you take.

 

[Electro]

:rofl: Thank God I have the right concepts on my mind. The problem Jcsd is that I have to submit the answer online and it is not the same as submitting the answer as 53 deg. and 143 deg. This is my dilemma. Anyway I will chose the one with respect to the West East line because it says "Determine the direction of the displacement vector". So it doesn't specify the North line.
Thanks Again