# Unit Vector

1. Oct 13, 2015

### jeff12

1. The problem statement, all variables and given/known dat

A car travels 20 mi at 60 degrees north of west, then 35 mi at 45 degrees north of east.

Express each displacement vector in unit vector notation. Take the +x-axis due east and the +y-axis due north. Use the component method to obtain the resultant displacement vector in unit vector notation.

2. Relevant equations
$$\ A = \ A_x i + \ A_y j$$
$$\ B = \ B_x i + \ B_y j$$
$$\ C = \ C_x i + \ C_y j$$

3. The attempt at a solution

A= -20sin60i+35cos60j
B=35cos45i+20sin45
C=?

How do I find C, since no lengths were given. Did I do A and B correctly?

Last edited: Oct 13, 2015
2. Oct 13, 2015

### Staff: Mentor

The solutions look incorrect so far. For A, both the x and y components involve the 20 mile displacement, and for B, both the x and y components involve the 35 mile displacement. Also, can you show a sketch that shows how you determine when to use the cos() and when to use the sin() to find the components on the x & y axes?

Finally, you are indeed given the lengths of 20 and 35 miles, so you will be able to calculate the sum vector C = A + B.

3. Oct 13, 2015

### Staff: Mentor

Also, you can't reduce the vector components down to a single scalar number "7.5". Vectors have a magnitude and a displacement, and can be expressed as that two number combination, or in rectangular component form. Are you familiar with how to convert between those two forms?

4. Oct 13, 2015

### jeff12

Oh yes you are right. But are the sin and cos right?

I do not know to convert to rectangular component form. I think we are suppose to put it in unit vector notation.

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5. Oct 13, 2015

### Staff: Mentor

6. Oct 18, 2015

### jeff12

Okay I redid everything and found:

A=20cos60i+20sin60j
B=35cos45i+35sin45j
C=(-20cos60+35cos45)i+(20sin60+35sin45)j

Did I do all the trig stuff right? I am suppose to find the magnitude too which I found was 97.62.

7. Oct 19, 2015

### jeff12

Bump.

8. Oct 20, 2015

### Staff: Mentor

I believe there is a sign error in this equation...

9. Oct 20, 2015

### Mister T

Would it make any difference in your solution if the direction of $\mathrm{\vec{A}}$ were instead 60° north of east?

Last edited: Oct 20, 2015
10. Oct 20, 2015

### jeff12

Oh I fixed it. I think I understand everything now. But I want to make sure, C is the resultant right? Because C is technically R.

Yes because that would change the problem entirely.

11. Oct 20, 2015

### Staff: Mentor

Could you show your work for the final answer for C? Thanks, that will help us answer...

12. Oct 20, 2015

### Staff: Mentor

He's just seconding my comment about the sign error...

13. Oct 20, 2015

### Mister T

The problem statement doesn't specify. It was you who chose $\mathrm{\vec{A}}$ to represent one of the two given vectors and $\mathrm{\vec{B}}$ the other. Likewise you are free to choose $\mathrm{\vec{C}}$ or $\mathrm{\vec{R}}$ as their sum.

If $\mathrm{\vec{C}}$ is your choice, then you'd write $\mathrm{\vec{A}+\vec{B}=\vec C}$.

If $\mathrm{\vec{R}}$ is your choice, then you'd write $\mathrm{\vec{A}+\vec{B}=\vec R}$.

The term "resultant" has the same meaning as "vector sum".

14. Oct 20, 2015

### Mister T

No, you are not supposed to find the magnitude. You are supposed to express the result in unit vector notation.

15. Oct 20, 2015

### jeff12

There was no C. I figured that out. I was confused because in his example he kept saying R=A+B+C. But since I was only given two vectors it would be R=A+B. I found R=14.75i+42.07j. The magnitude is 44.58 mi at $70.68^o$.