Posted by **Anonymous** on Sunday, February 10, 2008 at 1:32am.

Can you please help me correct my answers for the following two questions?

1) A tour boat travels 25 km due east and then 15 km S50°E. Represent these displacements in a vector diagram, then calculate the resultant displacement.

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My Work:

I drew the vectors and connected them head to tail to form a triangle.

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Then, I found the resultant displacement:

|r|² = 25² + 15² - 2(15)(25)cos130

√|r|² = √1332.1

|r| = 36.5 km

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Now, I'm having some problems finding the direction of the resultant displacement:

(sin50/36.5) = (sinC/25)

31.6° = C

So, I get 36.5 km S31.6°E

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The textbook answer is 36.5 km S54°E

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2) Vectors a and b have magnitudes 2 and 3, respectively. If the angle between them is 50°, find the vectors "5a - 2b", and state its magnitude and direction.

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Here's my work for this question:

I drew the vectors and connected them head to tail.

-----

I tried to find the magnitude by:

5a - 2b

= 5(2a) - 2(3b)

= 10a - 6b

|r|² = 10² + 6² - 2(10)(6)cos130

√|r|² = √213

|r| = 14.6

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Then I tried to find the direction by:

(sinx/3) = (sin130/14.6)

x = 9.1°

So, I get 14.6, 9.1° to vector a

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Textbook answer is 7.7, 37° to vector a

- Discrete Math - Vectors -
**Reiny**, Sunday, February 10, 2008 at 9:06am
I will do the second question first, since I have a question about your interpretation of S50E from the first.

I drew the 2 unit vector to run east, then the 3 unit angle downwards to form the 50 degree angle.

so when you construct 5a - 2b, you would draw a horizontal line 10 units long for the first part, then you must go into the opposite direction of the second vector for 6 units

so the magnitude equation would be

|r|² = 10² + 6² - 2(10)(6)cos50

and r = 7.67

now if x is the angle between the resultant and the first vector

sinx/6 = sin50/7.67

I got x = 36.8 degrees

back to your first problem,

I was always under the impression and taught that a direction like your

S50°E meant:

face south then 50 degrees towards the east, so I thought that the end part of your equation should have been ....cos140

you had....cos130

but the ....cos130 produced the answer supplied by your text, so I am confused.

I am using the Canadian interpretation of S50°E, is it different where you are???

Perhaps some of the other math or physics expert could help out here

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