Find the magnitude and direction of vectorA - vector B?

  • Thread starter bigcookie
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  • #1
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Homework Statement


Vector A⃗ has a magnitude of 20m and makes an angle of 30∘ above the positive x axis. Vector B⃗ has a magnitude of15m and is oriented 60∘ to the left of the y axis.
Find the magnitude of A⃗ −B⃗ .
Find the direction of A⃗ −B⃗ .

Homework Equations


sin 150 = sin 30 = 0.5
cos 150= - cos 30
Angle of direction = tan-1 (y/x)

The Attempt at a Solution


so far
Ax [/B]= 20 cos 30 =17.3 m
Ay = 20 sin 30 = 10 m

Bx = 15cos 150 = -12.99 m
By=15 sin 150 = 7.5 m
add Ax+Bx= 4.31
Ay+By= 17.5
sqrt of [(4.31)^2+(17.5)^2]= 18.0
but i dont know where to go from here ?
 
Last edited:

Answers and Replies

  • #2
Buzz Bloom
Gold Member
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Vector A⃗ has a magnitude of 20m and makes an angle of 30∘ above the positive x axis. Vector B⃗ has a magnitude of15m and is oriented 60∘ to the left of the y axis.
Hi bigcookie:

This problem statement does not state a problem. What is the question whose answer is sought?

Regards,
Buzz
 
  • #3
Buzz Bloom
Gold Member
2,365
433
add Ax+Bx= 4.31
Ay+By= 17.5
Hi @bigcookie:
I see that you edited the problem statement . I apologize for not checking on the thread more promptly.

I think you have calculated the x y coordinates for A and B correctly, but I don't understand why you ADD the coordinates. The problem statement is interested in A-B.

Regards,
Buzz
 

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