Find the magnitude and direction of vectorA - vector B?

In summary, vector A⃗ has a magnitude of 20m and is oriented 30∘ above the positive x axis, while vector B⃗ has a magnitude of 15m and is oriented 60∘ to the left of the y axis. To find the magnitude of A⃗ -B⃗ , we can use the Pythagorean theorem to find the length of the resulting vector, which is equal to 18.0m. To find the direction of A⃗ -B⃗ , we can use trigonometric functions to calculate the angle, which is equal to 150∘.
  • #1
bigcookie
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0

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 don't know where to go from here ?
 
Last edited:
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  • #2
bigcookie said:
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
bigcookie said:
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
 

1. What is the formula for finding the magnitude of vectorA - vector B?

The magnitude of vectorA - vector B can be found using the Pythagorean theorem, where the magnitude is the square root of the sum of the squares of the individual components of the vector.

2. How do you determine the direction of vectorA - vector B?

To determine the direction of vectorA - vector B, you can use the inverse tangent (arctangent) function to find the angle between the vector and the positive x-axis.

3. Can you find the magnitude and direction of vectorA - vector B using only their individual magnitudes and directions?

Yes, you can use the law of cosines to find the magnitude of vectorA - vector B and the law of sines to find the direction.

4. Are there any special cases when finding the magnitude and direction of vectorA - vector B?

Yes, if vectorA and vectorB are parallel or anti-parallel, the magnitude will be 0 and the direction will be undefined (or 180 degrees), respectively.

5. Can you find the magnitude and direction of vectorA - vector B in any coordinate system?

Yes, the formula for finding the magnitude and direction of vectorA - vector B remains the same in any coordinate system. However, the individual components of the vectors may need to be converted into the appropriate coordinate system.

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