Few Vector Addition questions, Not sure if correct

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Homework Statement



1) Two men are near a hole, One man walks 12m east and then 12m North from a hole, the other walks 15m west and then 11m north, Find the dot product of their net displacements from the tree
_______

2) Vector A = -i +2j - 5k, Vector B = 3i +2j -2k
Find the magnitude and direction of the vector difference A -B

Homework Equations



|A||B|cos(t) = A(x)B(x) + A(y)B(y) = |A||B|

The Attempt at a Solution



1 )I didn't calculate the resultant as I think you don't need it, I used the components given

A = 12i + 12j
B = -15i +11j

|A||B| = (12*-15) + (12*11) = -48.

_____________________

2) A - B = -4i -3k

|A-B| = [itex]\sqrt{-4^2 + 0^2 + -3^2}[/itex] = 5.0

arctan([itex]\frac{0}{-4}[/itex] = 0

So the magnitude is 5.0 and the angle is 0 degrees? This seems wrong to me..

If anyone can tell me if I did these right or give me a tip in the right direction, it'd be very much appreciated. Thanks in advance
 

Answers and Replies

  • #2
vela
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Homework Statement



1) Two men are near a hole, One man walks 12m east and then 12m North from a hole, the other walks 15m west and then 11m north, Find the dot product of their net displacements from the tree
_______

2) Vector A = -i +2j - 5k, Vector B = 3i +2j -2k
Find the magnitude and direction of the vector difference A -B

Homework Equations



|A||B|cos(t) = A(x)B(x) + A(y)B(y) = |A||B|

The Attempt at a Solution



1 )I didn't calculate the resultant as I think you don't need it, I used the components given

A = 12i + 12j
B = -15i +11j

|A||B| = (12*-15) + (12*11) = -48.
Don't forget the units.
2) A - B = -4i -3k

|A-B| = [itex]\sqrt{-4^2 + 0^2 + -3^2}[/itex] = 5.0

arctan([itex]\frac{0}{-4}[/itex] = 0

So the magnitude is 5.0 and the angle is 0 degrees? This seems wrong to me..
The formula you used for the angle is for a vector that lies in the xy-plane. You actually have a three-dimensional problem here, so you'll need to think a bit more about how to describe the direction of the vector.
 
  • #3
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Thanks a lot for your reply, For the second one would it actually be arctan z/x? Since there's no y coordinate the vector lies in the xz plane, that would make it arctan z/x?
 
  • #4
vela
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Yes, that'll work. Just make sure you describe exactly what angle you're calculating.
 
  • #5
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Ok, I've got 36.9 degrees south of west, I think that seems right seeing as this vector is in the 3rd quadrant of the xz axis, from the +x it'd be 216.9 degrees. Have I done anything wrong?
 
  • #6
vela
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Sounds good.
 
  • #7
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Alright vela, Thanks a ton for your help, I really appreciate it.
 

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