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The vector product of vectors A⃗ and B⃗ has magnitude 12.0 m2 and is in the +z-direction.Vector A⃗ has magnitude 4.0 m and is in the −x-direction. Vector B⃗ has no x-component.Part A: What is the magnitude of vector B⃗ ? (I solved this question the answer is 3 and it's correct)Part B: What is the direction angle θ of vector B⃗ measured from the +y-direction to the +z-direction? (This is the part that I didn't get it correct)

Answer :

Answer:

θ=180°

Explanation:

The problem says that the vector product of A and B is in the +z-direction, and that the vector A is in the -x-direction. Since vector B has no x-component, and is perpendicular to the z-axis (as A and B are both perpendicular to their vector product), vector B has to be in the y-axis.

Using the right hand rule for vector product, we can test the two possible cases:

  • If vector B is in the +y-axis, the product AxB should be in the -z-axis. Since it is in the +z-axis, this is not correct.

  • If vector B is in the -y-axis, the product AxB should be in the +z-axis. This is the correct option.

Now, the problem says that the angle θ is measured from the +y-direction to the +z-direction. This means that the -y-direction has an angle of 180° (half turn).

The vector combination of two vectors is orthogonal to the two vectors; in this case, its product is now in the +z-direction, so vectors A and B are in the x-y plane.

  • Now, since vector A is in the -x -direction but vector B has no x-component, vector B is also parallel towards the x-axis.
  • The product has always been in the +z-direction, vector A has been in the -x -direction, and vector B has been parallel to both the product and vector A.
  • As per the right-hand thumb rule vector B would be in either the +y or -y-direction. B would now be in the -y-direction.
  • Since AB is the vector product of [tex](-x)\times (-y) =(+z)[/tex]. Since vector B is in the -y-direction, [tex]\angle \theta[/tex] [tex]= 180^{\circ}[/tex].

Therefore, the final answer is "[tex]180^{\circ}[/tex]".

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