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Homework Help: 3-D vector into its component form

  1. Oct 13, 2012 #1
    **refer to attached picture**
    I'm having some trouble understanding what is happening here. I'm trying to write this vector into its component form so I can solve an equilibrium problem. But I don't understand how the book is splitting this vector. I only drew the vector that I am having trouble with so this isn't the entire problem.

    1. The problem statement, all variables and given/known data
    Write the vector in cartesian vector form i.e split it up into its components. i,j,k are unit vectors in the direction x,y,z respectively. F is the force.

    2. Relevant equations

    pythagoras theorem

    3. The attempt at a solution

    To find the x & y components you consider the projected vector which lies on the x-y plane. x would be 3/5F and y would be -4/5F. z can be found by looking at the original vector which comes out to be 4/5F.
    So F = 3/5F i - 4/5F j + 4/5F k
    But that is not what the book says, the way it's written in the book is as follows
    F = 3/5(3/5F) i - 4/5(3/5F) j + 4/5F
    I don't understand why the book is doing this.

    Attached Files:

  2. jcsd
  3. Oct 13, 2012 #2
    There is a small triangle in the xy plane.

    There is a big triangle with one leg parallel to the z axis.

    You've marked a 345 triangle in the xy plane, the hypotenuse of this triangle also being the short leg of the larger triangle.

    How can this leg by the 5 in one triangle and the 3 in the other?

    Which of the lengths in this diagram is F? It's the 5 in the big triangle, isn't it? What does this tell you about the lengths of all other sides in the problem?
  4. Oct 14, 2012 #3
    I think you are reading the diagram wrong. My bad I'm not very good with 3-d drawing as I have a hard time understanding them. F would be the somewhat thicker line with the arrow at the end of it. That is the vector that I am trying to split. The axes have been labeled. Everything else defines the vector. 3 4 5 triangles simply state the relative sides of the vector and its projection on the x-y plane. This lengths aren't the actual lengths of the vector. I have to multiply the magnitude with these lengths to get the actual components of the vector. Since I didn't post the entire problem here I am just taking the magnitude to be F. And I don't understand what you mean by you last question.
  5. Oct 14, 2012 #4
    With respect, I'm reading the diagram just fine. Let me phrase the questions in a slightly different way, then. Look only at the big triangle first. The force vector has magnitude [itex]F[/itex]. We've established that [itex]F_z = 4F/5[/itex] and so the part lying in the xy plane must have magnitude [itex]3F/5[/itex].

    Now look at the small triangle. The hypotenuse is only [itex]3F/5[/itex]. So while this is a 345 triangle, all sides must be in proportion to this length, a length which is not [itex]F[/itex] like you originally supposed.
  6. Oct 14, 2012 #5
    I don't understand what the z component has to do with the projected vector. The hypotenuse for both the original and the projected vector is 5. Of all the problem I've done with position vectors I've never had to use a number like the 3 in the big triangle unless I'm using it to find an angle. I find and x and y components using the triangle that lies in the r-y plane and the z component using the triangle that exists with the original vector and its worked for me everytime. I don't understand where you are getting the 3/5 F from. What I'm trying to understand is why I am getting different components than the book.
  7. Oct 14, 2012 #6
    Your problem is that you implicitly assume the projected vector has magnitude [itex]F[/itex] when it doesn't. This is why you compute the x and y components incorrectly. I'm trying to explain to you that the projected vector has magnitude [itex]3F/5[/itex].

    This is not right. All you know from the 345 triangle is that the x component is [itex]3/5[/itex] of the magnitude of the projected vector, not [itex]3/5[/itex] of [itex]F[/itex]. These magnitudes are not the same.
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