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Vector IN the yz- plane?

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Two vectors are given as A=3x+y-2z and B=x-z. A third vector C lying in the yz-plane is perpendicular to A and its magnitude is 7 units. Find the angle between C and B.


    2. Relevant equations
    cross and dot product


    3. The attempt at a solution
    I can't imagine of way, a vector lying on the yz plane can be perpendicular to a 3d vector which has 3 nonzero unit vector components. An explanation on this is much appreciated.
     
  2. jcsd
  3. Sep 23, 2013 #2

    HallsofIvy

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    A vector in the "yz-plane" has the form <0, y, z>. Further, two vectors are perpendicular if and only if there dot product is 0. <a, b, c> will be perpendicular to <0, x, y> if and only if by+ cz= 0. For example, <a, z, -y> is perpendicular to <0, y, z>.
     
  4. Sep 23, 2013 #3

    Ray Vickson

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    A = 3x+y-2z and B=x-z are numbers, not vectors! So, my first question is: what do you mean? Have you copied the questions exactly as they were given to you?
     
  5. Sep 23, 2013 #4
    I'm not capable of writing exactly as they are given, they are vectors.
     
  6. Sep 23, 2013 #5

    Ray Vickson

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    Then I am not capable of helping you.
     
  7. Sep 23, 2013 #6

    Mark44

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    I'm guessing they're something like this:
    a = 3i + j - 2k
    b = i - k

    They could also be written like this:
    a = <3, 1, -2>
    b = <1, 0, -1>
     
  8. Sep 23, 2013 #7
    yeah, i guess i couldn't think of writing it in that way. they were just meant to be x, y and z's with hats. i'll probably be more thoughtful next time.
    HallsofIvy, I have found a result using dot product but what really bothered me with this question is not the solution but some things that i couldn't grasp in it. i will look into it tomorrow. thanks for your answer.
    Ray Vickson, i hope you will be capable to help me with questions that are well written in the future.
     
  9. Sep 23, 2013 #8

    Ray Vickson

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    Yes, no problem.

    Even if you can't figure out how to write something symbolically, you can explain in words what you mean; for example, you could say that x stands for the unit vector along the x-axis, etc (but I would discourage that choice of notation, for reasons I won't go into here). Or, you could write e_x for the unit vector along the x-axis (making sure to explain your notation). Or, you could say that x really means <1,0,0>, where those are the three components of the vector. Or, you could forget about using x,y and z, and just write what they mean: <1,0,0>, <0,1,0> and <0,0,1>. There are lots of ways to do it; just not the way you tried!
     
  10. Sep 24, 2013 #9
    I see now that i've had some misconceptions about the problem. I've tried to solve the problem but i am not sure about the result.

    Vector A = 3i + j - 2k and vector C is perpendicular to it and it is also lying on the yz plane.

    A.C = 0 = 3.Ci + 1.Cj -2.Ck Ci=0 since it is on they yz plane
    Cj - 2.Ck= 0 Cj=2Ck
    Vector C has the magnitude of 7:
    C= 2Ck.j + Ck.k

    7² = 49 = 4Ck² + Ck²
    Ck = 7/**√5

    C= 14/√5.j + 7/√5.k I need the angle between B and C.

    B= i - k it is given. It has the magnitude of √2

    B.C = 1.0 + 0. 14/√5 -1.7/√5 = √2.7.cosθ

    -1/√10 = cosθ
    arccos(-1/√10) = 108.4

    Sorry for the messy math, do you think this is the answer?
     
  11. Sep 24, 2013 #10

    Mark44

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    Vector A is very clear, but it would be clearer if you wrote C in a similar form before you start working with it.

    The components you give below for vector C are difficult to understand, as Ci could be interpreted as C * i and similarly for Cj and Ck. To make things clearer, let's define C this way:
    C = c2j + c3k

    I'm using the given information that this vector lies in the y-z plane.
    The notation above is awful. On the right side, it looks like C * k2.
    ??? What does this (above) mean?
    This is correct, but it's difficult to follow your reasoning, since you're leaving so much out, and your notation is so unclear.

    1.0 and -1.7 appear to be the numbers 1.0, and -1.7, rather than the products 1 * 0 and -1 * 7. It's common practice to use * to indicate multiplication.
    What units for 108.4? Radians, degrees?
     
  12. Sep 24, 2013 #11
    I've done some remediations on it; it should be clearer now.
    Vector A= 3i + j - 2k
    Vector C = C2j + C3k
    I'm using the given information that this vector lies in the y-z plane.

    A.C = 0 = 3*0 + 1*C2 - 2*C3
    C2 - 2.C3= 0
    C2=2C3
    C= 2C3j + C3.k
    Using the given magnitude of the vector C, which is 7.

    7² = 49 = 4C32 +C32

    C= 14/√5*j + 7/√5*k
    I need the angle between B and C.

    B= i - k it is given. It has the magnitude of √2

    B.C = 1*0 + 0*14/√5 -1*7/√5 = √2*7*cosθ

    -1/√10 = cosθ
    arccos(-1/√10) = 108.4°
     
  13. Sep 24, 2013 #12

    Mark44

    Staff: Mentor

    Looks good. Thanks for making the extra effort to make it more readable!
     
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