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Homework Help: Vector equation questions

  1. Nov 12, 2008 #1
    Hello guys, I've got two questions here which I'm really unsure about on how to actually tackle. As a result, maybe it would be better if I could possibly be linked to some resources where I could read up on how to actually solve them.

    The problem statement, all variables and given/known data

    1. Find the equation of the plane in 3 dimensions through the point (1, 0, 1) perpendicular to the vector i + j - k

    2. What is the parametric form of the line joining the points (5, 0, 1) and (7, 4, 7)

    (Sorry if I posted this in the wrong forum).


    (I think I might have worked out question 1.)
    Last edited: Nov 12, 2008
  2. jcsd
  3. Nov 12, 2008 #2


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    You haven't posted in the wrong but you also haven't shown any work! We won't do your work for you. Both of those look to me like simple applications of the definitions. What formulas does your textbook give for equations of line and plane?
  4. Nov 12, 2008 #3
    If you want us to check your answer to the first question, we will willingly do so.

    As for the second one: What is the general form of the parametric form for a line? You will need a point on the line and a vector pointing in the direction of the line, won't you?
  5. Nov 12, 2008 #4
    I was hoping I could be linked to some resources to find the answer actually.

    Equation of a plane takes the form of Ax + By + Cz + D = 0

    Point = (1, 0, 1)
    Normal point = (i + j - k) = (1, 1, -1)

    Point Normal Form
    = (1, 1, -1) x ((x, y, z) - (1, 0, 1)
    = 1(x -1) + 1(y-0) -1(z-1)
    = x - 1 + y - z + 1

    Equation of Plane
    = x + y - z = 0

    I don't really know if that's right though.

    From my notes, it's OP(t) = (1-t)OA + (t)OB

    I think I should maybe take a look through my notes again, see if I can figure this out. A check on the first question would be great though.

    Thanks for the responses guys.
  6. Nov 12, 2008 #5
    Correct, well done.

    Yes, or equivalently

    OP(t)=OA + t BA

    In both equations, A,B are two points on the line.

    Good idea. Just post your answer if you'd like to have it checked.
  7. Nov 12, 2008 #6
    Thanks a lot

    Yeah, just got back, I'll take a stab at this now and post my result :)
  8. Nov 12, 2008 #7
    Ok, well, I'm really not sure what I'm doing but I did this:

    OP(t) = OA + (t)(BA) = (1-t)OA + (t)OB

    (1-t)(5,0,1) + (t)(7,4,7)
    (-6t + 6) + (7t + 4t + 7t)
    12t + 6

    OP(t) = 12t + 6

    Surely there's more to it then that?
  9. Nov 12, 2008 #8
    Never mind, to err is human:smile:
    How did you get from the first line to the second?? In the first line you add two multiples of 3-vectors whule in the secind line you have just a singe real number!!

    Remember that OP(t) should be a three-component vector. You're essentially done after the first line you wrote. There's no more to the question than inserting the given points into the formula. (If you are only interested in the result, that is. From the mistakes you made I wouldn't be surprised if there is a lot more work to do of you want to get a thorough understanding of what you acutally do, not only how)
    Last edited: Nov 12, 2008
  10. Nov 12, 2008 #9


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    Yes, there is! OP(t) is a vector, not a number! (1-t)(5, 0, 1) does NOT mean 1-t times the sum 5+ 0+ 1, it is the vector (5(1-t), 0, 1-t) and tOB= (7t, 4t, 7t). Add those.
  11. Nov 12, 2008 #10
    ah, I see the error of my ways, ok:

    OP(t) = OA + (t)(BA) = (1-t)OA + (t)OB

    (1-t)OA = (5 -5t, 0, 1-t)
    (t)OB = (7t, 4t, 7t)

    OP(t) = (5 -5t, 0, 1-t) + (7t, 4t, 7t)
    OP(t) = (5 + 2t, 4t, 1 + 6t)

    So that's the answer? I think I'll have to do more research on this regardless?
  12. Nov 12, 2008 #11
    It's not so much about research as about practicing, studying your notes, textbook and whatever else you can find and not forgetting to think about what you're actually doing (for example what should be a vector, what should be a number, what does an equation mean geometrically and so on)
  13. Nov 12, 2008 #12
    Thankyou both very much for the help

    Yeah, I completely understand where my problem is, I'll definitely do this. Thanks again.
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