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Homework Help: Help required w/ vectors: general equations, intersection points

  1. Apr 18, 2014 #1
    1. The problem statement, all variables and given/known data

    In this question we consider the following six points in R3:
    A(0,10,3) B(4,18,5) C(1,1,1) D(1,0,1) E(0,1,3) F(2,6,2)

    a) Find a vector equation for the line through the points A and b
    b) Find general equations for the line from a
    c) Find a vector equation for the plane through the points C, D and E

    2. Relevant equations

    3. The attempt at a solution

    Question a:
    r = (0, 10, -3) + t(-4 - 0, 18 - 10, -5 - -3)
    r = (0, 10, -3) + t(-4, 8, 2)
    r = (0 - 4t, 10 + 8t, -3 + 2t)

    Is this right?

    Question b:
    I have no idea how to do this. I've only done questions where the they give you two vector equations not one.

    Question c:
    x = 1 + t(-1 - 1) + t2(0 - 1)
    y = 1 + t(0 - 1) + t2(1 - 1)
    z = 1 + t(1 - 1) + t2(3 - 1)

    x = 1 - 2t -t2
    y = 1 - t
    z = 1 + 2t2

    Is this right?

    Please just help me out straight away instead of giving me guidance... It takes like 2 days to solve these small questions here...
  2. jcsd
  3. Apr 18, 2014 #2

    Simon Bridge

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  4. Apr 18, 2014 #3


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  5. Apr 18, 2014 #4

    Simon Bridge

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    That's what I thought at first.
    Then I figured it did mean "a" (not A) as in "part (a) of the question".
    Since there are infinite possible lines from point A and only one line in part (a)...

    I suspect that by "general equation" they mean to put the vector equation calculated in part (a) into standard form.
    But that's just a guess - need OP to verify.
  6. Apr 19, 2014 #5
    Yeah I thought the "a" from the question meant the vector A not "a" as for the question "a". Dumb mistake.

    No wonder why I couldn't solve it.
    Last edited: Apr 19, 2014
  7. Apr 19, 2014 #6

    Simon Bridge

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    That would explain it all right :)
    Be reassured that quite experienced people made the same mistake.
  8. Apr 20, 2014 #7
    Could someone please explain when to minus the vector 2 with vector 1?

    This is what I found in the course book:

    Example 1:
    Find a general equation of a line in R2 that passes through the point (2,3) and is parallel to the vector (-3,-2).

    (x,y) = (2,3) + t(-3,-2)

    For the vector equation, this example did not minus the second vector with the first vector.

    Example 2:
    Find the general equation of the line in R3 that passes through the points (2,-1,0) and (1,1,1).

    The vector equation is
    (x,y,z) = x0 + tv
    (x,y,z) = x0 + t(x1 - x0)
    (x,y,z) = (2,-1,0) + t(-1,2,1)

    I don't understand why the second example used v and did x1 - x0

    The only difference I can spot with the example 1 and 2 is that second example asks points not a point.
    So I am assuming that when its asking for more than 1 point we have to work with minus the vectors?

    If so did I get the first and last question wrong?
  9. Apr 20, 2014 #8


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    That's because there is NO "first vector": (2,3) is a point, not a vector.

    You need to distinguish between "points" and "vectors". A vector requires two points to determine it: a "start" and an "end". The vector from the point [itex](x_0, y_0, z_0)[/itex] to [itex](x_1, y_1, z_1)[/itex] is [itex](x_1- x_0, y_1- y_0, z_1- z_0)[/itex]. Unfortunately we are using the same notation, "( )", to mean two different things. Some textbooks write a vector with "< >" rather than "( )" to distinguish them: (a, b, c) is a point while <a, b, c> is a vector. Of course, the more formal "coordinate vectors", [itex]\vec{i}[/itex] is the unit vector in the direction of the x-axis, [itex]\vec{j}[/itex] is the unit vector in the direction of the y-axis, and [itex]\vec{k}[/itex] in the direction of the z-axis, make it very clear. [itex]a\vec{i}+ b\vec{j}+ c\vec{k}[/itex] is obviously a vector, not a point.

    which "first and last question" are you talking about? From the previous posts?
    Last edited by a moderator: Apr 20, 2014
  10. Apr 20, 2014 #9

    Simon Bridge

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    A point is a point, but it's location can be represented by a position vector.
    This makes it easy to confuse different roles a vector can play.

    i.e. points ##A## and ##B## in post #1 can be represented by vectors ##\vec a = (0,10,3)^t = 10\hat\jmath + 3\hat k## and ##\vec b=(4,18,5)^t = 4\hat\imath + 18\hat\jmath + 5\hat k##.

    In terms of HallsofIvy's last post;
    The "initial point" for these vectors is the coordinate system origin: point O which is (0,0,0).

    notice that the ^t indicates a "transpose" because I am using column vectors, but I'm too lazy to write in columns. I'm also using a hat rather than an over-arrow to indicate a unit vector.

    We can say that vector ##\vec a## "points to" ##A##, but so does the unit vector ##\hat a = \vec a/a##.

    I am still being careful to distinguish between point A, and it's position vector.
    This can look like splitting hairs but it does help to think this way.

    Hold on to your hat:
    The position vector that points from A to B is ##\vec r_{AB}=\vec b - \vec a## and the vector from B to A is ##\vec r_{BA}=\vec a - \vec b = -\vec r_{AB}##
    ... got that?

    1. so you can get a vector that shows a distance between two positions, as well as the direction to go from one to the other by subtracting position vectors.

    2. Which order you subtract is "finish minus start".


    1. you are told to find the line through given points P and Q.
    The vector equation of a line has to be a position + a scalor times a direction.
    Here, "position" and "direction" are different roles that vectors can play in an equation.

    So you can pick either of these points for the position part.

    Pick ##P##, position ##\vec p## then the line must be in the direction from P to Q, so the direction is given by the vector ##\vec v = \vec q-\vec p## and the equation is ##\vec r = \vec p + \lambda\vec v##

    Easy to see in 2D.
    P is (1,1) and Q is (2,3) then ##\vec v = (1,2)^t## and ##(x,y)^t=(1,1)^t+\lambda(1,2)^t##
    Sketch the points on a graph in and you can see that this has to be the case.

    2. you are told to find the line through point ##P## and parallel to vector ##\vec v##
    - in this example the direction has already been determined so you don't need to subtract two points to find it, you can just write it down directly.

    You can recover the point Q if you like by putting ##\lambda = 1##

    ##\vec q = (1,1)^t+(\lambda=1)(1,2)^t=(2,3)^t##

    ... any point on the line can be found by varying the value of ##\lambda##.

    When ##\lambda## is time, the vector ##\vec v## is the change in position in a unit of time, i.e. the "velocity". (strictly - the average velocity.)

    So the answer to your question is: you subtract the vectors when it makes sense to do so. This depends on the roles the vectors are playing in the problem. In the examples above, you subtract positions of points to get directions but you don't subtract a position and a direction.
  11. Apr 20, 2014 #10
    Thank you @HallsofIvy and @Simon Bridge for both really helpful answers. Unfortunately, I am still am having some trouble understanding though... :(

    For clarification:
    Question: Find a vector equation for the line through the points A and B
    A(0,10,-3), B(-4,18,-5)

    Since I have to calculate the vector only using the points, I must calculate the length by subtracting the "end" with "start". This will be the direction (I am assuming?)
    So, vector equation is a v, therefor: v = (position) + (scalar)(direction)

    This is what I've got:
    v = [itex]\vec{AB}[/itex] = (x,y,z) = (0,10,-3) + t(-4,8,-2)
    Which can be simplified (also be written) into: (x,y,z) = (0,10,-3) + t(-4,8,-2)

    Position (origin) = (0,-10,-3)
    Scalar = t
    direction = (-4,8,-2)

    Based on my understanding of the help received, I think this is right with a not a lot of confidence...
    Last edited: Apr 20, 2014
  12. Apr 20, 2014 #11

    Simon Bridge

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    That is the equation of a line all right - it is the line through (0,10,-3) in the direction of (-4,8,-2)
    Each time t increases by +1, you advance a distance of |(-4,8,-2)|=√84 units along the line.

    The reasoning is correct - your problem is: "how can I be confident I have the correct line?"

    You can check to see if you have the correct line by seeing if the required points lie on it.

    point A has to be on the line:
    When t=0, (x,y,z)=(0,10,-3) so point A is on the line: excellent.

    point B has to be on the line as well:
    If you can find the value of t which makes (x,y,z)=(-4,18,-5) then point B is also on the line, and you can be 100% confident you have the correct equation.
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