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Thread: Vector perpendicular to two skew lines

  1. #1

    Line perpendicular to two skew lines

    I've been given these two lines and need to find the equation of a l line that is perpendicular to BOTH . What i've done so far:

    r : x = 2+3a , y = 2 , z= 3-2a Checked and they aren't parallel and don't have any commom points.
    s: x = 1-2b , y =2+b , z= b

    Also, the normal vector of their planes is (2,1,3) . The plan of r is 2(x-2) +1(y-2) + 3(z-3) = 0 and the plan of s is 2(x-1) + 1(y-2) + 3(z) = 0.

    Thinking of a vector that is parallel to l , it is (l1,l2,l3) and its dot product with the vectors parallel to the other 2 lines equals 0

    (l1,l2,l3) * (3,0,-2) =0
    (l1,l2,l3) * (-2,1,1) = 0

    3l1 - 2l3 = 0
    -2l1 + l2 + l3 = 0

    Here's where i'm stuck .
    Last edited by hellawowser; 11-24-2017 at 08:54 PM.

  2. #2
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    Quote Originally Posted by hellawowser View Post
    I've been given these two lines and need to find the equation of a l line that is perpendicular to BOTH . What i've done so far:

    r : x = 2+3a , y = 2 , z= 3-2a Checked and they aren't parallel and don't have any common points.
    s: x = 1-2b , y =2+b , z= b

    Also, the normal vector of their plans is (2,1,3) . The plan of r is 2(x-2) +1(y-2) + 3(z-3) = 0 and the plan of s is 2(x-1) + 1(y-2) + 3(z) = 0.

    Thinking of a vector that is parallel to l , it is (l1,l2,l3) and its dot product with the vectors parallel to the other 2 lines equals 0

    (l1,l2,l3) * (3,0,-2) =0
    (l1,l2,l3) * (-2,1,1) = 0

    3l1 - 2l3 = 0
    -2l1 + l2 + l3 = 0

    Here's where i'm stuck .
    You said, "the normal vector of their planes is (2,1,3)". I don't know what you mean by "their planes"; each of the two lines is in infinitely many planes, and since they do not intersect, there is no plane that they both lie in. But the vector (2,1,3) is the vector product of the direction vectors of the lines, so it is perpendicular to both of them. (It is the normal vector to any plane containing two lines parallel to the given lines, and is the direction vector of the line you want to find.)

    What you call "the plane of r" and "the plane of s" are parallel planes containing each line. I don't see why you need to do any of that. What you need is the equation of a line whose direction vector is the normal you found, that passes through both given lines. One way to do this is to find values of a and b such that the vector joining the corresponding points on lines r and s is parallel to (2,1,3).

  3. #3
    Quote Originally Posted by Dr.Peterson View Post
    One way to do this is to find values of a and b such that the vector joining the corresponding points on lines r and s is parallel to (2,1,3).
    Sorry can you explain again? I still don't know how i gonna find which points of each lines that are part of the third line

  4. #4
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    Try something! Often you don't know what to do until you start working on the problem.

    But the specific approach I had in mind is to write a system of equations in the variables a, b, and k that says that the vector from a point on line r (specified in terms of a) to a point on line s (specified in terms of b) is k times the vector (2,1,3).

    If you don't see what I mean, make an attempt and show me what you've done, so I can guide you in the right direction.

  5. #5
    Like this Xr -Xs = 2 * (k) ?

    2+3a -(1- 2b) = 2k
    2 - (2+b) = k
    3-2a - (b) = 3k

    Then i've found b = k = 11/12 and a = -1/3. But that a line perpendicular only to r and doesnt intersect s

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    Quote Originally Posted by hellawowser View Post
    Like this Xr -Xs = 2 * (k) ?

    2+3a -(1- 2b) = 2k
    2 - (2+b) = k
    3-2a - (b) = 3k

    Then i've found b = k = 11/12 and a = -1/3. But that a line perpendicular only to r and doesnt intersect s
    Those are the equations I get (though your vector equation doesn't quite make sense to me as written; but b = -k, not k, so your solution can't be right.

    Check and correct your solution to the equations, THEN check whether you have the right line! Also, tell me specifically the equation of the line you think you get as the answer.

  7. #7
    Quote Originally Posted by Dr.Peterson View Post
    Those are the equations I get (though your vector equation doesn't quite make sense to me as written; but b = -k, not k, so your solution can't be right.

    Check and correct your solution to the equations, THEN check whether you have the right line! Also, tell me specifically the equation of the line you think you get as the answer.
    Same thing. The equations i've found is:

    x =11/3+2c, y=2+c, z=17/9+3c , is you use a point of r. a=5/3
    x = 7/3 + 2c , y = 4/3 + c , z = -4/6 + 3c , if you use s , (b = -4/6)

    So which vector should i use ? it makes sense to me using (2,1,3) because it is perpendicular to two planes that contain the lines,so any line that uses the vector as direction,and intersect both lines will be perpendicular to both.

  8. #8
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    Quote Originally Posted by hellawowser View Post
    Same thing. The equations i've found is:

    x =11/3+2c, y=2+c, z=17/9+3c , is you use a point of r. a=5/3
    x = 7/3 + 2c , y = 4/3 + c , z = -4/6 + 3c , if you use s , (b = -4/6)

    So which vector should i use ? it makes sense to me using (2,1,3) because it is perpendicular to two planes that contain the lines,so any line that uses the vector as direction,and intersect both lines will be perpendicular to both.
    Did you check your solution to the system?

    The equations were

    2+3a -(1- 2b) = 2k
    2 - (2+b) = k
    3-2a - (b) = 3k

    which simplify to

    3a + 2b - 2k = -1
    b + k = 0
    a + b + 3k = 3

    You got a = 5/3, b = -2/3, k = 2/3. Putting those into the equations, we get

    3(5/3) + 2(-2/3) - 2(2/3) = -1
    (-2/3) + (2/3) = 0
    (5/3) + (-2/3) + 3(2/3) = 3

    So, yes, these work.

    So the point on line r is

    r : x = 2+3(5/3) = 7, y = 2 , z= 3-2(5/3) = -1/3

    and the point on line s is

    s: x = 1-2(-2/3) = 7/3, y =2+(-2/3) = 4/3, z= -2/3

    It looks like you got the right point on s, but made some mistake on r.

    Check your work and see what went wrong. (It's possible I'm wrong!)

    Note that there is no question that the direction vector you want is (2,1,3); the problem lies in your equations for the line. When you get them right, they will both represent the same line, just with different starting points.

  9. #9
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    Correcting a mistake

    I wrote the simplified equations incorrectly; they should be

    3a + 2b - 2k = -1
    b + k = 0
    2a + b + 3k = 3

    With your solution, a = 5/3, b = -2/3, k = 2/3, we get

    3(5/3) + 2(-2/3) - 2(2/3) = -1
    (-2/3) + (2/3) = 0
    2(5/3) + (-2/3) + 3(2/3) = 14/3 which is not 3

    So your solution was wrong; in fact, you may have done just what I did with the third equation.

    Try again.

  10. #10
    Quote Originally Posted by Dr.Peterson View Post
    I wrote the simplified equations incorrectly; they should be

    3a + 2b - 2k = -1
    b + k = 0
    2a + b + 3k = 3

    With your solution, a = 5/3, b = -2/3, k = 2/3, we get

    3(5/3) + 2(-2/3) - 2(2/3) = -1
    (-2/3) + (2/3) = 0
    2(5/3) + (-2/3) + 3(2/3) = 14/3 which is not 3

    So your solution was wrong; in fact, you may have done just what I did with the third equation.

    Try again.
    a= 5/7 and b = -11/14

    l : x = 2+ 15/7 + 2c , y = 2 , z = 3 - 10/7 + 3c

    I think its right now but i'll try to see if there's a different approach

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