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Math Help - Distence bewteen a pt. and plane

  1. #1
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    Distence bewteen a pt. and plane

    Find the distance between the point M(-2,3,4) and the line
    L: x-3/1 = y+1/2 = z-1/1.
    I was wondering if someone can start me off with this problem.
    I wanted to take the Line: x-3/1 = y+1/2 = z-1/1 and turn it to a "Parametric Equation"
    This is what I have for the PE: x= -2 +t, y = 3+2t, z= 4+t
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  2. #2
    Senior Member BAdhi's Avatar
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    I think the parametric equation should be x=3+t,y=2t-1,z=t+1 (from \frac{(x-3)}{1}=\frac{(y+1)}{2}=\frac{(z-1)}{1}=t, if so the point (3,-1,1) (call it \vec{OA}) is on the line. So the \vec{AM} is (-5,4,3).

    if the vector parallel to the line L is \vec{PQ}

    Now you'll have two vectors that cross each other and one vector starts from point A to point M

    take the definition of \vec{PQ}\times\vec{AM}=|PQ||AM|\sin \theta

    if you draw the vector's you'll find |AM|\sin \theta is the distance of from the point to the required line
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  3. #3
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    May I ask what is the AM.

    IF you can, can you use the following and Ill take from there and get back you. I appreciate help!!!

    U = (1, 2, 1), which is the direction vector for the line, correct???
    P = (3, -1, 1) point on the line, correct???

    and Ill take P and U to get PQ, correct???

    You dont seem 100% sure on the parametrics equation that you have. LOL

    I use this formula correct???

    D= ll PQ x u ll / ll u ll
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  4. #4
    Senior Member BAdhi's Avatar
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    U = (1, 2, 1), which is the direction vector for the line, correct???
    P = (3, -1, 1) point on the line, correct???
    correct, here the P is the equal to the point A that i've used

    and Ill take P and U to get PQ, correct???
    wrong, what is Q here it, shouldn't it be M? So use P and M to get PM i

    I use this formula correct???

    D= ll PQ x u ll / ll u ll
    yes but it has to be M
    Attached Thumbnails Attached Thumbnails Distence bewteen a pt. and plane-untitled.bmp  
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  5. #5
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    What I would do is this: the plane x+ 2y+ z= C is perpendicular to the line for all C. Find the value of C such that the plane contains the point (-2, 3, 4) and then determine where the line intersects that plane. The distance from the point to the line is the distance between those two points.
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  6. #6
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    where is the x+ 2y+ z= C coming from??? I dont see that. It seems there may be more than one way to solve this problem.
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  7. #7
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    Nevermind
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  8. #8
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    Question:

    Instead of using a paramtric equation, what are the other ways to solve this problem.

    now I am thinking of using the Standard Equation of a Plane in space.

    Now I am confusing my self.

    Im lost.
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  9. #9
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    Given a plane Ax+By+Cz+D=0 and a point (p,q,r) the distance between the two is:
    \dfrac{|Ap+Bq+Cr+D|}{\sqrt{A^2+B^2+C^2}}.
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  10. #10
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    Alright BAdhi,

    This is what I got. By the way I didnt get your image, error???
    ------------------------------------------------------------------

    Parametric Eqaution: x = 3 + t, y = -1 + 2t, z = 1 + t

    Direction Vector: u = (1,2,1)

    To find the point, let t = 0 and get

    P = (3,-1,1)

    So,

    PQ= (-2 - (3), 3 - (-1), 4 - 1) = (-5, 4, 3)

    PQ x u =
    i j k
    -5 4 3
    1 2 1

    = (-2, -8, -14)

    D= ll PQ x ull / ll u ll

    finale answer = sqrt(264) / sqrt(6)
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  11. #11
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    Would my answer below be correct if I do it that way???
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