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Math Help - distance from the line to the plane (which one is correct?)

  1. #1
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    distance from the line to the plane (which one is correct?)

    find the distance from the line x=2+t, y=1+t, z= -(1/2)-(1/2)t to the plane x+2y+6z =10

    my first attempt ::

    by setting t = 0, we get a point on the line, say P :<2,1,-1/2>. the point on the plane, say, S we get if we set y=0, z=0 ==> x= 0. thus the point S on the plane is <10,0,0>.

    thus the vector PS = S - P =8i-j+1/2k

    vector normal to the plane, n = i+2j+6k

    so the distance point P to the plane = |PS.n|/|n| = 9/sqrt(41) unit


    my second attempt ::

    the vector normal to the line, n1 = i+j-1/2k

    the vector normal to the plane, n2 = i+2j+6k

    n1.n2 = 0 ==> the two normal are perpendicular ===> the line and the plane are parallel

    thus the distance for the line to the plane = sqrt(2^2+1^2+(-1/2)^2) = sqrt(21)/2 unit



    WHICH ATTEMPT IS CORRECT SINCE I GOT TWO DIFFERENT ANSWERS??? PLEASE HELP ME..
    Last edited by mr fantastic; August 11th 2010 at 04:56 AM.
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  2. #2
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    Quote Originally Posted by bobey View Post
    find the distance from the line x=2+t, y=1+t, z= -(1/2)-(1/2)t to the plane x+2y+6z =10

    my first attempt ::

    by setting t = 0, we get a point on the line, say P :<2,1,-1/2>. the point on the plane, say, S we get if we set y=0, z=0 ==> x= 0. thus the point S on the plane is <10,0,0>.

    thus the vector PS = S - P =8i-j+1/2k

    vector normal to the plane, n = i+2j+6k

    so the distance point P to the plane = |PS.n|/|n| = 9/sqrt(41) unit
    This is incorrect. That gives |PS| which is the distance from P to S, not the distance from P to the nearest point on plane.

    my second attempt ::

    the vector normal to the line, n1 = i+j-1/2k

    the vector normal to the plane, n2 = i+2j+6k

    n1.n2 = 0 ==> the two normal are perpendicular ===> the line and the plane are parallel
    Well, yes, of course. Otherwise, the two would intersect so the distance would be 0. There was really no reason to do this.

    thus the distance for the line to the plane = sqrt(2^2+1^2+(-1/2)^2) = sqrt(21)/2 unit
    What? This is simply the distance from the the given point to (0, 0, 0). What does that have to do with the plane?


    WHICH ATTEMPT IS CORRECT SINCE I GOT TWO DIFFERENT ANSWERS??? PLEASE HELP ME..
    I don't see why either attempt should be correct. Both are based on incorrect logic.

    As you say, a perpendicular vector to the plane is i+ 2j+ 6k and a point on the line is (2, 1, -1/2). A line through (2, 1, -1/2) in the direction of i+ 2j+ 6k is x= 2+ t, y= 1+ 2t, z= -(1/2)+ 6t. Since that line passes through (2, 1, -1/2) and is perpendicular to the plane, it will intersect the plane at the point closest to (2, 1, -1/2). That line will intersect the given plane when (2+ t)+ 2(1+ 2t)+ 6(-(1/2)+ 6t)= (2+ 2- 3)+ (t+ 4t+ 36t)= 1+ 41t= 10 or t= 9/41. Put t into the equation of the perpendicular line to determine that point and then find the distance between those two points.
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