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Math Help - Very Urgent: (distances of vectors)

  1. #1
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    Very Urgent: (distances of vectors)

    1) Find the coordinates of the foot of the perpendicular from Q(3,2,4) to the line r= (-6,-7,-3) +t(5,3,4)

    2) The common perpendicular of two skew lines with the direction vectors d1 and d2 is the line that intersects both the skew lines and has direction vector n= d1 x d2. Find the points of intersection fo the common perpendicular with each of the lines (x,y,z)=(0,-1,0) + s(1,2,1) and
    (x,y,z)= (-2,2,0) + t(2,-1,2)

    These were assigned tonight and this content will be on tomorrows test I have no idea what to do. We've been studying distance between vectors,planes and points, but I can't do these at all.

    Thanks!
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  2. #2
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    Quote Originally Posted by white_cap View Post
    1) Find the coordinates of the foot of the perpendicular from Q(3,2,4) to the line r= (-6,-7,-3) +t(5,3,4)

    ...
    1. Create a plane perpendicular to the line containing Q:

    [5, 3, 4] \cdot (r - [3, 2, 4]) = 0

    2. Calculate the point of intersection between the line and the plane:

    [5, 3, 4] \cdot ([-6,-7,-3] +t[5,3,4] - [3, 2, 4]) = 0

    [5, 3, 4] \cdot ([-9,-9,-7] +t[5,3,4] ) = 0

    -100+50t = 0~\implies~ t = 2

    3. Plug in this value into the equation of the line. You'll get F(4, -1, 5)
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  3. #3
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    Quote Originally Posted by white_cap View Post
    ...

    2) The common perpendicular of two skew lines with the direction vectors d1 and d2 is the line that intersects both the skew lines and has direction vector n= d1 x d2. Find the points of intersection fo the common perpendicular with each of the lines (x,y,z)=(0,-1,0) + s(1,2,1) and
    (x,y,z)= (-2,2,0) + t(2,-1,2)

    ...
    1. Calculate the normal vector to the 2 direction vectors:

    \vec n = [1, 2, 1] \times [2, -1, 2] = [5, 0, -5]

    2. Calculate the equation of a line through a point of l_1 in the direction of \vec n:

    l_n: [x, y, z] = \underbrace{[0,-1,0] + s \cdot [1,2,1]}_{\text{ arbitrary point on the first line }} + r \cdot [5, 0, -5]

    3. Calculate the point of intersection l_n \cap l_2:

    [0,-1,0] + s \cdot [1,2,1] + r \cdot [5, 0, -5] = [-2,2,0] + t \cdot [2,-1,2]

    4. You have now a system of simultaneous equations in 3 variables. I don't know how you solve such a system but you should get:

    r=-\frac15 ~\wedge~ s=1~\wedge~t=1

    5. Plug in s into the equation of l_1 to get A(1, 1, 1)
    and plug in t into the equation of l_2 to get B(0, 1, 2)
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