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Math Help - vectors distance

  1. #1
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    vectors distance

    could you pls help me on these?

    find the distance of:
    (a) point P(1,3,1) to the surface of 3x+7y-5z=0
    (b) point P(7,6,3) to the straight line of r=6j+k-t(i-j-k)
    (c) L1:r=5k+t(3j+k) to L2:r=i+j-3k+t(i+3j+k)


    thanks! =)
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  2. #2
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    Quote Originally Posted by ailing418 View Post
    could you pls help me on these?

    find the distance of:
    (a) point P(1,3,1) to the surface of 3x+7y-5z=0
    (b) point P(7,6,3) to the straight line of r=6j+k-t(i-j-k)
    (c) L1:r=5k+t(3j+k) to L2:r=i+j-3k+t(i+3j+k)


    thanks! =)
    to a)

    The equation 3x + 7y - 5z = 0 describes a plane with it's normal vector \vec n = (3,7,-5). Transform the equation such that the normal vector is an unit vector. Plug in the coordinates of the staionary vector of P. The result is the distance of P from the plane:

    d = \dfrac{3 \cdot 1 + 7 \cdot 3 - 5 \cdot 1}{\sqrt{3^2+7^2+5^2}}

    to b)

    Determine the equation of a plane which contains P and which is perpendicular to the line. Calculate the coordinates of the point of intersection between the line and the plane. Calculate the distance between these two points.

    p: -x+y+z-2=0

    point of intersection Q\left(-\tfrac53 , \tfrac{13}3 , -\tfrac23  \right)

    d = \sqrt{\frac{35}3}

    to c)

    Determine the equation of a plane which contains line L1 and is parallel to line L2. Then take any point of L2 and calculate it's distance to the plane as you've done in a).

    p: \left((0,3,1) \times (1,3,1)\right) ((x,y,z) - (0,0,5))=0 with P(1,1,-3) \in L_2
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