Calculus

Find an equation of the plane.
The plane that passes through
(8, 0, −2)and contains the line
x = 6 − 2t, y = 3 + 5t, z = 3 + 2t

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  1. we have two points in that plane
    (8, 0, -2) and (6, 3, 3)

    a vector through those points has direction
    (6-8)i + 3 j + (2+2)k
    = -2 i + 3 j + 4 k
    so we have two vectors parallel to plane, their cross product is normal to the plane
    i j k
    -2 5 2 the given line direction
    -2 3 4 the line direction we found

    i( 20-6) + j(-4+8) + k (-6+10)
    = 14 i + 4 j + 4 k
    so plane is
    14 x + 4 y + 4 z = constant
    but (8,0,-2) is in it so
    112 -8 = constant
    14 x + 4 y + 4 z = 104
    check my arithmetic !

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    posted by Damon

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