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  • 5 years ago

find the equation of the line that contains the center ellipsoid 2x^2+4x+3y^2-3y+z^2-4z=0 and the line is perpendicular to the plane 2x+3y+4z=2 please help i have a test tomorrow!!!!!

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  1. anonymous
    • 5 years ago
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    The set of parametric equations that represents this line is given by x = x1 + at, y = y1 + bt, and z = z1 + ct, where (x1, y1, z1) is the center of the ellipsoid and <a, b, c> is the vector representing the direction of the line. The vector perpendicular to the plane is given by the coefficients of the equation of the plane, so <2, 3, 4>. To find the center of the ellipsoid, we have to complete the square for the x, y, and z components. \[2x^2+4x+3y^2-3y+z^2-4z=0\] \[2(x^2+2x+1)+3(y^2-y+1/4)+z^2-4z+4\] \[=2+3/4+4\] \[2(x+1)^2+3(y-1/2)^2+(z-2)^2=27/4\] Now, we could multiply both sides by 4/27 to get the relationship between the axes of the ellipsoid, but we don't need to do this to find the center. The center is at (-1, 1/2, 2). Our line is: In parametric equations: x = -1 + 2t y = 1/2 + 3t z = 2 + 4t Symmetric form: (x+1)/2 = (y-1/2)/3 = (z-2)/4

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