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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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!!!!!

Mathematics
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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 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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