(Solved) Find a point on the surface x^2+2y^2+3z^2=12 where the tangent plane is perpendicular to the line


Question: Find a point on the surface \({{x}^{2}}+2{{y}^{2}}+3{{z}^{2}}=12\) where the tangent plane is perpendicular to the line \(x=1+2t,y=3+8t,z=2-6t\)

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Solution: The downloadable solution consists of 1 pages
Deliverable: Word Document

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