The problem is to find the parametric equation of the line that is tangent to the line of intersection between the plane $x+2y+3z=6$ and the surface $x^2+y^2=2$ and passes through the point $(1,1,1)$.
How would I solve this problem?
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2 Answers
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Normal vector of the given plane $(1, 2, 3)$
Normal vector of the given surface $(2x, 2y, 0)$ or $(x, y, 0)$ or
$(1, 1, 0) \,$ at point $(1, 1, 1)$ which is on the plane and also on the given surface.
The cross product of the normal vectors $(1, 2, 3)$ and $(1, 1, 0)$ will be the vector of the line tangent to the intersection of both the given plane and the surface which comes to $(3, -3, 1)$.
So parametric equation of the line passing through point $(1, 1, 1)$
is $(1, 1, 1) + t (3, -3, 1)$
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If the surfaces are defined by $f(x,y,z)=0$ and $g(x,y,z)=0$, then $\nabla f \times \nabla g$ is a vector tangent to the line of intersection
