Find a point on the surface of z = 8- 3x^2 - 2y^2 at which the tangent plane is perpendicular to the line x = 2 - 3t, y = 7 + 8t, z= 5 - t. Find the equation of the tangent line at that point.
I don't know how to find the point. I only know how to find the equation of the tangent plane to the surface at a certain point.
Quote:
Originally Posted by tdat1979
Find a point on the surface of z = 8- 3x^2 - 2y^2 at which the tangent plane is perpendicular to the line x = 2 - 3t, y = 7 + 8t, z= 5 - t. Find the equation of the tangent line at that point.
I don't know how to find the point. I only know how to find the equation of the tangent plane to the surface at a certain point.
I think I can help you but first you have to tell how will you get tangent plane and what is the equation of tangent plane you got.
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I don't know if if doing it right but i got the point (2,7,5) and vector
(-3,8,-1) from the equation of the line. So the tangent plane would be
-3(x-2) + 8(y-7) - (z-5) = 0 which would give me -3x + 8y - z = 45. If I'm wrong please tell me what I did wrong.
So we want to know when the gradient of the surface is parallel to the tangent of the line.
The tangent vector to the line is
We know that 2 vectors are parallel if and only if for some scalar (or if their cross product is 0)
Now we need to solve the vector equation
Note that our only choice for c is from the z component.
so this implies that and
Now just plug these back into the function to find z.