Solving Hyperboloid Problem: Finding Points of Parallel Normal Line

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SUMMARY

The discussion focuses on solving the hyperboloid equation x² - y² + 2z² = 1 to find points where the normal line is parallel to the vector joining the points (3, -1, 0) and (5, 3, 6). The gradient vector of the hyperboloid is correctly identified as <2x, -2y, 4z>, which must be a scalar multiple of the vector <2, 4, 6>. Participants clarify that the solution requires finding (x, y, z) that satisfies both the hyperboloid equation and the condition of parallelism with the specified vector.

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Homework Statement


Hello, can someone tell me where I went wrong? So, I am supposed to find the points on the hyperboloid x^2-y^2+2z^2 = 1 where the normal line is parallel to the line that joins the points (3,-1,0) and (5,3,6).


Homework Equations


I think I'm supposed to find the gradient vector of the hyperboloid, because that has the direction of the normal line.


The Attempt at a Solution


So, if I'm right, the gradient of the hyperboloid is <2x, -2y, 2z>, and that is supposed to have the same direction as the vector <2, 4, 6>. So, wouldn't I just be looking for values of x, y, and z that is in multiples of <2, 4, 6>? That seems wrong to me. Thanks!
 
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Wouldn't the gradient be <2x, -2y, 4z>? Then you'd need to find (x,y,z) such that <2x, -2y, 4z> is a scalar multiple of <2, 4, 6> AND (x,y,z) must lie on the given hyperboloid, i.e. it must satisfy x2 - y2 + 2z2 = 1.
 

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