Given \(\displaystyle x^{2}-2y^{2}+x^{2}=27\)
\(\displaystyle x=3t-5, \;\ y=2t+7, \;\ z=1-\sqrt{2}t\)
Partials:
\(\displaystyle f_{x}=3x_{0}^{2}\)
\(\displaystyle f_{y}=-4y_{0}\)
\(\displaystyle f_{z}=2z_{0}\)
So, \(\displaystyle 3x_{0}^{2}i-4y_{0}j+2z_{0}k\) is normal to the surface at \(\displaystyle (x_{0},y_{0},z_{0})\) on the surface.
The coefficients of the t's in the parametric line equations indicate the vector which the line is parallel to.
Thus, the normal must be parallel to the given line and, therefore, to the vector.
\(\displaystyle (3, \;\ 2, \;\ -\sqrt{2})\) is normal to the plane, which is parallel to the line.
So, we have:
\(\displaystyle 3x_{0}^{2}=3\)
\(\displaystyle -4y_{0}=2\)
\(\displaystyle 2z_{0}=-\sqrt{2}\)
Solve for x, y, and z for the point needed.
