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I have been asked to find parametric equations for the tangent like to the cruve of intersection of the surfaces $ x^2 + y^2 + z^2 = 4 $ and $ z^2 = x^2 + y^2 $ at $ (1,1,-\sqrt2) $

My solution;

I let

$ f = x^2 + y^2 + z^2 $

and let

$ g = x^2 + y^2 - z^2 $

then i computer the gradients and evaluated at point P

gradient f $ <2x, 2y, 2z> $ gradient g $ <2x, 2y, -2z> $

so these gradients evaulated at $ (1,1,-\sqrt2) $

$$ \begin{aligned} \nabla f &= <2,2,-2\sqrt2>\\ \nabla g &= <2,2,2\sqrt2> \end{aligned} $$

Now I need to calculate the cross product of these two vectors and then find the parametric equations for the tangent line. However I am not quite sure how to find the cross product, but once I have the cross product I just have to take $ (1,1,-\sqrt2) + t(\text{values for cross product}\ x,y,z) $

i evaluated my parametric equations for the tangent line to be

$$ \begin{aligned} x&= 1 + 8\sqrt2t \\ y&= 1+ 8\sqrt2t \\ z&= 0 \end{aligned} $$ I am not sure if my z component is 0 or 4, since the cross product of z is 0 but then the original point of z was $ -\sqrt2 $ so should it be $ z = -\sqrt2 + 0t = -\sqrt2 $

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  • $\begingroup$ I tried to improve the formatting, getting the $\sqrt2$ in there, and aligning some formulas that I guessed you intended to be aligned. I am uncertain about your preferred vector notation. You can use $(x,y,z)$, but if you are used to angle brackets, that's fine, too. If you use \langle in place of $<$ and \rangle in place of $>$, the result will be more pleasing to the eye :-) $\endgroup$ – Jyrki Lahtonen Jul 24 '16 at 22:34
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$$(a,b,c)\times (d,e,f)=(bf-ce,cd-af,ae-bd)$$

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You could always make a substitution in your surface equations to get the curve of intersection. Then parameterize this curve and take the derivative of it. This will give you a direction vector which you can then use to construct a tangent line with the given point. If you do this with gradients as you have tried, the gradient will give you a vector normal to the level curve. If you do this with both surfaces you can take the cross product of both of these vectors to get a vector which is now normal to both of these vectors and thus tangent to the curve of intersection. You can then use this vector to find your directional numbers. Remember the equation of a line parametrically is x=(x value from the point you have)+(directional number from tangent vector)t. (so your directional value for z is 0 but the point value is not. thus you have parametrically z=(z value of point given).)This is the same format for x,y and z. Hope this helps

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