# A word problem in Vector Calc

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 $z = -\sqrt2 + 0t = -\sqrt2$

• 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 :-) – Jyrki Lahtonen Jul 24 '16 at 22:34

$$(a,b,c)\times (d,e,f)=(bf-ce,cd-af,ae-bd)$$