Question

Find a vector function for the curve of intersection of surfaces x^2+y^2+z^2=34 and x-z = 6 (question on the solution regarding this).

Answer 1

At the end of the solution, how is \[x = 3 + 2\sqrt{2}cost\]? I can see how z and y is obtained, but not x.

Answer 2

You understand how z and y were obtained?
Hmm ok that's the part I'm confused on lol.

Getting x from z is really straight forward though,\[\large\rm \color{royalblue}{z=-3+2\sqrt2 \cos t}\]We earlier wrote x in terms of z,\[\large\rm x=\color{royalblue}{z}+6\]\[\large\rm x=\color{royalblue}{-3+2\sqrt2 \cos t}+6\]

Answer 3

Ah I see! :) I wasn't looking at the original question of using the information " x = z + 6 ".