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Determine the parametric equation for the tangent line at the point $P = (1,1,1)$ to a curve which is described by the solution set of the following equations:

$x^2 + y^2 + z^2 = 3$, $3x + 4y + 5z = 12$

I dont have a clue where to start. How can I differentiate functions which are determined by a solution set of equations?

Seirios
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dreamer
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2 Answers2

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Hints: to find the solution set, you can set the equations equal to one another, or add them, to find the solution set:

$x^2 + y^2 + z^2 = 3$,
$+$
$3x + 4y + 5z = 12$

$$x^2 + 3x + y^2 + 4y + z^2 + 5z \color{blue}{\bf{-15}} = 0\tag{1}$$ $$x^2 + 3x \color{blue}{\bf -4} + y^2 + 4y \color{blue}{\bf -5} + z^2 + 5z \color{blue}{\bf - 6} = 0$$ $$\implies (x +4)(x - 1) + (y+5)(y-1) + (z+6)(z-1) = 0\tag{2}$$

The curve of interest and the solution set of the two given equations are the solutions to $(1)$ = $(2)$. (Note that $P$ is among the solutions.) This curve is what you want to differentiate.

amWhy
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Find normals to both surface by taking partial derivatives: $(1,1,1)$ and $(3,4,5)$. Their cross-product $(1,-2,1)$ is tangent to both surfaces and, therefore, to their intersection. This is the direction of the tangent you are looking for. The parametric equation is then

$x=1+t, y=1-2t, z=1+t.$

Alexander Bogomolny
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