How Does Rolle's Theorem Apply to Intersecting Planes in Vector Calculus?

[monea83]

Given is a one-parameter family of planes, through

$$x \cdot n(u) + p(u) = 0$$

with normal vector n and base point p, both depending on the parameter u.

Two planes with parameters $$u_1$$ and $$u_2$$, with $$u_1 < u_2$$, intersect in a line (planes are assumed to be non-parallel). This line also lies in the plane

$$x \cdot (n(u_1) - n(u_2)) + p(u_1) - p(u_2) = 0$$

Now, the book I am reading claims that, "by Rolle's theorem, we get:"

$$x_1 n_1'(v_1) + x_2 n_2'(v_2) + x_3 n_3'(v_3) + p'(v_4) = 0$$ with $$u_1 \leq v_i \leq u_2$$.

However, I don't see how the theorem applies here... for starters, I don't see anything of the form $$f(a) = f(b)$$, as required by the theorem.

 

[LCKurtz]

Given is a one-parameter family of planes, through

$$x \cdot n(u) + p(u) = 0$$

with normal vector n and base point p, both depending on the parameter u.

I'm trying to understand your notation. If n and p are vectors, I supose x is a vector dotted into n, which gives a scalar?? How do you add a scalar to a vector? Is the 0 on the right side a scalar or vector? And you say you have a one parameter family of planes through

$$x \cdot n(u) + p(u) = 0$$

What do you mean by that?