Gradient and line

[Yann]

Sorry to ask yet another question, but it's very hard to contact my teacher.

Homework Statement

$$f(x,y,z) = x^2-yz+z^2$$

a = (0,1,1), b = (1,3,2). Find a point c on the line joining a and b such that;

$$f(b)-f(a)=\nabla f(c)\bullet(b-a)$$

2. The attempt at a solution

f(b) = 1-3*2+2^2=-1
f(a) = 0-1*1+1 = 0

$$\nabla f(c) = (2x)i - (z)j + (2z-y)k$$

(b - a) = i + 2j + k

$$-1 = [(2x)i - (z)j + (2z-y)k]\bullet[i + 2j + k]$$

$$-1 = 2x -2z + 2z-y$$

$$0 = 2x -y +1$$

Let L be the line between a and b

$$L = ti+(1+2t)+(1+t)k$$

By substitution to find the point on the line that is also on the plane;

$$0 = 2t -1-2t +1$$

$$0 =$$

So, it seems to be impossible

 

[gammamcc]

Your result is 0=0 so EVERY point on the line would work (unless there is a mistake somewhere).

 

[Yann]

Exactly. But the question seem to imply there's only 1 point (find THE point).

 

[gammamcc]

A quick check: Mathematica or Maple graphs.