Gradient and line

  • Thread starter Yann
  • Start date
  • #1
48
0
Sorry to ask yet another question, but it's very hard to contact my teacher.

Homework Statement



[tex]f(x,y,z) = x^2-yz+z^2[/tex]

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

[tex]f(b)-f(a)=\nabla f(c)\bullet(b-a)[/tex]

2. The attempt at a solution

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

[tex]\nabla f(c) = (2x)i - (z)j + (2z-y)k[/tex]

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

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

[tex]-1 = 2x -2z + 2z-y[/tex]

[tex]0 = 2x -y +1[/tex]

Let L be the line between a and b

[tex]L = ti+(1+2t)+(1+t)k[/tex]

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

[tex]0 = 2t -1-2t +1[/tex]

[tex]0 = [/tex]

So, it seems to be impossible
 

Answers and Replies

  • #2
150
0
Your result is 0=0 so EVERY point on the line would work (unless there is a mistake somewhere).
 
  • #3
48
0
Exactly. But the question seem to imply there's only 1 point (find THE point).
 
  • #4
150
0
A quick check: Mathematica or Maple graphs.
 

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