## mathstina Group Title Show that every plane that is tangent to the cone x^2+y^2=z^2 passes through the origin. one year ago one year ago

1. mathstina Group Title

hw? i need guidance.

2. TuringTest Group Title

let $g(x,y,z)=x^2+y^2-z^2=0$then the normal vector to the surface is given by$\nabla g(x,y,z)$further we know that the point Po=(0,0,0) is on the cone. The equation of the tangent plan at any point x,y,z is then given by$\nabla g\cdot(\vec P-\vec P_0)=0$where $$\vec P=\langle x,y,z\rangle$$ pluf in the pieces and what do you get?

3. mathstina Group Title

is it (2,2,-2)?

4. TuringTest Group Title

$\nabla g(x,y,z)=\langle2x,2y,2z\rangle$

5. mathstina Group Title

yes, how to get the P?

6. piscez.in Group Title

well the function is x2+y2-z2= 0

7. TuringTest Group Title

oops$\nabla g(x,y,z)=\langle2x,2y,-2z\rangle$

8. piscez.in Group Title

im guessing the tangent at any point would be 2x, 2y,-2z

9. TuringTest Group Title

not the tangent, the normal vector/gradient

10. TuringTest Group Title

P=(x,y,z) is a general point on the plane

11. piscez.in Group Title

gradient and tangent are the same thing arent they

12. mathstina Group Title

ok. hw do i ans to this qn?

13. TuringTest Group Title

no because the gradient is normal to the surface in this case it would be the tangent vector to the surface if it were$\nabla z(x,y)$

14. TuringTest Group Title

sorry, I messed up, let $$\vec P_0=\langle l,m,n\rangle$$

15. TuringTest Group Title

so then what is $$\vec P-\vec P_0$$ ???

16. piscez.in Group Title

i think gradient is the rate of change in genera, a scalar quantity, and tangent by definition is the rate of change of a function and its a vector quantity, normal, on the other hand is the vector thats perpendicular to the tangent? no?

17. mathstina Group Title

P= <-2,-2,2> ??

18. TuringTest Group Title

for the surface $$f(x,y,z)$$ the gradient $$\nabla f$$ represents a vector normal to the surface http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx @mathstina no, there should be no pure numbers here...

19. piscez.in Group Title

i think i got it thankk you @TuringTest

20. TuringTest Group Title

general point$\vec P=\langle x,y,z\rangle$specific point$\vec P_0=\langle l,m,n\rangle$gradient at specific point$\nabla g(l,m,n)=2\langle l,m,n\rangle$since $$\vec P-\vec P_0$$ lies in the plane, then the gradient is perpendicular to it, so the equation for a tangent plane at a point (l,m,n) is$\nabla g(l,m,n)\cdot(\vec P-\vec P_0)=0$

21. TuringTest Group Title

$\nabla g(l,m,n)\cdot(\vec P-\vec P_0)=0$$2\langle l,m,n\rangle\cdot\langle x-l,y-m,z-n\rangle=2l(x-l)+2m(y-m)-2n(x-n)=0$at the origin we have (x,y,z)=(0,0,0) so$-2l^2-2m^2+2n^2=0\implies l^2+m^2-n^2=0$and because of the equation of the cone, we know that whatever point on it we pick we will have x^2+y^2-z^2=0, so the above is always true regardless of where we draw out tangent plane (l,m, and n don't matter). therefore every tangent plane to the cone contains the origin

22. mathstina Group Title

ok, i m trying to grasp.

23. TuringTest Group Title

It is a tricky one, I know. I used the concept of the plane equation heavily here. If you are not good with plane equations this is a tough thing to grasp. Let me know any specific parts you don't understand.

24. TuringTest Group Title

oh sorry, I did that same typo again :P should be$\nabla g(l,m,n)=2\langle l,m,-n\rangle$but the rest is correct...

25. mathstina Group Title

why ve to multiply by 2 for 2<l,m,n>?

26. TuringTest Group Title

$g(x,y,z)=x^2+y^2-z^2\implies\nabla g(x,y,z)=\langle 2x,2y,-2z\rangle=2\langle x,y,-z\rangle$

27. mathstina Group Title

ok. thanks. time for me to sleep. good bye.

28. TuringTest Group Title

welcome, g'night!

29. piscez.in Group Title

@TuringTest your solution is spot on, thank you

30. TuringTest Group Title

Let me try to elaborate so it is more understandable for @mathstina . To get the surface of the cone written as a function we need to get all variables on the same side$z^2=x^2+y^2\implies g(x,y,z)=x^2+y^2-z^2=0$Now we pick a point (actually a direction vector) on the surface of the cone $$\vec P_0=\langle l,m,n\rangle$$and find the gradient at that point$\nabla g(x,y,z)=2\langle x,y,-z\rangle\implies \nabla g(l,m,n)=2\langle l,m,-n\rangle$now let $$\vec P=\langle x,y,z\rangle$$ be a point in the tangent plane to the curve at $$\vec P_0$$. This implies that the vector $$\vec P-\vec P_0$$ is in the tangent plane. Any vector in the tangent plane should be perpendicular to the normal vector (the gradient), so their dot product is zero.$\nabla g(l,m,n)\cdot(\vec P-\vec P_0)=0$$2\langle l,m,-n\rangle\cdot\langle x-l,y-m,z-n\rangle=0$$2l(x-l)+2m(y-m)-2n(z-n)=0$Plugging in the origin to see if it satisfies both the plane and surface equations, at $$x=y=z=0$$ we see that the equation of the plane gives$-2l^2-2m^2+2n^2=0$$l^2+m^2-n^2=0$We can see that this means that these numbers are a solution to $$g(x,y,z)$$ and so are on the surface of the cone. Since l, m, and n are arbitrary, this means that every tangent plane to the cone contains the origin.