Find all points where the level surface tangent plane is parallel

[Addez123]

Homework Statement

Given the level surface $$x^2 + y^2 + z^2 = 5$$ find all points where the tangentplane is parallel to the plane: $$x - 2y + 3z = 13$$

Relevant Equations

Normal vector = grad(curve)
$$x^2 + y^2 + z^2 = 5$$
$$x - 2y + 3z = 13$$

First I find the normal vector given any position:
$$w(x, y, z) = x^2 + y^2 + z^2$$
$$∇w(x, y, z) = (2x, 2y, 2z)$$

Normal vector of plane:
$$w_2 = x - 2y + 3z$$
$$∇w_2 = (1, -2, 3)$$

$∇w = ∇w2$ => point where planes are parallel = (1/2, -1, 3/2)

This is completely off, but I can't find any help on how to solve this anywhere on youtube.

 

[member 587159]

Some questions for you:

- What is the tangent plane at a different point $(x_0,y_0,z_0)$ on your surface?
- When are two planes with standard equation parallel?

 

[Addez123]

Some questions for you:

- What is the tangent plane at a different point $(x_0,y_0,z_0)$ on your surface?
- When are two planes with standard equation parallel?

1. Assuming x_0, y_0, z_0 is on the level surface, the equation is:
$$ 2(x - x_0) + 2(y - y_0) + 2(z - z_0) = 5 $$
2. They are parallel when their normal vectors are scalars of each other.

 

[member 587159]

1. Assuming x_0, y_0, z_0 is on the level surface, the equation is:
$$ 2(x - x_0) + 2(y - y_0) + 2(z - z_0) = 5 $$
2. They are parallel when their normal vectors are scalars of each other.

Well, think about it geometrically. $x^2 + y^2 + z^2 = 5$ is the equation of the surface of a sphere at the origin with radius $\sqrt{5}$. You can easily visualise how a tangent plane at a sphere looks like. For example, without even calculating anything, you see that the equation of the tangent plane at $(0,0, \sqrt{5})$ must be a translate of the $x,y$-plane. Thus the tangent plane at $(0,0,\sqrt{5})$ is given by the equation $z= \sqrt{5}$.

However, the formula you gave says that the tangent plane at this point is $2x + 2y +2(z-\sqrt{5}) = 5$, which is not the same thing.

This is a good illustration that shows that thinking geometrically can tell you quickly whether you are on the right track or not.

Now, we identified where your mistake lies. You calculated wrongly the equation of the tangent plane!

Can you correct yourself?

 

[Addez123]

I can see that it's incorrect but not how to solve it.
For one, the result should be equal to zero, because were doing the dot product of vectors within the plane against the normal vector.
Secondly, our normal vector at point $$(x_0, y_0, z_0)$$ is $$2(x_0, y_0, z_0)$$
Then my equation becomes
$$2x_0(x−x_0)+2y_0(y−y_0)+2z_0(z−z_0)=0 $$

But honestly I can't see why I need to do this at all. The only thing I need to do is align the normal vectors and I got both:
$$∇w=(2x,2y,2z)$$
$$∇w_2=(1,−2,3) $$

Why can't I just set them equal to each other?