Find all points on the surface where the normal line passes through the origin I have $$x^4+y^4+z^4=3$$
I need to find all points on this surface that passes through the origin.
Here's what I did:
$$\nabla f = \langle 4x^3,4y^3,4z^3 \rangle$$
Let any point on this surfaec be $P(x_0,y_0,z_0)$, and define vector $\vec{OP}=\langle x_0,y_0,z_0 \rangle$ where $O$ represents the origin
Since the gradient vector is normal to the surface, I was thinking to find a line such that $\langle x,y,z\rangle = \vec{OP}+t\nabla{f(P)}$ that pass through origin, while satisfying the constraint $x^4+y^4+z^4=3$
I made some guess like $\langle 1,1,1 \rangle$, but not sure how to generally approch this question
What should I do?

Since the normal vector passes through $(0,0,0)$ and $(x, y, z)$ then
$( 4 x^3, 4 y^3 , 4 z^3 ) = \alpha (x, y, z) $ for some $\alpha$
Hence, $ \alpha = 4 x^2 = 4 y^2 = 4 z^2 $
From which $|x| = |y| = |z| $
The solution of which is $(x, y, z) = \beta (\pm 1, \pm 1, \pm 1) $
Plugging this into $x^4 + y^4 + z^4 = 3 $ we deduce that $\beta = \pm 1$
Hence there are $8$ solutions for $(x,y,z)$ and they are $(\pm 1, \pm 1, \pm 1)$
