MHB Field lines of 3D vector functions

[brunette15]

My question regards finding the field lines of the 3D vector function F(x,y,z) = yzi + zxj + xyk.

I was able to compute them to be at the curves x^2 - y^2 = C and x^2 -z^2 = D, where C and D are constants.

From my understanding the field lines will occur at the intersection of these two curves, however I am struggling to understand why the field lines occur at the intersection. If anyone can please just clarify or give me more of an insight as to what is actually happening with the field lines of this vector function that would be very helpful :)

 

[I like Serena]

My question regards finding the field lines of the 3D vector function F(x,y,z) = yzi + zxj + xyk.

I was able to compute them to be at the curves x^2 - y^2 = C and x^2 -z^2 = D, where C and D are constants.

From my understanding the field lines will occur at the intersection of these two curves, however I am struggling to understand why the field lines occur at the intersection. If anyone can please just clarify or give me more of an insight as to what is actually happening with the field lines of this vector function that would be very helpful :)

Hi brunette15! ;)

If we pick a point on the intersection curve and consider it part of a field line, we can try and see in which direction that field line has to go.
The normal of the first surface is given by the gradient of $x^2 - y^2$, which is $2x\boldsymbol{\hat\imath} - 2y\boldsymbol{\hat\jmath}$.
That just happens to be perpendicular to the given direction $\mathbf F(x,y,z)$ of a field line, which we can tell from their dot product, which is zero.
The same holds for the second surface.
In other words the direction of the intersection curve is parallel to the direction of the field line.
That means that the field line coincides with the curve of intersection.

 

[brunette15]

Hi brunette15! ;)

If we pick a point on the intersection curve and consider it part of a field line, we can try and see in which direction that field line has to go.
The normal of the first surface is given by the gradient of $x^2 - y^2$, which is $2x\boldsymbol{\hat\imath} - 2y\boldsymbol{\hat\jmath}$.
That just happens to be perpendicular to the given direction $\mathbf F(x,y,z)$ of a field line, which we can tell from their dot product, which is zero.
The same holds for the second surface.
In other words the direction of the intersection curve is parallel to the direction of the field line.
That means that the field line coincides with the curve of intersection.

That helps a lot! Thankyou so much! :D