animatrix_
Hi,

You can do it like this:

vector getPerpendicularVector ( vector v )
{
    vector vn = normalize ( v );
    vector vu = abs ( vn.x ) < abs ( vn.z ) ? { 1, 0, 0 } : { 0, 0, 1 };

    return normalize ( cross ( vn, vu ) );
}

vikus
This should also work

vector n = normalize(set(p.z, 0, -p.x));
addpoint(0, p + n);

vector n = normalize(set(@P.z, 0, -@P.x));
addpoint(0, @P + n);

vikus
It normal of a 2D vector. Fliping the axis of P and negate one of them. Hope the image explains .

Image Not Found

Works in 3D as well you can build stable rotaion on that.

Kareeem
Thanks that definitely makes more sense now!

vikus

Kareeem
Thanks that definitely makes more sense now!

Glad to help.

Here is an older file with rotation, scale where I lern the stuff:

Image Not Found

Kareeem
This actually works, thanks! I needed to fix it up a bit though

vector n = normalize(set(@P.z, 0, -@P.x));

I´ll try to figure out what you are doing here ... I mean I get what you are doing but how on earth would one think of that?

vector n = normalize( cross( @P, {0,-1,0} ) );

tamte

Kareeem
This actually works, thanks! I needed to fix it up a bit though

vector n = normalize(set(@P.z, 0, -@P.x));

I´ll try to figure out what you are doing here ... I mean I get what you are doing but how on earth would one think of that?

you can also write the same thing as :

vector n = normalize( cross( @P, {0,-1,0} ) );

but once you expand the math of the cross product that's what's left

Tanto
There is an infinity of vectors perpendicular to your direction. Cross product is indeed what you need. If any perpendicular vector will do, just use any vector as the second argument of your cross product. If you need a specific one, try to figure out on which plane you want it to be and use the normal of that plane as your second cross product vector.

This [en.wikipedia.org] might be of help.

Kareeem

animatrix_
Hi,

You can do it like this:

vector getPerpendicularVector ( vector v )
{
    vector vn = normalize ( v );
    vector vu = abs ( vn.x ) < abs ( vn.z ) ? { 1, 0, 0 } : { 0, 0, 1 };

    return normalize ( cross ( vn, vu ) );
}

Wow, this seems advanced. I am having trouble understanding even the first line. What is this syntax? A short version for something?
Thanks for taking the time though, much appreciated!

animatrix_
It's ternary operator [en.m.wikipedia.org].

Kareeem

animatrix_
It's ternary operator [en.m.wikipedia.org].

Hey thanks for clarifying! Although I did not know that term I was aware of the"short version" of the if statement.
But I am still not sure what you are doing in the first line ... Are you creating a custom function here?

animatrix_
Hi,

You can do it like this:

vector getPerpendicularVector ( vector v )
{
    vector vn = normalize ( v );
    vector vu = abs ( vn.x ) < abs ( vn.z ) ? { 1, 0, 0 } : { 0, 0, 1 };

    return normalize ( cross ( vn, vu ) );
}

// calc dir from start to end
vector start = point(0, "P", 0);
vector end = point(0, "P", 1);
vector dirn = normalize(end-start);
vector diru;


// calc math magic
if (abs(dirn.x) < abs(dirn.z)) {
    diru = {1,0,0};
}
else{
    diru = {0,0,1};
}

// export @
v@N = normalize(cross(dirn, diru));

animatrix_
The abs line picks the axis that is least aligned with vn, ensuring the cross product produces a well-conditioned perpendicular vector with a stable magnitude.

If vn is already close to { 1, 0, 0 } (X-axis), then { 1, 0, 0 } is a bad choice for vu because the cross product would produce a very small (almost zero) vector, leading to numerical issues.

If vn is closer to { 0, 0, 1 } (Z-axis), then { 0, 0, 1 } is a bad choice for the same reason.