how to find a plane perpendicular to a line?

X = [126.3798 126.3818]
Y = [37.5517 37.5495]
Z = [539.5 531.5]

line(X,Y,Z)
view(45,40)

I want to find a plane or line perpedicular to a blue line.

I thinks i shoud use normal vector.

how to do this?

Thanks.

Answer 1

[Torsten]

[-126.3798+126.3818,-37.5517+37.5495,-539.5+531.5]*[x;y;z] = [-126.3798+126.3818,-37.5517+37.5495,-539.5+531.5]*[126.3818;37.5495;539.5]

if the indicated point in your plot lies in the plane.

[Sierra]

I found my question is wrong. I uploaded another question.

Sorry Torsten

https://kr.mathworks.com/matlabcentral/answers/1832348-how-to-find-a-line-perpendicular-to-another-line

Answer 2

[John D'Errico]

What you need to understand is a plane is defined by TWO things. The normal vector to the plane, AND a point on the plane. That is, if you know only the normal vector, then there are infinitely many planes normal to that line, since you could slide the plane along the line.

So having provided only the line, perhaps as defined by two points on the line, then you already have the normal vector! The line itself defines that vector, since the plane is normal to the line. If your points are given by:

X = [126.3798 126.3818];
Y = [37.5517 37.5495];
Z = [539.5 531.5];

then the normal vector is just

N = [X(2) - X(1), Y(2) - Y(1), Z(2) - Z(1)]

N = 1×3

0.0020 -0.0022 -8.0000

You can arbitrarily pick any point (x,y,z) on the line, and that will then define a specific plane. We might pick, for example, the first of those two points, as

P = [X(1), Y(1), Z(1)]

P = 1×3

126.3798 37.5517 539.5000

In conjunction with the normal vector itself, that gives you the plane, as the locus of points normal to the line, and passing through that point on the line. Essentially, the plane is then the set defined by the equation

dot(XYZ - P,N) == 0

[Sierra]

I found my question is wrong. I uploaded another question.

Sorry John

https://kr.mathworks.com/matlabcentral/answers/1832348-how-to-find-a-line-perpendicular-to-another-line