Error with dot product? Dot product of cross product vectors is not 0

I am attempting to find a vector normal to a plane. Provided two vectors, and in the plane, a vector normal to both is found using the cross product or . Therefore, and . Have I entered something incorrectly?

%gps reported positions (xSec_DDMMYYYY)
x0_18081997 = [3325396.441 5472597.483 -2057129.050];
x3_18081997 = [3309747.175 5485240.159 -2048664.333];
x0_19081997 = [4389882.255 -4444406.953 -2508462.520];
x3_19081997 = [4402505.030 -4428002.728 -2515303.456];

%determine vector normal to orbital plane
n_18 = cross(x0_18081997,x3_18081997);
check_18 = dot(n_18, x0_18081997);
if ~check_18==0
    error('n_18 is not orthogonal to plane.')
end

[Christopher Simpson]

There's some odd behavior where if one of the three elements in either vector is 0 the cross product returns an orthogonal vector.

x0 = [x0_18081997(1) x0_18081997(2) x0_18081997(3)]; 
x1 = [0              x3_18081997(2) x3_18081997(3)];
ab = cross(x0,x1);
dotab = dot(ab,x0)

But if we use the original vector elements, the cross product no longer returns an orthogonal vector.

x0 = [x0_18081997(1) x0_18081997(2) x0_18081997(3)]; 
x1 = [x3_18081997(1) x3_18081997(2) x3_18081997(3)];
ab = cross(x0,x1);
dotab = dot(ab,x0)

It is assumed that you've included the script in the original question.

Answer 1

[KSSV]

How about trying the same with unit vectors?

%gps reported positions (xSec_DDMMYYYY)
x0_18081997 = [3325396.441 5472597.483 -2057129.050];
x3_18081997 = [3309747.175 5485240.159 -2048664.333];
x0_19081997 = [4389882.255 -4444406.953 -2508462.520];
x3_19081997 = [4402505.030 -4428002.728 -2515303.456];
%determine vector normal to orbital plane
x0_18081997 = x0_18081997/norm(x0_18081997) ;
x3_18081997 = x3_18081997/norm(x3_18081997) ;

n_18 = cross(x0_18081997,x3_18081997);
check_18 = dot(n_18, x0_18081997);
if ~abs(check_18)>=10^-6
    error('n_18 is not orthogonal to plane.')
end

[Christopher Simpson]

Sure, but it still doesn't equal 0 when you take the dot product of orthogonal vectors. You even introduced a new check for an relative error of in your suggested code.

%gps reported positions (xSec_DDMMYYYY)
x0_18081997 = [3325396.441 5472597.483 -2057129.050];
x3_18081997 = [3309747.175 5485240.159 -2048664.333];
x0_19081997 = [4389882.255 -4444406.953 -2508462.520];
x3_19081997 = [4402505.030 -4428002.728 -2515303.456];

%determine vector normal to orbital plane
n_18 = cross(x0_18081997/norm(x0_18081997),x3_18081997/norm(x3_18081997));
check_18 = dot(n_18, x0_18081997);
if ~check_18==0
    error('n_18 is not orthogonal to plane.')
end

The only difference between this code and the original is the check value.

Error using p2_script (line 31)
n_18 is not orthogonal to plane.
 
>> check_18

check_18 =

  -7.6398e-11

When not unit vectors:

Error using p2_script (line 31)
n_18 is not orthogonal to plane.
 
>> check_18

check_18 =

        4064

[KSSV]

-7.6398e-11 this is zero ......

[Christopher Simpson]

Is it an analytical function or numerical?

[Christopher Simpson]

I get that you're trying to say that it's basically 0.

• Why are we left with some trace value?
• Why only when we reduce the magnitude are we able to escape the error?
• Am I going to need to reduce every vector I use a cross product for to a unit vector first?
• I thought this was an analytical function. Why would we get some machine remainder for 0?