Above-Below Test in 3D: Solving for Line L1

[totototo]

Hi;
If I have a line L1 first end point of L1(a1,a2,a3) and the second end point of L1(b1,b2,b3) in 3-dimensions.
From them I could get the eqaution of the line in 3-dimensions which is :
(X-a1/b1)=(y-a2/b2)=(z-a3/b3) if it is wrong please do let me know
I have an example:
first end point of L1(2,0,0) & the second end point of L1(-2,0,0)
note:(a1,a2,a3) for the first end point and (b1,b2,b3) for the second end point
I want to test the point(2,-2,2)(which refers to (x,y,z)) if it is above or below the line L1
so I pluged in the values in the above eqaution and I got negative sign ,I tried another point which is on the other side(-2,2,-2) but it gave negative sign which means I have something wrong could you please help me in this?


Thanks
totototo

 

[pat_connell]

Your equation for the line in 3-dimensions is not quite correct. The equation should be: (x-a1)/(b1-a1)=(y-a2)/(b2-a2)=(z-a3)/(b3-a3). Using the example with first end point of L1(2,0,0) and the second end point of L1(-2,0,0): For the point (2,-2,2) you get (2-2)/(-2-2)=(-2-0)/(0-0)=(2-0)/(0-0), which gives -1 = -1 = 1, so this point is on the line. For the point (-2,2,-2) you get (-2-2)/(-2-2)=(2-0)/(0-0)=(-2-0)/(0-0), which gives 0 = 1 = -1, so this point is not on the line.

 

[tacman]

Hi totototo,

Your approach to solving for line L1 in 3D is correct. The equation you have provided, (X-a1/b1)=(y-a2/b2)=(z-a3/b3), is known as the parametric form of a line and can be used to represent a line in 3D space.

For the example you have provided, the equation for line L1 would be: (X-2/-4)=(y-0/0)=(z-0/0). However, there seems to be a mistake in your calculation for the second end point of L1. The coordinates you have provided, (-2,0,0), do not match the equation you have given. The correct equation for the second end point would be (-2/4,0/0,0/0).

To test if a point is above or below a line in 3D, we can use the dot product of the vector formed by the two end points of the line and the vector formed by the two end points and the given point. If the dot product is positive, the point is above the line. If it is negative, the point is below the line.

Using this method, we can see that the point (2,-2,2) is indeed below the line L1, as the dot product is negative. The point (-2,2,-2) is also below the line, as the dot product is also negative. This means that your calculations are correct and there is no mistake in your approach.

I hope this helps clarify any confusion. Let me know if you have any further questions or concerns. Good luck with your 3D calculations!