Finding Endpoints on a 3D Vector | Simple Method and Formula

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To locate endpoints on a 3D vector defined by two points, the midpoint can be calculated as [(a,b,c)+(d,e,f)]/2. The equation for any point along the line is given by x = (d - a)s + a, where 's' is a scalar. To find the endpoints of a linear object centered at the midpoint and oriented along the vector, use the formula C ± k(D - A), where C is the midpoint and k is a scalar multiple. This method allows for both algebraic and geometric approaches to determine the endpoints. Understanding these concepts will facilitate finding the desired points on the vector.
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I'm trying to find the simplest way to locate points on a 3D vector.

I have 2 points (a,b,c) (d,e,f) which define a 3D vector. I know the midpoint between those points. [(a,b,c)+(d,e,f)]/2

I have a linear "object" with a known length L and I want to find the endpoints (u,v,w),(x,y,z) of that object centered at the midpoint and oriented along the vector.

Thanks!
 
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It sounds like you want to find points on the line defined by the two points you mention first. In that case, note that every point x = (x, y, z) on that line satisfies the equation x = (d - a)s + a for some number s, where a = (a, b, c) and d = (d, e, f). After some thought you should see why. You should also then see that the constant vector a in the above equation can be any position vector on the line, including the "midpoint" you mentioned previously. Does this help with your question?
Of course, you can also do this geometrically, without referring to the algebraicization. Just draw the triangle created by the position vectors a and d and the rest should follow.
 
fenpark15 said:
I have a linear "object" with a known length L and I want to find the endpoints (u,v,w),(x,y,z) of that object centered at the midpoint and oriented along the vector.

Hi fenpark15! :smile:

Hint: If the centre is C, and the endpoints are A and D, then the new endpoints will be C ± a multiple of (D - A). :wink:
 

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