I have a line in the vector form: (x,y,z)=t(1,1,1)+(0,0,1) and I need to find a vector that represents all the points whose distance from this line equals 1. I started calculating the distance between a point (a,b,c) and this line using the vetorial product... and I got a big equation in terms of a, b and c. I think this equation describes a cylinder whose centerline is (x,y,z)=t(1,1,1)+(0,0,1). But I don't know how to get rid of this equation and obtain the vector...

Claire Calderon

Claire Calderon

Open question

2022-08-22

I have a line in the vector form: ( x , y , z ) = t ( 1 , 1 , 1 ) + ( 0 , 0 , 1 ) and I need to find a vector that represents all the points whose distance from this line equals 1. I started calculating the distance between a point ( a , b , c ) and this line using the vetorial product... and I got a big equation in terms of a, b and c. I think this equation describes a cylinder whose centerline is ( x , y , z ) = t ( 1 , 1 , 1 ) + ( 0 , 0 , 1 ). But I don't know how to get rid of this equation and obtain the vector...
Can anyone help me?

Answer & Explanation

muilasqk

muilasqk

Beginner2022-08-23Added 10 answers

Actually, there are infinitely many points as 1 unit far from the line. For those points, namely having coordinates (p,q,r), there exists a unit vector (a,b,c) normal to the vector of line (1,1,1) and a point (x,y,z) on the line such that
( p , q , r ) = ( x , y , z ) + ( a , b , c ) .
The unicity of ( a , b , c ), along its normality implies
a 2 + b 2 + c 2 = 1 ( a , b , c ) ( 1 , 1 , 1 ) = a + b + c = 0 ,
which yields
c = a b a 2 + b 2 + a b = 1 2 b = a ± 2 3 a 2 2 c = a 2 3 a 2 2 .
Since the point ( x , y , z ) is arbitrary, the equation of all points far from the line as much as 1 is
( t + a , t + a ± 2 3 a 2 2 , 1 + t + a 2 3 a 2 2 )
for | a | 2 3

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