What is the distance between the line with Cartesian equation x−4=y−5=z and the line with Cartesian equation x−2=y−5=z−1? I am not sure how to do it. I think the lines are parallel with the same direction but I'm not sure.

opositor5t

opositor5t

Answered question

2022-08-11

What is the distance between the line with Cartesian equation x−4=y−5=z and the line with Cartesian equation x−2=y−5=z−1?
I am not sure how to do it. I think the lines are parallel with the same direction but I'm not sure.

Answer & Explanation

Barbara Klein

Barbara Klein

Beginner2022-08-12Added 19 answers

HINT
The vector equation of the first line can be written as
L 1 ( a ) = ( a , a + 1 , a 4 ) = ( 0 , 1 , 4 ) + a ( 1 , 1 , 1 )
On the other hand, the vector equation of the second line can be written as
L 2 ( b ) = ( b , b + 3 , b 1 ) = ( 0 , 3 , 1 ) + b ( 1 , 1 , 1 )
Consequently the lines L 1 and L 2 are parallel.
In order to calculate the distance between them, let us take (0,3,−1)−(0,1,−4)=(0,2,3) .
Hence the distance d can be expressed as
d = ( 0 , 2 , 3 ) 1 3 ( 0 , 2 , 3 ) , ( 1 , 1 , 1 ) ( 1 , 1 , 1 )
Can you take it from here?
empalhaviyt

empalhaviyt

Beginner2022-08-13Added 1 answers

The direction vector of both lines is (1,1,1) and the unit direction vector
v ^ = 1 3 ( 1 , 1 , 1 )
Now take a point A on line 1 and B on line 2. The easiest from the equation of lines are A(4,5,0) and B(2,5,1).
Given both lines are parallel, the distance d between the lines is then simply | A B × v ^ | (this is because if the angles between vectors is θ, cross product will have the magnitude | A B sin θ | which is the perpendicular distance between the lines).
So d = 1 3 | ( 2 , 0 , 1 ) × ( 1 , 1 , 1 ) |

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