# Question about vectors 3D spaceWatch

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#1
I was wondering why when finding the vector with minimum distance (magnitude) between two skew lines in 3D space that the vector is always perpendicular to each line? I’m finding it difficult to visualise in my head and was wondering if there was a website to see this visually or an explanation that could help?
Last edited by Maximus 190; 1 week ago
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1 week ago
#2
(Original post by Maximus 190)
I was wondering why when finding the vector with minimum distance (magnitude) between two skew lines in 3D space that the vector is always perpendicular to each line? I’m finding it difficult to visualise in my head and was wondering if there was a website to see this visually or an explanation that could help?
just imagine a line passing near the origin.
Draw circles of increasing radius (centered on the origin) until they just touch the line (the line is a tangent)
That is the minimum distance from the origin and the line is perpendicular to the radius which represents the distance.
https://www.desmos.com/calculator/94f1dwt1np

Another way to think of it is to consider a point on the line and form a right angled triangle where the line connecting the origin to that point is the hypotenuse and the adjacent side connects the origin perpendicularly to the line (The opposite side is the line segment along the original line). By definition, the hypotenuse is always larger than the adjacent, unless they're equal.

Same for 2 lines except you have a "tangent" at each end.
Last edited by mqb2766; 1 week ago
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#3
(Original post by mqb2766)
just imagine a line passing near the origin.
Draw circles of increasing radius (centered on the origin) until they just touch the line (the line is a tangent)
That is the minimum distance from the origin and the line is perpendicular to the radius which represents the distance.
https://www.desmos.com/calculator/bdov40kkeo

Same for 2 lines except you have a "tangent" at each end.
Oh yeah that’s interesting. I like that! And obviously with a sphere in 3D space, the centre will always be the midpoint of the vector between the two lines with the vector between the two tangents being a diameter through the sphere. Thank you! 0
1 week ago
#4
(Original post by Maximus 190)
Oh yeah that’s interesting. I like that! And obviously with a sphere in 3D space, the centre will always be the midpoint of the vector between the two lines with the vector between the two tangents being a diameter through the sphere. Thank you! Its used in the definition of a line (2D) and plane (3D). Imagine a line in described by
ax + by = c
where a^2+b^2=1 (unit magnitude). Then c is the perpendicular distance of the line from the origin. Obviously, when the (a,b) coefficients are not unit magnitude, you can nomalise them (and the right hand side) by dividing by the value. Its a cheap way to determine the distance of a line from the origin. The vector (a,b) is the unit normal or the direction of the perpendicular connector.

Same with a plane in 3D
ax + by + cz = d
when a^2 + b^2 + c^2 = 1, then d is the perpendicular distance from the origin and as you say is the radius of a sphere which has the plane as a tangent. The vector (a,b,c) is the unit normal or the direction of the perpendicular connector.
Last edited by mqb2766; 1 week ago
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1 week ago
#5 By Pythagoras theorem the length of the hypotenuse in black is Now This means as that hypotenuse gets closer and closer to the black line its length gets smaller. In other words as the line gets closer to the normal (perpendicular line) it's size gets smaller. Now imagine this triangle spinning around a z axis, then the distance between two points can be thought of as calculating the distance on a 2D slice (plane), so the same rule applies in 3D.
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1 week ago
#6
if you didn't set off perpendicular to the line then you are making some component of the displacement along the line, so you could just start there instead and get an even shorter distance
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#7
(Original post by Meowstic)
if you didn't set off perpendicular to the line then you are making some component of the displacement along the line, so you could just start there instead and get an even shorter distance
Yeah, I guess if I accept that there will always be a vector perpendicular to each line then this is the simplest answer, thanks
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