Point on line1 which is closest to line2

[ronho1234]

the question asks me to find a point on the first line which is closest to line 2

what is the general procedure for this? do i get l ine2-line1=my shortest distance?

 

[Curious3141]

Are you working in 2d or 3d?

If in 2d, 2 lines are either parallel or intersecting. If it's the latter, just determine the intersection point. If it's the former, then all points lie equidistant.

In 3d, you have the above 2 possibilities, and then you have skew lines which are non-parallel yet non-intersecting (and hence non-coplanar). This is a little more complicated. I believe it's easier to find the shortest distance between skew lines than it is to find the point of closest approach, but I'm not positive on this.

 

[HallsofIvy]

The line shortest line to a line is always perpendicular to that line. So the simplest way to find the shortest distance between two skew lines is to find a plane that is perpendicular to both lines. To do that, write the equation of a general plane perpendiular to the first line, then set the constants so it is also perpendicular to the second plane.

 

[ehild]

The line shortest line to a line is always perpendicular to that line. So the simplest way to find the shortest distance between two skew lines is to find a plane that is perpendicular to both lines. To do that, write the equation of a general plane perpendiular to the first line, then set the constants so it is also perpendicular to the second plane.

Is there a plane that is perpendicular to both skew lines?

ehild

 

[ronho1234]

okay i have a normal which is perpendicular to both these lines: 1/3 (-i+2j-2k)

and i found out that the shortest distance between my two lines is 3 or something

what do i do now? do i leave line 1 in terms of xyz and line 2 in terms of ijk, take l2-l1 and make it =3?

 

[ehild]

You mean a vector by I1-I2, the vector that connects one point of I1 to a point of I2. Its length has to be 3.

Add the normal vector multiplied by the shortest distance to line 2. It must intersect line 1. Find the point of intersection.

What are the equations of your lines? ehild

 

[ronho1234]

ehild

the question tells me my lines are in parametric vector form where
l1 is represented as r1=i+j-3k-6(2i+2j+k) and
l2 is represented as r2=-2i+2k-1(-j-k)

my normal: 1/3 (-i+2j-2k)
and the shortest distance is 3

i don't understand what you mean by the normal multiplied by the shortest distance to line 2 which intersects line 1, could you please show the working out... I'm so confused, thankyou

 

[HallsofIvy]

Is there a plane that is perpendicular to both skew lines?

ehild

In general, no. I don't know what I was thinking.

 

[ehild]

ehild

the question tells me my lines are in parametric vector form where
l1 is represented as r1=i+j-3k-6(2i+2j+k) and
l2 is represented as r2=-2i+2k-1(-j-k)

I do not see the parameters. In this form, I1 and I2 are vectors, instead of lines.

If you shift one of the lines in the direction along the common normal vector with the distance of the skew lines, it will intersect the other line. That shift means adding or subtracting D=3n to the parametric form of the line. The point of intersection is that point which is at shortest distance from the other line. The point of intersection defines both parameter values, so you get the points on both lines which are closest to each other. I try to explain it with the drawing attached. The black D vector is the normal vector times distance.
ehild

 

[ehild]

In general, no. I don't know what I was thinking.

I think you meant a line perpendicular to both lines.

ehild

 

[HallsofIvy]

There still doesn't necessarily exist such a thing!

 

[ehild]

As there is no answer from the OP I try to give more help. The unit normal vector of both lines is (as given by the OP) n=1/3 (-i+2j-2k), and the distance of the lines is 3. When he parametric form of line 2 is r2=-2i+2k+s(j+k), adding (or subtracting) 3n results in a new line parallel with the former and at distance 3, which must intersect line 1. r3=r2+3n=-3i+(2+s)j+sk. If the equation of line 1 is r1=i+j-3k-p(2i+2j+k)=(1-2p)i+(1-2p)j-(3+p)k, it has to intersect r3:-3i+(2+s)j+sk=(1-2p)i+(1-2p)j-(3+p)k Comparing the components, -3=1-2p, 2+s=1-2p, s=-(3+p) This system of equation is solvable, so the lines intersect: we get the values for p and s, and substituting for p into the equation of line 1 we get the point which is closest to line 2.

ehild