Robot rail and laser safety curtain
A warehouse robot travels along a straight overhead rail modelled by r(s)=(4,1,2)+s(1,2,0), where distances are in metres and s is a real parameter. A laser safety curtain lies in the plane E: 2x - y + 2z = 6. The robot is approved to operate only where its perpendicular distance from the curtain is greater than 2 m. The current rail is fixed, but the entire rail can be translated along a normal vector of E without changing its direction. Maintenance also needs the point on the rail closest to the charging station at the origin, because a cable will be dropped vertically from that point. The rail is long enough that the parameter is not restricted by the endpoints. Determine whether the current path is approved, find the smallest normal translation that gives a 3 m clearance, and locate the closest approach to the origin.
1. Show that the rail is parallel to the plane E and compute its perpendicular distance from E. Is the current path approved?
Read the worked answer
A normal vector of E is n=(2,-1,2), and the direction vector of the rail is d=(1,2,0). Their scalar product is
d·n=1·2+2·(-1)+0·2=2-2+0=0.
Since d is perpendicular to n, the rail is parallel to E. Therefore every point on the rail has the same signed distance from E; use A=(4,1,2). The signed numerator is
2(4)-1+2(2)-6=8-1+4-6=5.
The length of n is sqrt(2^2+(-1)^2+2^2)=sqrt9=3, so the distance is |5|/3=5/3≈1.67 m. This is less than the required 2 m clearance, so the current path is not approved.
2. The rail is translated parallel to a normal vector of E. Find the smallest translation that gives a perpendicular distance of 3 m, and write the new line equation.
Read the worked answer
The signed distance must change from 5/3 to 3. The smallest change is therefore 3-5/3=4/3 in the direction of the normal. A unit normal is n/|n|=(2,-1,2)/3, so the translation vector is
(4/3)(2,-1,2)/3=(8/9,-4/9,8/9).
Applying this to the position vector (4,1,2) gives
(4+8/9, 1-4/9, 2+8/9)=(44/9,5/9,26/9).
So the new line is
r(s)=(44/9,5/9,26/9)+s(1,2,0).
Check: the numerator at the new point is 2(44/9)-5/9+2(26/9)-6 = (88-5+52-54)/9=81/9=9, and 9/3=3. The direction is unchanged, so the whole rail has distance 3 m.
3. Find the point on the original rail that is closest to the origin O=(0,0,0), and give the minimum distance.
Read the worked answer
For a point on the original rail, r(s)=(4+s,1+2s,2). Its squared distance from O is
D(s)=(4+s)^2+(1+2s)^2+2^2 =16+8s+s^2+1+4s+4s^2+4 =21+12s+5s^2.
Minimise D(s): D'(s)=12+10s, so s=-12/10=-1.2. Since D''(s)=10>0, this is a minimum. The closest point is
r(-1.2)=(4-1.2, 1-2.4, 2)=(2.8,-1.4,2).
The minimum squared distance is
D(-1.2)=21+12(-1.2)+5(1.44)=21-14.4+7.2=13.8,
so the minimum distance is sqrt(13.8)≈3.72 m. This point is still at the same 5/3 m distance from the laser plane, since the rail is parallel to the plane.