Question

Find the distance between the two skew lines

x=1+t, y=-2+3t, z=4-t
and
x=2s, y=2+s, z=-3+4s

Answer 1

gotta get my yard stick out of the closet for this one

Answer 2

my first thought is to create a plane from one of the lines; using the direction vector is its normal

Answer 3

x= 1+t
y=-2+3t ; <1,3,-1>
z =4-t


x=2s
y=2+s ; <2,1,4>
z=-3+4s

if we cross these im thinking we get a vector in a plane that is perp to them both

Answer 4

x 1 2 x= 12--1 = 13
y 3 1 -y=(4--2) = -6 ; <13,-6,-5>
z -1 4 z= 1-6 = -5


hmmmm

Answer 5

im thinking i would want to create a plane with this normal and a point from each line; create 2 planes

Answer 6

x= 1+t
y=-2+3t ; P (1,-2,4)
z =4-t

13(x-1) -6(y--2)-5(z-4) = 0


x=2s
y=2+s ; Q (0,2,-3)
z=-3+4s

13(x-0) -6(y-2)-5(z--3) = 0

Answer 7

plane1: 13x-6y-5z-5=0

plane2: 13x -6y-5z-3 = 0

Answer 8

since these planes are the same distance apart from each other at all points; we can construct a line from one plane to the other like we did in the last part ... and find t

Answer 9

x= 1+13t
y= -2 - 6t ; from plane 1
z= 4 - 5t

into plane 2

13(1+13t) -6(-2-6t) -5(4-5t) -3 = 0

13+13^2 t +12+6^2 t -20 +5^2 t -3 = 0

(13+12 -20-3) +t(13^2 +6^2 +5^2) = 0

t(13^2 +6^2 +5^2) = -(13-12-20-3)

t = -(13-12-20-3)/(13^2 +6^2 +5^2)

d = t* = (13-12-20-3)/sqrt(13^2 +6^2 +5^2)

Answer 10

the negative sign is spurious again; so ... 1.45 is my best attempt :)

Answer 11

see you got your parametrics

Answer 12

:) and then some

Answer 13

the distance from -5 to -3 gives us the vertical there if my planes are right: 2