Question

shortest distance between the lines
(x+1)/7=(y+1)/-6=z+1/1
and
(x-3)/1=(y-5)/-2=(z-7)/1

Answer 1

@amistre64 plz help!

Answer 2

if no one else comes along, ill be able to look at this after classes, which is in about 5 hours :/

Answer 3

okay! will await ur answer!! Big thanks in advance!

Answer 4

ok, im free again for the moment :)

Answer 5

the shortest distance is going to be when the line (a vector) between them is perp to both; to get a perp vector we have to cross the 2 given line vectors

Answer 6

(x+1)/7=(y+1)/-6=z+1/1; <7,-6,1>

(x-3)/1=(y-5)/-2=(z-7)/1; <1,-2,1>

x 7 1 x:-6--2
y -6 -2 -y: 7 - 1
z 1 1 z:-14--6

perp vector to both lines is: <4,6,8>, or reduced to <2,3,4> if need be

Answer 7

so, what im thinking now is that we can use this perp to define the planes that the lines live in; and then we can determine the distance between the planes as the distance between the lines

Answer 8

(x+1)/7
(y+1)/-6
(z+1)/1 :point is (-1,-1,-1)

(x-3)/1
(y-5)/-2
(z-7)/1; point is (3,5,7)

or as a side observation, it looks like we can just take those xyz parts and nestle them into the perp vector :)

2(x+1)+3(y+1)+4(z+1)=0
2x+3y+4z +10=0

2(x-3)+3(y-5)+4(z-7)=0
2x+3y+4z - 49 = 0

Answer 9

now, there is a formula floating out there that you can use if you want to to find the distance between the planes, but im not much for formulas/

Answer 10

one idea i have is to go about this similar to the last one i help you one, take the perp vector and a point from one of the planes, and see where it goes thru the other plane, and t will be the distance between them

Answer 11

oh, and t will be the scalar of the vector, so the length of the vector 2t,3t,4t will be the distance

x=-1+2t
y=-1+3t
z=-1+4t

2(-1+2t-3)+3(-1+3t-5)+4(-1+4t-7)=0

-8 + 4t -18 +9t-32+16t=0

t(29) = 58

t = 58/29

Answer 12

distance: sqrt(4t^2+9t^2+16t^2)

sqrt(t^2(4+9+16))

t sqrt(29)

Answer 13

you might wanna go thru and make sure my concepts are sound, and that i didnt mismath it along the way ;)