ima calculate distance between line L1=(x,y,z)=(3+t,1+t,2-2t) t is rreal number and line L2 its a linear equation x+2=0 is a top and bottom y=3

my progress:

L1=(3,1,2)+t(1,1,-2)

L2 z=s x=-s y=3 (x,y,z)=(0,3,0)+s(-1,0,1)

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- Dec 15th 2012, 08:37 AMPetrusdistance
ima calculate distance between line L1=(x,y,z)=(3+t,1+t,2-2t) t is rreal number and line L2 its a linear equation x+2=0 is a top and bottom y=3

my progress:

L1=(3,1,2)+t(1,1,-2)

L2 z=s x=-s y=3 (x,y,z)=(0,3,0)+s(-1,0,1) - Dec 15th 2012, 08:42 AMPlatoRe: distance
- Dec 15th 2012, 08:47 AMPetrusRe: distance
Here is a pic L2Attachment 26245

- Dec 15th 2012, 09:18 AMPlatoRe: distance
- Dec 15th 2012, 10:13 AMPetrusRe: distance
ik that Q-P=-3,2,-2 but i dont know how to calculate D x E

- Dec 15th 2012, 10:21 AMPlatoRe: distance
- Dec 15th 2012, 10:38 AMPetrusRe: distance
i get it to sqrt(17)/sqrt(3) is this wrong? it looks so for me :S

- Dec 15th 2012, 10:46 AMPlatoRe: distance
- Dec 15th 2012, 10:49 AMPetrusRe: distance
1i+2j+0k-(-1)k-1j-0i 1,1,1

- Dec 15th 2012, 10:56 AMPlatoRe: distance
- Dec 15th 2012, 10:59 AMPetrusRe: distance
- Dec 15th 2012, 11:33 AMPlatoRe: distance
- Dec 15th 2012, 11:39 AMPetrusRe: distance
-3,2-2 then i take sqrt(-3^2+2^2+-2^2)?

- Dec 15th 2012, 11:51 AMPlatoRe: distance
- Dec 15th 2012, 11:56 AMPetrusRe: distance
i am really sorry :( i totaly agree with u but now i got it :) -3*1+2*1+-2*1 = -3 and sqrt(-3^2) = 3 but how do i do when its 2x absolute value :S? the answer shall be sqrt(2)

btw that formula u gave me, where did u get it? what is the name of the formula :S? never seen it

i get the answer 3/sqrt(3)