1. ## distance

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)

2. ## Re: distance

Originally Posted by Petrus
line L2 its a linear equation x+2=0 is a top and bottom y=3

Sorry, but your description of $\displaystyle L_2$ makes no sense.

3. ## Re: distance

Here is a pic L2

4. ## Re: distance

Originally Posted by Petrus
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)
Now that is a bit clearer. You must prove that these are two skew lines.

If $\displaystyle \ell_1:P+tD~\&~\ell_2:Q+sE$ are a pair of skew lines then:

$\displaystyle D(\ell_1,\ell_2)=\frac{|(Q-P)\cdot(D\times E)|}{\|D\times E\|}$

5. ## Re: distance

ik that Q-P=-3,2,-2 but i dont know how to calculate D x E

6. ## Re: distance

Originally Posted by Petrus
ik that Q-P=-3,2,-2 but i dont know how to calculate D x E

$\displaystyle D\times E=\left| {\begin{array}{rrr} i & j & k \\ 1 & 1 & { - 2} \\ { - 1} & 0 & 1 \\ \end{array} } \right|$

7. ## Re: distance

i get it to sqrt(17)/sqrt(3) is this wrong? it looks so for me :S

8. ## Re: distance

Originally Posted by Petrus
i get it to sqrt(17)/sqrt(3) is this wrong? it looks so for me :S
It is not what I get. What is $\displaystyle D\times E~?$

9. ## Re: distance

1i+2j+0k-(-1)k-1j-0i 1,1,1

10. ## Re: distance

Originally Posted by Petrus
1i+2j+0k-(-1)k-1j-0i 1,1,1

$\displaystyle |(Q-P)\cdot(D\times E)|=3$

11. ## Re: distance

Originally Posted by Plato
$\displaystyle |(Q-P)\cdot(D\times E)|=3$
is my D x E wrong? and how u get that to 3? im kinda confuse sorry but wanna ty u for helping me so far
and that llD x Ell idk how u calculate with 2 absolutevalue if im going to be honest

12. ## Re: distance

Originally Posted by Petrus
is my D x E wrong? and how u get that to 3? im kinda confuse sorry but wanna ty u for helping me so far
and that llD x Ell idk how u calculate with 2 absolutevalue if im going to be honest

$\displaystyle <-3,2,-2>\cdot <1,1,1>=~?$

13. ## Re: distance

-3,2-2 then i take sqrt(-3^2+2^2+-2^2)?

14. ## Re: distance

Originally Posted by Petrus
-3,2-2 then i take sqrt(-3^2+2^2+-2^2)?

Sorry, I am done with this.
You have absolutely no idea what any of this is about.
Good luck.

15. ## Re: 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