distance

**Petrus** · December 15th 2012, 07:37 AM

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)

**Plato** · December 15th 2012, 07:42 AM

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 $L_2$ makes no sense.

Please try again.

**Petrus** · December 15th 2012, 07:47 AM

Here is a pic L2

**Plato** · December 15th 2012, 08:18 AM

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

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

**Petrus** · December 15th 2012, 09:13 AM

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

**Plato** · December 15th 2012, 09:21 AM

Originally Posted by Petrus

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

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

**Petrus** · December 15th 2012, 09:38 AM

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

**Plato** · December 15th 2012, 09:46 AM

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 $D\times E~?$

**Petrus** · December 15th 2012, 09:49 AM

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

**Plato** · December 15th 2012, 09:56 AM

Originally Posted by Petrus

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

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

**Petrus** · December 15th 2012, 09:59 AM

Originally Posted by Plato

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

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

**Plato** · December 15th 2012, 10:33 AM

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

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

**Petrus** · December 15th 2012, 10:39 AM

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

**Plato** · December 15th 2012, 10:51 AM

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.

**Petrus** · December 15th 2012, 10:56 AM

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)

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