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)
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)
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)