12 Mar 04
4 edits
Originally posted by geniusWhat do you mean by (1/3)(x+2y+3z)? The standard way of expressing a line is x= a + kb, or equivalently b^x = b^a, where x is the position vector, a and b are constant vectors, b is not zero, k is a real parameter and ^ is the vector product. The line formed is invariant under nonzero scalar multiplication of b, and under addition of cb to a for any real c, so the equation is not unique.
why can these have more than one equation? for instance, a vector that passes through the points A(1,2,3) and B(4,5,6)
AB=b-a=3i+3j+3k = the direction vector
thus, the equation of the line=(1/3)(x+2y+3z) or (1/3)(4x+5y+6z)
?
12 Mar 04
1 edit
Originally posted by Acolyteo.k.-got confused-we only did it today...😞
What do you mean by (1/3)(x+2y+3z)? The standard way of expressing a line is [b]x= a + kb, or equivalently b^x = b^a, where x is the position vector, a and b are constant ...[text shortened]... b]b[/b] to a for any real c, so the equation is not unique.[/b]
t=x/3=2y/3=3z/3 or t=4x/3=5y/3=6z/3
i think 😛
13 Mar 04
Originally posted by geniusOk, if you're putting it in that form it could be, for example, x-1=y-2=z-3 (the equations you give don't work - try putting x=1,y=2,z=3 into them). If you have an equation of the form (x,y,z) = (a,b,c) + k(d,e,f), you can re-write it as (x-a)/d=(y-b)/e=(z-c)/f, and vice versa. Once again of course, this formulation is not unique, as you can add a constant to each bit, or multiply them all by a non-zero constant, and the line would not be changed.
o.k.-got confused-we only did it today...😞
t=x/3=2y/3=3z/3 or t=4x/3=5y/3=6z/3
i think 😛