Problem regarding equation of line in space

Question

I have doubt regarding similarity in equation of plane and equation of line in space. I know that equation of form Ax +By+Cz+D=0 where A,B,C are nonzero represents a plane i.e we require 3 variables to get a specific eq of a plane. Now suppose we have a cartesian eq of line $\frac {x-1}{2} $ = $\frac {y-2}{3} $ =$ \frac{z}{4}$ we can reduce the above equation to x=2k+1, y=3k+2 , z=4k and eliminating k we get x+2y-2z=5 . Does this mean that equation of line has been converted to equation of plane. If not then what does the converted form of line represent ?

Answer 1

You started with the equations $$ x = 2k+1, \quad y = 3k + 2, \quad z = 4k. $$ By taking a linear combination of these equations, you found $$ x + y - 2z = (2k + 1) + 2(3k + 2) - 2(4k) = 5. $$ You can repeat this process with any $a,b,c$ to get the equation $$ ax + by + cz = a(2k + 1) + b(3k + 2) + c(4k). $$ We will have eliminated the constant on the right hand side if an only if we have $$ 2a + 3b + 4c = 0. $$ In other words, we will get an equation of the form $ax + by + cz = d$ (for some constant $d$) if and only if the vector $(a,b,c)$ is perpendicular to $(2,3,4)$, which is the direction of the line.