How would I pick $a,b,c$ to create a line that is parallel to $z$-axis and intersects $x$-axis at point $x=k, y=0, z=0$?

$$ ax+by+cz=d $$

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    $\begingroup$ Your equation $ax+by+cz=d$ is for a plane $\endgroup$ – J. W. Tanner Mar 12 at 13:06
  • $\begingroup$ Possible duplicate of What is the equation for a 3D line? $\endgroup$ – Brian Mar 12 at 13:07
  • $\begingroup$ I'm asking what's the special case for a line equation that's orthogonal to both x-axis and y-axis. and k units distance from z-axis. $\endgroup$ – DiscreteMath Mar 12 at 13:10
  • $\begingroup$ $ax + by + cz = d$ is the equation of a plane in $\mathbb{R}^3$. To define a line you need two linear equations - the line is then the intersection of the two planes. In this case the two planes are $x=k$ and $y=0$.. $\endgroup$ – gandalf61 Mar 12 at 13:13
  • $\begingroup$ So for example the following: $$(x,y, z) = (1/2, 0, t)$$ is a line orthogonal to xy-plane and parallel to z-axis? $\endgroup$ – DiscreteMath Mar 12 at 13:17

Since, line passes through (k,0,0) and is parallel to z-axis so, the equation of line is $(x,y,z) = (k,0,0) + \alpha$(0,0,1), $\alpha \epsilon \mathbb{R}$

  • $\begingroup$ If i'm understanding correctly, its a "parametric vector equation" because otherwise you need a system of "n - 1" equations to describe a line in n-dimensions.. But, really a "parametric vector equation" is really a vector represents of 3 equations, where 1 of the equations is linearly dependent, and 2 are linearly independent, leading to a parametric variable "$\alpha$" in your case. (but, more commonly the variable "t") $\endgroup$ – DiscreteMath Mar 12 at 13:26

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