0
$\begingroup$

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

$$ ax+by+cz=d $$

$\endgroup$
  • 2
    $\begingroup$ 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
1
$\begingroup$

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), $\alpha \epsilon \mathbb{R}$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.