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I have a vector $n$ and I seek a parametric equation for a line that is orthogonal to $n$ and passes through a point $(a,b,c)$.

I got the equation of the plane formed by the normal vector and that contains the point using $\mathbf{n}.(\mathbf{r-r_o})=0$. But, I can't figure out how to get a general equation. Eventually, I wish to arrive at a parametric equation that I can use to get lines spanning $360^\circ$.

I could start with an arbitrary direction vector $l$ that is perpendicular to $n$ and passes through $(a,b,c)$ and get the equation using $(a,b,c)+t(l_1,l_2,l_3)$, but that will not give me all the vectors that I'm looking for.

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  • $\begingroup$ No it should work. What vector do you miss? $\endgroup$ – mvw Jan 27 '16 at 1:12
  • $\begingroup$ For example, if I start with $n=(1,0,0)$ and say, I choose the arbitrary vector $l=(0,1,0)$. Now, for whatever value of my $t$, I will not be able to get the vector $(0,0,1)$. $\endgroup$ – Ajax Jan 27 '16 at 1:19
  • $\begingroup$ What was your $(a,b,c)$? $\endgroup$ – mvw Jan 27 '16 at 1:21
  • $\begingroup$ I took some time to check the exact numbers that I had. If I have $(a,b,c) = (1,2,3)$, I wouldn't get the vector $(0,0,1)$ irrespective of my value of $t$, right? $\endgroup$ – Ajax Jan 27 '16 at 1:43
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Letting $\mathbf r_0$ be the vector from the origin to the point $(a,b,c)$, the line you’re looking for is the orthogonal projection of the line $s\mathbf r_0$ onto the plane $P$ defined by $\mathbf n$ and $\mathbf r_0$. This line is given by the parametric equation $\mathbf r(t)=\mathbf r_0+t\,\pi\mathbf r_0$, where $\pi\mathbf r_0$ is the orthogonal projection of $\mathbf r_0$ onto the plane through the origin with normal $\mathbf n$.

Alternatively, the point where the line $k\mathbf n$ intersects $P$ is on the line you seek. Since this point is on the plane $P$, this amounts to finding the value of $k$ for which $\mathbf n\cdot(k\mathbf n-\mathbf r_0)=0$, after which you have $\mathbf r(t)=\mathbf r_0 + t(k\mathbf n-\mathbf r_0)$ as a parametrization of the line. Solving for $k$ is the same as finding the orthogonal projection of $\mathbf r_0$ onto $\mathbf n$, which in turn is the orthogonal rejection of $\mathbf r_0$ with respect to the projection $\pi$, i.e., $k\mathbf n = (1-\pi)\mathbf r_0$, so these two approaches are really equivalent.

Can you take it from here?

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