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Find a line $s$, perpendicular to and intersecting $r: x=y=\frac{z+3}{-2}$ that goes through ${(1,5,-2)}$ and then find their intersection. So I have found a vector that satisfies $a+b-2c=0$ e.g. $(1,1,1)$ and called my line $s: (1,5,-2)+t(1,1,1)$ which is obviously going through $(1,5,-2)$ but then when I set them to be each other, they don't cross.

I have tried like twenty vectors in the form $a+b+2c=0$ but the lines never cross and I'm really freaking confused.

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    $\begingroup$ there are infinitely many lines perpendicular to the given line $r$ and passing through the given point. Is there some missing information, such as the lines are meant to be co-planar? $\endgroup$ – David Quinn Mar 3 '18 at 18:26
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    $\begingroup$ Since there are infinitely many directions perpendicular to the line $r$, it is very unlikely that you will find the one that also passes through the point by guessing. $\endgroup$ – Tob Ernack Mar 3 '18 at 18:26
  • $\begingroup$ I dont see what you are trying to do. The easy way is to assume a point on original line. Then find vector joining this with 1,5,-2. Now use dot product. $\endgroup$ – King Tut Mar 3 '18 at 18:26
  • $\begingroup$ You can, on the other hand, parametrize $r$ and select an arbitrary point $(t, t, -2t-3)$ on the line. Then the line perpendicular to $r$ and passing through the point $(1, 5, -2)$ will be the one that minimizes the distance from $(1, 5, -2)$ to $(t, t, -2t-3)$. $\endgroup$ – Tob Ernack Mar 3 '18 at 18:28
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First of all, what is the line $r$ in parametric form? $$t=x=y=(z+3)/(-2)$$ $$r:(t,t,-2t-3)\qquad t\in\mathbb R$$ Draw a vector from the given point $(1,5,-2)$ to the point on $r$ with parameter $t$. If this is perpendicular to $r$, its dot product with the direction vector $(1,1,-2)$ is zero: $$((1,5,-2)-(t,t,-2t-3))\cdot(1,1,-2)=0$$ $$(1-t,5-t,2t+1)\cdot(1,1,-2)=0$$ $$1-t+5-t-4t-2=0$$ $$t=\frac23$$ This corresponds to a point of intersection at $(t,t,-2t-3)=(2/3,2/3,-13/3)$ and a perpendicular line of $(1,5,-2)+s(1-t,5-t,2t+1)=(1,5,-2)+s(1/3,13/3,7/3)$, $s\in\mathbb R$. We can check that the lines are perpendicular: $(1/3,13/3,7/3)\cdot(1,1,-2)=0$.

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  • $\begingroup$ Right so the distance between r&P would be the distance between the point of intersection and P. I already know this, I just had to type something mathematical in order for the site to let me say thank you so thank you! $\endgroup$ – Nobody Mar 3 '18 at 19:05
  • $\begingroup$ @Nobody Yes. More specifically, with the two lines perpendicular and intersecting, that gives the shortest distance as I worked out above. $\endgroup$ – Parcly Taxel Mar 3 '18 at 19:07
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Perpendicular and orthogonal are two different things:

  • Orthogonal lines have orthogonal directions (dot product vanishes) There are infinitely many lines that are orthogonal to a given line and going through a given point.
  • Perpendicular lines are orthogonal and cross. There is only one line perpendicular to a given line $L$ and going through a given point that is not on $L$.

So you first need to find the orthogonal projection of your point $A=(1,5,-2)$ onto your line, call it $B$, and then find out the equation of the line $(AB)$.

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  • $\begingroup$ I'd never heard of such a sharp distinction between orthogonal and perpendicular. I googled it, and was led back to this question on this site: math.stackexchange.com/questions/1568937/… $\endgroup$ – Matthew Leingang Mar 3 '18 at 18:37
  • $\begingroup$ @MatthewLeingang, that's the thing, me neither and all this time I'd been using the same method to find perpendicular lines and it just happened to have worked by luck until this one time! I'm so glad though because I would have gone through life not knowing! $\endgroup$ – Nobody Mar 14 '18 at 10:07
  • $\begingroup$ @Nobody: If you read the linked question, you'll find that there's quite a bit of historical usage that treats them synonymously. $\endgroup$ – Matthew Leingang Mar 16 '18 at 18:04
  • $\begingroup$ @MatthewLeingang we have this very clear distinction in French, and I never realised that it was a particularity of ours. But I do subscribe to it, I think it is a very relevant distinction to make. $\endgroup$ – Arnaud Mortier Mar 16 '18 at 19:16
  • $\begingroup$ @ArnaudMortierot it is definitely a relevant distinction and to be quite frank, I'm shocked that it is not more universally used! $\endgroup$ – Nobody Mar 20 '18 at 14:05
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For starters, you should check the calculations you're using to get a vector orthogonal to the direction of $r.$ (That is what $a+b-2c=0$ is meant to do, right?)

Next, consider a plane through $(1,5,-2)$ perpendicular to the line $r.$ Any vector parallel to that plane will be orthogonal to the direction of $r.$ But there is only one point where $r$ intersects that plane, and only one direction from from $(1,5,-2)$ to that point. Trying to guess that vector is probably not a good way to go.

If you can express $r$ in the form $(x_0,y_0,z_0) + t(x_v,y_v,z_v),$ you can subtract $(1,5,-2)$ to obtain a displacement vector from $(1,5,-2)$ to an arbitrary point on the line $r$ (depending on the value of $t$). Make that vector orthogonal to the direction of $r$ and you have the desired vector.

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