Why do my "perpendicular'' lines not cross?

Question

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.

Answer 1

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: $(1/3,13/3,7/3)\cdot(1,1,-2)=0$.

Answer 2

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 $(AB)$.

Answer 3

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.