Find the equation of the line

Question

I have to find the line that passes through $(3, 1, -2)$ and intersects under right angle the line $x=-1+t, y=-2+t, z=-1+t$.

(HINT: If $(x_0, y_0, z_0)$ the intersection point, find the coordinates.)

I have done the following:

The line that passes through $(3, 1, -2)$ is of the form $$\overrightarrow{l}(t)=(3, 1, -2)+t\overrightarrow{u}$$ where $\overrightarrow{u}$ is a vector parallel to the line.

Let $(x_0, y_0, z_0)$ be the intersection point of the line $\overrightarrow{l}$ and the line $x=-1+t, y=-2+t, z=-1+t$, then we have the folowing:

$$x_0=3+tu_1=-1+t, \\ y_0=1+tu_2=-2+t, \\ z_0=-2+tu_3=-1+t$$

Is it correct so far??

How could I continue??

Do we have to use the fact that the two lines intersect under right angle??

Answer 1

You are on the right track, but you want to use another parameter besides t for the second line.

Using s instead, we have the equations

$x_0=3+su_1=-1+t,\;\; y_0=1+su_2=-2+t,\;\; z_0=-2+su_3=-1+t$.

Since the lines are perpendicular, we have that $\langle1,1,1\rangle\cdot\langle u_1,u_2,u_3\rangle=u_1+u_2+u_3=0$.

Adding the 3 equations gives $2=-4+3t$, so $t=2$ and the lines intersect at $(1,0,1)$.

Thus we can take $\langle u_1,u_2,u_3\rangle=\langle2,1,-3\rangle$, so the second line has parametric equations

$x=3+2s,\;\; y=1+s,\;\; z=-2-3s$.

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Alternate solution:

The plane which is perpendicular to the given line and passes through $(3,1,-2)$ has equation

$\;\;\;x+y+z=2$, so the plane and line intersect where

$\;\;\;(-1+t)+(-2+t)+(-1+t)=2$, so $3t=6$ and $t=2$.

Then $(1,0,1)$ is the point of intersection, so the perpendicular line we want is parallel to

$\vec{u}=\langle3-1,1-0,-2-1\rangle=\langle2,1,-3\rangle$; and therefore it has parametric equations

$x=3+2s,\;\; y=1+s,\;\; z=-2-3s$.

Answer 2

Yes, you can use those equations to find the direction vector $\vec{u}$. Since there are four variables, $u_1$, $u_2$, $u_3$, and $t$, you will need one more equation relating them. For this, note that the two lines will be perpendicular to each other, so $\vec{u}\cdot\langle 1, 1, 1 \rangle = 0$. In other words, $u_1 + u_2 + u_3 = 0$. Then you can solve the system and substitute t into the original line equations to find the point of intersection. (You could solve the original equations for each component of u, then substitute them into the dot product equation and solve for t.)

Answer 3

Your biggest mistake is that you've used $t$ in the equations for both lines. This causes problems when you try to find intersection points, since it's not certain that at the intersection point the $t$ for one line is the same as the $t$ for the other line. For that reason you should use $s$ or something else for one of the lines.

Assume they intersect at $(x_0, y_0, z_0)$. Then you know two points of the line $\vec l$, so use those to parametrize it: $$ \vec l (s) = s\cdot (3, 1, -2) + (1-s)\cdot (x_0, y_0, z_0) = (x_0, y_0, z_0) + s\cdot (3-x_0, 1-y_0, -2-z_0) $$ What do we know about $(x_0, y_0, z_0)$? Well, we know that it lies on the other line. Therefore there is a $t_0$ such that $(x_0, y_0, z_0) = (-1+t_0, -2 + t_0, -1 + t_0)$ (that is what it means to say that that point lies on that line). If we insert this into our equation for $\vec l$, we get $$ \vec l (s) = (-1+t_0, -2 + t_0, -1 + t_0) + s\cdot (3-(-1+t_0), 1-(-2 + t_0), -2-(-1 + t_0)) \\ = (-1+t_0, -2 + t_0, -1 + t_0) + s\cdot (4-t_0, 3- t_0, -1- t_0) $$ Now we need to decide $t_0$, and we do that by using the right-angle condition. If there were no right-angle condition, then we would be done here, and any $t_0$ would work and give us a line that goes through the point we want it to, and intersects the line we want it to intersect.

But we have to find the right $t_0$ to make $(1, 1, 1)$ stand at a right angle to $(4-t_0, 3- t_0, -1- t_0)$, which we can express as $$ (1, 1, 1)\cdot (4-t_0, 3- t_0, -1- t_0) = 0\\ 4-t_0+ 3- t_0+ -1- t_0 = 0\\ 6 = 3t_0\\ t_0 = 2 $$ Using this we finally get that the line $\vec l$ is given by $$ \vec l (s) = (-1+2, -2 + 2, -1 + 2) + s\cdot (4-2, 3- 2, -1- 2)\\ = (1, 0, 1) + s\cdot (2, 1, -3) $$