Find an equation for the line that goes through the point and intersects line1 and line2

Question

Let $L_1$ be a line in $\mathbb R^3$ that is defined by $(x,y,z)=(2,2,0)+t(3,0,2)$

a) Find the plane that includes $L_1$ and the point $A=(9,2,3)$

b) The line $L_2$ is defined by $(x,y,z)=(5,1,0)+t(2,1,1)$. Find an equation for the line that goes through $A$ and intersects both $L_1$ and $L_2$.

I understand that the plane is found by taking two points on the line and using that. together with $A$ to find a plane. This results in $y=2$ which is the plane.

What I don't understand is how to find b). I know that the line I'm to find has to be in the plane I found in a), i.e. $y=2$. By (a) any line in the plane will go through $L_1$. But how do I include the second line?

Answer 1

Let's call the line $L_{3}$. Its equation is of the form $(x, y, z) = (9, 2, 3) + \lambda (v_{1}, v_{2}, v_{3}), \lambda \in \mathbb{R}$ for some direction vector $(v_{1}, v_{2}, v_{3})$ which we aim to find.

Any vector from a point on $L_{1}$ to $A$ has the form \begin{equation*} (9 - 2 - 3t, 2 - 2, 3 -2t) = (7 - 3t, 0, 3 -2t) \end{equation*} for some $t \in \mathbb{R}$ while any vector from $A$ to a point on $L_{2}$ is of the form \begin{equation*} (5 + 2s - 9, 1 + s - 2, s - 3) = (2s -4, s- 1, s - 3) \end{equation*} for some $s \in \mathbb{R}.$

We can find a direction vector for our line by finding values of $s$ and $t$ such that the corresponding vectors are parallel. This requires that $s = 1$ by comparing the second components, and so a direction vector is $(-2, 0, -2).$

Answer 2

For b), the intersection of the plane $y=2$ and $L_2$ is at $t=1$, or $B(7,2,1)$. Then, the line that connects $A$ and $B$, which by construction also intersects $L_1$ and $L_2$, is,

$$(9,2,3)+t(A-B)=(9,2,3)+t(2,0,2)$$

Answer 3

Basically find the intersection of plane (from part a) with L2 and build a line passes through point A, because the plane in the part a not fully includes the line L2 which means that the lines both passing through A and line L1 are not spans L2 and there is a single intersection with that line so find the intersection of plane with L2 then there will be a single line that connects A and intersection including a point from L1.