Find a line that passes through a point A and cuts two other lines.

Question

Let $L_1$ be the straight line in $R^3$ given by (x, y, z) = (2, 2, 0) + t(3, 0, 2). A plane containing the line $L_1$ and the point A = (8, 2, 3)is given by (x,y,z) = (2,2,0) + s(6,0,3) + t(3,0,2)

The line $L_2$ is given by (x, y, z) = (5, 1, 0) + $\tau$(2, 1, 1). Determine an equation for a line that passes through the point A = (8, 2, 3) and meets both $L_1$ and $L_2$.

What I've understood is that the $L_3$ has to be on the plane since it goes thru two points on it. So far I have tried solving the problem by making $L_3 = L_2$ and $L_3 = L_1$ with gaussian elimination but it didn't lead me nowhere. At this point I am clueless.

Answer 1

$L_1: (2, 2, 0) + t (3,0,2) $

$L_2: (5,1,0) + s(2,1,1) $

$ A = (8,2,3) $

The vector from $A$ to a point on $L_1$ must be a multiple of the vector from $A$ to a point on $L_2$. Thus define the first vector as follows

$ V_1 = (2, 2, 0) + t (3, 0, 2) - (8, 2, 3) = (-6, 0, -3) + t (3, 0, 2) $

and the second vector

$ V_2 = (5,1,0) + s(2,1,1) - (8,2,3) = (-3, -1, -3) + s (2,1,1) $

If $V_1 $ is along $V_2$ then $V_1 \times V_2 = 0 $

Now,

$ V_1 \times V_2 = \begin{vmatrix} \mathbf{i} && \mathbf{j} && \mathbf{k} \\ -6 + 3 t && 0 && -3 + 2 t \\ -3 + 2 s && -1 + s && -3 + s \end{vmatrix} \\= (-(-1+s)(-3+2t) , (-3+2t)(-3+2s) - (-3+s)(-6+3t) , (-6+3t)(-1+s)) $

Simplifying, this becomes

$ V_1 \times V_2 = ( -3 + 3 s + 2 t - 2 t s , -9 +3 t + t s , 6 - 3 t - 6 s + 3 t s ) $

Now we want to find $t, s$ such that all the above components are $0$, i.e.

$ -3 + 3 s + 2 t - 2 t s = 0 $

$ -9 + 3 t + t s = 0 $

$ 6 - 3 t - 6 s + 3 t s = 0 $

Solving, gives the following values for $t$ and $s$:

$ t = 2.25 , s = 1 $

Therefore, the points of intersection of $L_3$ with $L_1$ and $L_2$ are

$ P_1 = (2, 2, 0) + 2.25 (3, 0, 2) = (8.75, 2, 4.50) $

$ P_2 = (5, 1, 0) + (2, 1, 1) = (7, 2, 1 ) $

Constructing $L_3$ from these two points, gives us

$ L3: (7, 2, 1) + r (1.75, 0, 3.50) = (7, 2, 1) + r' (1, 0, 2) $

with $r' = 1$ we get our point $A = (8,2,3) $ on $L_3$.

Answer 2

One point on $L_1$ is $P_1=(2,2,0)$. A second point on $L_1$ is $P_2$ found by putting $t=1$ in the equation for $L_1$. Let your desired line be $\ell.$ Let $\Pi_1 $ be theplane formed by $\ell$ and $L_1.$ The normal to $\Pi_1$ is $$\mathbf u=AP_1 \times AP_2.$$ I changed the parameter for $L_2$ to $\tau$. Similarly, we can find the vector $\mathbf v $ that is normal to the plane $\Pi_2$ containing $\ell$ and $L_2.$ Since $\ell$ is perpendicular to both $\mathbf u$ and $\mathbf v$, $\ell$ is in the direction of $\mathbf w=\mathbf u \times \mathbf v$. The parametric equation of $\ell$ is $$(x,y,z)=A + s\mathbf w.$$