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

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.

• The plane is off topic. $L_3$ does not have to be on it. Hint: $L_3$ has an equation of the form $(x,y,z)=(8,2,3)+s(a,b,c)$ where $a,b,c$ must be such that $\exists s,t\quad (8,2,3)+s(a,b,c)=(2,2,0)+t(3,0,2)$ and $\exists u,v\quad(8,2,3)+u(a,b,c)=(5,1,0)+v(2,1,1).$ Commented Dec 21, 2022 at 9:37
• More hint: ultimately, you will find that these conditions on $a,b,c$ are equivalent to $b=0,c=2a.$ Commented Dec 21, 2022 at 10:01
• Your first hint is what I mean by I tried doing $L_3= L_1 and L_3 = L_2$. I also found that b has to zero since it is supposed to be parallell with the plane (the only time the plane was actually useful). And when I did gaussian elimination, I get no solution since it equals to 0+0= x. I don't know if that is what you meant tho. Commented Dec 21, 2022 at 10:17
• Sorry I just realised something, I used the same parameter variable for $L_3$ when I made it equal to $L_1 and L_2$. Anyways I still don get how you got c=2a. Is it after you solved the equation you got that? Commented Dec 21, 2022 at 10:19
• "I also found that b has to zero since it is supposed to be parallell with the plane": no, it is not supposed to. This is a result of solving the equations. "And when I did gaussian elimination, I get no solution since it equals to 0+0= x. I don't know if that is what you meant tho. " No, I did not mean that. I did not use Gaussian elimination. " c=2a. Is it after you solved the equation you got that?" It is after eliminating the four parameter variables that I got both $b=0$ and $c=2a.$ Commented Dec 21, 2022 at 10:24

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 $$(x,y,z)=A + s\mathbf w.$$

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

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

$$A = (8,2,3)$$

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 $$L_3$$.

• Why did you use cross product, aren't $V_1$ and $V_2$ parallell? Commented Dec 21, 2022 at 17:12
• I want to find $t$ and $s$ that will make them parallel. Commented Dec 21, 2022 at 18:26