1
$\begingroup$

I'm studying some algebra. Consider this assignment:

Find the straight line which passes through a given point, $M$ and which intersects two given straight lines, $p, q$.

$$M=[-1, -7, 4]$$ $$p: X_1=\left[2, -5, 1\right]+t_1(3, -2, -1), t_1 \in R$$ $$q: X_2=\left[6, 5, 11\right]+t_2(2, 3, 5), t_2 \in R$$

Now the standard approach seems to be to make a plane that shares one of the lines and the point $M$. But it's late and I wanted to have some fun, so I tried to solve this task without any planes, just using relationships given by their equations.

Let $\vec{p} = (3, -2, -1)$ be the directional vector of line $p$ and let $\vec{q} = (2, 3, 5)$ directional vector of line $p$. Let's call the line we're looking for $r$ with $\vec{r}$ being it's directional vector.

Further let $P \in p$ and $Q \in q$ be points used in equations above.

We know the following, assuming the line we seek exists:

  • $M \in r$
  • $r \cap p \neq \emptyset$
  • $r \cap q \neq \emptyset$

This means there is a point which is in $q$ and $r$ as well another different point that lies in $p$ and $r$. For some (not any) $a_1, a_2, a_3, a_4 \in R$, following equations should have a solution:

  • $M+a_1\vec{r} = P + a_2 \cdot \vec{p}$
  • $M+a_3\vec{r} = Q + a_4 \cdot \vec{q}$

I can move points on one side of the equation and vectors with their coeficients on the other.

$$a_1\cdot \vec{r} - a_2 \cdot\vec{p} = P-M$$ $$a_3\cdot \vec{r} - a_4 \cdot\vec{q} = Q-M$$

Then I'll expand this to equations for individual coordinates. I'll use $r_1, r_2, r_3$ for individual coordinates of vector $\vec{r}$, I'll put numbers I know instead of the rest:

$$a_1\cdot r_1 - a_2\cdot 3 = 6$$ $$a_1\cdot r_2 - a_2\cdot (-2) = 0$$ $$a_1\cdot r_3 - a_2\cdot (-1) = -2$$ $$a_3\cdot r_1 - a_4\cdot 2 = 10$$ $$a_3\cdot r_2 - a_4\cdot 3 = 0$$ $$a_3\cdot r_3 - a_4\cdot 5 = 8$$

Now my question is: What can I do now? This is some bunch of equations. And since there is multiplications between coeficients, I don't know how to solve this using standard matrix of equations.

So, the question: Are my calculations correct so far? Does this lead to solution? If yes, how? If not, why?

  • 0
    As far as I can see, your calculations are correct. You have seven variables and 6 equations, so there is one degree of freedom in your solution (because the length of $r$ is not determined). Without loss of generality, you can use this freedom to set $a_1=1$. Use the first three equations to express $r_1$, $r_2$ and $r_3$ in terms of $a_2$ and eliminate $r_1$, $r_2$ and $r_3$ in the last three equations. With the substitution $c=a_2a_3$, you will get a linear system of equations with the variables $c$, $a_3$ and $a_4$. – 2017-01-30
  • 0
    If you want to challenge yourself try to solve this problem with Plucker line coordinates and homogeneous point coordinates. – 2017-01-31

0 Answers 0