Answer:
The point of intersection is:
[tex] \displaystyle\left( \frac{2}{5}, 1, 0\right) [/tex]
Step-by-step explanation:
Let us combine the vectors to get each equation in the format of a single vector, and we should use a different parameter for the second line, I will use k:
[tex]\vec{r}(t)=\left< 1-3t, 5t, 0\right>, \vec{s}(k)=\left< -2k,1,1+5k\right>[/tex]
Then we set the x,y and z components of the equations, equal to each other:
[tex]\begin{cases}1-3t=-2k\\5t=1\\0=1+5k\end{cases}[/tex]
We have to solve that system of equations:
Solving the second and last for t and k we get:
[tex]\begin{cases}1-3t=-2k\\t=\displaystyle\frac{1}{5}\\k=\displaystyle-\frac{1}{5}\end{cases}[/tex]
We plug them into the first equation and we get:
[tex]\begin{cases}\displaystyle 1-3\cdot\frac{1}{5}=-2\cdot\left(-\frac{1}{5}\right)\\t=\displaystyle\frac{1}{5}\\k=\displaystyle-\frac{1}{5}\end{cases}[/tex]
Once we simplify:
[tex]\begin{cases}\displaystyle \frac{2}{5}=\frac{2}{5}\\t=\displaystyle\frac{1}{5}\\k=\displaystyle-\frac{1}{5}\end{cases}[/tex]
So, the system actually has those solutions we have found for t and k. We can now use any of the equations of the two lines. Plugging [tex]t=\frac{1}{5}[/tex] into the equation of the first line we get:
[tex]\vec{r}(t)=\left< 1-3\left(\frac{1}{5}\right), 5\left(\frac{1}{5}\right), 0\right>=\left< \frac{2}{5}, 1, 0\right>[/tex]
Therefore the point of intersection is:
[tex] \displaystyle\left( \frac{2}{5}, 1, 0\right) [/tex]
