Linear system in a $3D$ space.

Question

Solve the following linear system in 3-space. $$\begin{aligned}l_1&:(x, y, z) = (6, 0, 5)+ s (1, 2, 3)\\\\l_2&: (x, y, z) = (-1, 7, 14) + t (-3, 1, 1)\end{aligned}$$

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My attempt:

First, I converted $l_1$ and $l_2$ to the parametric form, which is:

$$\begin{aligned}l_1&:\begin{cases}x_1=6+s\\y_2=2s\\z_1=5+3s\end{cases}\\\\l_2&:\begin{cases}x_2=-1-3t\\y_2=7+t\\z_2=14+t\end{cases}\end{aligned}$$

Then I set $x_1 = x_2$ and got $s + 3t = -19$.

Then did the same for $y_1$ and $y_2$ $(2s-t=7)$, and $z_1$ and $z_2$ $(3s-t=9)$

From there I got stuck. Any help?

Answer 1

It is $$s+3t=-7$$ $$2s-t=7$$ $$3s-t=9$$

Answer 2

Setting components equal, $$\color{red}6+s=-1-3t\tag1$$ $$2s=7+t\tag2$$ $$5+3s=14+t\tag3.$$

Multiply $(1)$ by $2$ and subtract $(2)$ to get $12=-9-7t,$ or $t=-3$, and then use $(2$) to find $s=2$.

Observe that $(3$) also holds with these $s$ and $t$ values.

You mistakenly said $\color{red}{18}+s=-1-3t$.