Dawson Downs

2022-07-15

Solve the following linear system in 3-space.
$\begin{array}{rl}{l}_{1}& :\left(x,y,z\right)=\left(6,0,5\right)+s\left(1,2,3\right)\\ \\ {l}_{2}& :\left(x,y,z\right)=\left(-1,7,14\right)+t\left(-3,1,1\right)\end{array}$
My attempt:
First, I converted ${l}_{1}$ and ${l}_{2}$ to the parametric form, which is:
$\begin{array}{rl}{l}_{1}& :\left\{\begin{array}{l}{x}_{1}=6+s\\ {y}_{2}=2s\\ {z}_{1}=5+3s\end{array}\\ \\ {l}_{2}& :\left\{\begin{array}{l}{x}_{2}=-1-3t\\ {y}_{2}=7+t\\ {z}_{2}=14+t\end{array}\end{array}$
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?

Do you have a similar question?

coolng90qo

Expert

It is
s+3t=−7
2s−t=7
3s−t=9

Still Have Questions?

Stephanie Hunter

Expert

Setting components equal,
$\begin{array}{}\text{(1)}& 6+s=-1-3t\end{array}$
$\begin{array}{}\text{(2)}& 2s=7+t\end{array}$
$\begin{array}{}\text{(3)}& 5+3s=14+t.\end{array}$
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 $18+s=-1-3t$

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