Question

determine whether the lines intersect, and if so, find the point of intersection. (if an answer does not exist, enter dne.)
\\(\frac{x}{3}=\frac{y - 2}{-1}=z + 1,\frac{x - 1}{4}=y + 2=\frac{z + 3}{-3}\\)
if the lines intersect, find and the angle between the lines. (round your answer to one decimal place. if an answer does not exist, enter dne.)
\\(\theta =\\)
\\((x,y,z)=(\\) \\()\\)

Explanation:

Step1: Write the parametric equations of the lines

The first line has symmetric - equations $\frac{x}{3}=\frac{y - 2}{-1}=\frac{z + 1}{1}$. Let $\frac{x}{3}=\frac{y - 2}{-1}=\frac{z + 1}{1}=t$. Then $x = 3t$, $y=-t + 2$, $z=t - 1$.
The second line has symmetric - equations $\frac{x - 1}{4}=\frac{y+2}{1}=\frac{z + 3}{-3}$. Let $\frac{x - 1}{4}=\frac{y + 2}{1}=\frac{z + 3}{-3}=s$. Then $x = 4s+1$, $y=s - 2$, $z=-3s - 3$.

Step2: Set the $x$, $y$, and $z$ - coordinates equal to each other

Set up the system of equations:
$3t=4s + 1$ (Equation 1 for $x$ - coordinates), $-t + 2=s - 2$ (Equation 2 for $y$ - coordinates), $t - 1=-3s - 3$ (Equation 3 for $z$ - coordinates).
From Equation 2, we can express $t$ in terms of $s$: $t=4 - s$.

Step3: Substitute $t$ into Equation 1

Substitute $t = 4 - s$ into $3t=4s + 1$.
$3(4 - s)=4s + 1$.
Expand the left - hand side: $12-3s=4s + 1$.
Move the terms with $s$ to one side: $12 - 1=4s+3s$.
$7s = 11$, so $s=\frac{11}{7}$.

Step4: Find the value of $t$

Substitute $s=\frac{11}{7}$ into $t = 4 - s$.
$t=4-\frac{11}{7}=\frac{28 - 11}{7}=\frac{17}{7}$.

Step5: Check if the values of $t$ and $s$ satisfy the third equation

Substitute $s=\frac{11}{7}$ and $t=\frac{17}{7}$ into the $z$ - coordinate equation $t - 1=-3s - 3$.
Left - hand side: $t - 1=\frac{17}{7}-1=\frac{17 - 7}{7}=\frac{10}{7}$.
Right - hand side: $-3s - 3=-3\times\frac{11}{7}-3=\frac{-33 - 21}{7}=\frac{-54}{7}$.
Since the left - hand side and the right - hand side are not equal, the lines do not intersect.

Answer:

The lines do not intersect.