Question

determine whether the lines l1 and l2 are parallel, skew, or intersecting. l1: x = 12 + 8t, y = 16 - 4t, z = 4 + 12t l2: x = 1 + 4s, y = 3 - 2s, z = 4 + 5s parallel skew intersecting if they intersect, find the point of intersection. (if an answer does not exist, enter dne.)

Explanation:

Step1: Check for parallel lines

The direction vectors of the lines are found from the coefficients of the parameters. For $L_1$, the direction vector $\vec{v}_1=\langle8, - 4,12
angle$ and for $L_2$, the direction vector $\vec{v}_2=\langle4,-2,5
angle$. If the lines are parallel, then $\vec{v}_1 = k\vec{v}_2$ for some scalar $k$. We check the ratios: $\frac{8}{4}=2$, $\frac{-4}{-2} = 2$, but $\frac{12}{5}
eq2$. So the lines are not parallel.

Step2: Check for intersecting lines

Set the $x$, $y$, and $z$ - coordinates equal to each other:
$x$ - coordinates: $12 + 8t=1 + 4s$ which can be rewritten as $8t-4s=-11$;
$y$ - coordinates: $16-4t = 3-2s$ which can be rewritten as $-4t + 2s=-13$. Multiply this equation by 2, we get $-8t + 4s=-26$.
Add the two new - formed equations: $(8t-4s)+(-8t + 4s)=-11+( - 26)$. The left - hand side is $0$ and the right - hand side is $-37$. Since $0
eq - 37$, there are no solutions for $s$ and $t$. So the lines do not intersect.
Since the lines are not parallel and not intersecting, they are skew.

Answer:

skew
DNE