Question

determine whether the lines (l_1) and (l_2) are parallel, skew, or intersecting.(l_1:\frac{x - 1}{1}=\frac{y - 1}{-2}=\frac{z - 12}{-3})(l_2:\frac{x - 2}{1}=\frac{y + 6}{3}=\frac{z - 13}{-7})parallel skewintersctingif they intersect, find the point of intersection. (if an answer does not exist, enter dne.)((x,y,z)=(quad))

Explanation:

Step1: Write parametric equations for lines

For line $L_1$: Let $\frac{x - 1}{1}=\frac{y - 1}{-2}=\frac{z - 12}{-3}=t$. Then $x=t + 1,y=-2t+1,z=-3t + 12$. For line $L_2$: Let $\frac{x - 2}{1}=\frac{y + 6}{3}=\frac{z - 13}{-7}=s$. Then $x=s + 2,y=3s-6,z=-7s + 13$.

Step2: Set up a system of equations

Set the $x$, $y$, and $z$ - coordinates equal:
$t + 1=s + 2$ (Equation 1 for $x$ - coordinates), which simplifies to $t-s=1$.
$-2t + 1=3s-6$ (Equation 2 for $y$ - coordinates), which simplifies to $2t+3s=7$.
From Equation 1, we have $t=s + 1$. Substitute $t=s + 1$ into Equation 2:
$2(s + 1)+3s=7$.

Step3: Solve the system of equations

Expand the left - hand side of the equation: $2s+2 + 3s=7$.
Combine like terms: $5s+2=7$.
Subtract 2 from both sides: $5s=5$, so $s = 1$.
Substitute $s = 1$ into $t=s + 1$, we get $t=2$.

Step4: Check the $z$ - coordinates

For the $z$ - coordinates, for $L_1$ with $t = 2$, $z=-3t + 12=-3\times2+12=6$.
For $L_2$ with $s = 1$, $z=-7s + 13=-7\times1+13=6$.
Since the values of $t$ and $s$ satisfy all three equations, the lines intersect.
To find the point of intersection, substitute $t = 2$ into the equations of $L_1$ (or $s = 1$ into the equations of $L_2$).
$x=t + 1=2 + 1=3$.
$y=-2t+1=-2\times2+1=-3$.
$z=-3t + 12=-3\times2+12=6$.

Answer:

$(3,-3,6)$