Answer to Question #307583 in Analytic Geometry for Cee

Question

Determineif the 2 lines are parallel, skew or intersecting:

⟨x,y,z⟩=⟨4,2,3⟩+⟨3,3,2⟩t and

⟨x,y,z⟩=⟨-4,10,3 ⟩+⟨ 20,4,2⟩s

*If parallel (type the number 1).

*If intersecting (type the number 2).

*If skew (type the number 3).

Answer 1

Answer

The given lines can be written as

$\left\langle {x,y,z} \right\rangle = \left\langle {4,2,3} \right\rangle + \left\langle {3,3,2} \right\rangle t\\$

$\left\langle {x,y,z} \right\rangle = \left\langle {4 + 3t,2 + 3t,3 + 2t} \right\rangle$

$\frac{{x - 4}}{3} = \frac{{y - 2}}{3} = \frac{{z - 3}}{2} = t$ .... (1)

The direction vector of line (1) is

${r_1} = \left\langle {3,\,3,\,2} \right\rangle$

And

$\left\langle {x,y,z} \right\rangle = \left\langle { - 4,10,3} \right\rangle + \left\langle {20,4,2} \right\rangle s\\$

$\left\langle {x,y,z} \right\rangle = \left\langle { - 4 + 20s,10 + 4s,3 + 2s} \right\rangle$

$\frac{{x + 4}}{{20}} = \frac{{y - 10}}{4} = \frac{{z - 3}}{2} = s$ .... (2)

The direction vector of the line (2) is

${r_2} = \left\langle {20,\,4,\,2} \right\rangle$

Since we can see that the two vectors ${r_1} = \left\langle {3,\,3,\,2} \right\rangle$ and ${r_2} = \left\langle {20,\,4,\,2} \right\rangle$ are not the same and even can't be written as linear combination of one another therefore, the two lines are not parallel.

Now, when not parallel, then there are two possibilities.

They may intersect or may not (skew)

In case they intersect, then for some values of $t$ and $s$, we can write,

$4 + 3t = - 4 + 20s$ ... (3)

$2+3t=10+4s$ ... (4)

$3+2t=3+2s$ ... (5)

Solving (3) and (4)

$4 -2+ 3t-3t = - 4-10 + 20s-4s$

$2=-14+16s\\ s=1$

replacing this value in (4)

$2+3t=10+4(1)\\t=4$

Checking these two values of $s=1$ and $t=4$ in equation (5)

$3+2(4)=3+2(1)\\ 11=5$ FALSE

Hence we can't find the two values of $s$ and $t$, therefore, the two lines don't intersect.

The given two lines are skew.