Question

Determine if the lines $L_1$ and $L_2$ are parallel, intersecting, or skew. If they are skew, find the shortest distance between them.

The lines are given by: $L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (\hat{i} - 3\hat{j} + 2\hat{k})$ $L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu (2\hat{i} + 3\hat{j} + \hat{k})$

Answer 1

The lines are skew.
The shortest distance between them is $\frac{3\sqrt{19}}{19}$ units.

Answer 2

1. Identify points and direction vectors:
   For $L_1$: $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b_1} = \hat{i} - 3\hat{j} + 2\hat{k}$.
   For $L_2$: $\vec{a_2} = 4\hat{i} + 5\hat{j} + 6\hat{k}$ and $\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}$.

2. Check for parallelism:
   Compare direction vectors $\vec{b_1}$ and $\vec{b_2}$.
   $\vec{b_1} = \hat{i} - 3\hat{j} + 2\hat{k}$
   $\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}$
   Since $\frac{1}{2} \neq \frac{-3}{3} \neq \frac{2}{1}$, the direction vectors are not proportional. Therefore, the lines are not parallel.

3. Check for intersection (coplanarity):
   Calculate $\vec{a_2} - \vec{a_1}$:
   $\vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.

Calculate $\vec{b_1} \times \vec{b_2}$:

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}((-3)(1) - (2)(3)) - \hat{j}((1)(1) - (2)(2)) + \hat{k}((1)(3) - (-3)(2))$$ $$= \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 + 6) = -9\hat{i} + 3\hat{j} + 9\hat{k}$$

Calculate the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$$(3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = (3)(-9) + (3)(3) + (3)(9)$$ $$= -27 + 9 + 27 = 9$$

Since the scalar triple product is $9 \neq 0$, the lines are not coplanar. As they are neither parallel nor coplanar, the lines are skew.

4. Calculate the shortest distance between skew lines:
   The formula for the shortest distance $d$ between two skew lines is:

$$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$

We have already calculated the numerator: $|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})| = |9| = 9$.

Now calculate the magnitude of $\vec{b_1} \times \vec{b_2}$:

$$|\vec{b_1} \times \vec{b_2}| = |-9\hat{i} + 3\hat{j} + 9\hat{k}| = \sqrt{(-9)^2 + (3)^2 + (9)^2}$$ $$= \sqrt{81 + 9 + 81} = \sqrt{171}$$ $$= \sqrt{9 \times 19} = 3\sqrt{19}$$

Substitute these values into the distance formula:

$$d = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}}$$

Rationalize the denominator:

$$d = \frac{3}{\sqrt{19}} \times \frac{\sqrt{19}}{\sqrt{19}} = \frac{3\sqrt{19}}{19}$$