Question

In Exercises 97 and \(98,\) verify that the lines are parallel, and find the distance between them. $$ \begin{aligned} &L_{1}: x=2-t, \quad y=3+2 t, \quad z=4+t\\\ &L_{2}: x=3 t, \quad y=1-6 t, \quad z=4-3 t \end{aligned} $$

Short answer

The lines L1 and L2 are indeed parallel and the shortest distance between them is \(\sqrt{3.5}\).

Step-by-step solution

Check if Lines are Parallel

To check if the lines are parallel, look for the proportionality in the directional vectors of two lines. For L1, the directional vector \( \vec {m1} \) can be found from coefficients of t, thus (‎-1, 2, 1). Similarly, for L2, the directional vector \( \vec {m2} \) is (3, -6, -3). The lines are parallel if the ratio of respective coordinates of the directional vectors are equal. From the given exercise, \(-1/3 = 2/(-6) = 1/(-3)\), so the lines are indeed parallel.

Calculate Distance between Lines

Since the lines are parallel, use the formula for the shortest distance, \(D\), between two parallel lines which is \[D=\frac{|\vec {m1}\times \vec {r}|}{|\vec {m1}|}\]. In this formula, \(\vec {m1}\) is the directional vector of L1 and \(\vec {r}\) is a vector formed by subtracting any point on L1 from a point on L2, say (3,1,4) - (2,3,4) = (1,-2,0). Now you just need to compute the required quantities in the formula.

Compute Vector Quantities

First calculate \(\vec {m1}\times \vec {r}\) which turns out to be 2i -j -4k. Its norm, \(|\vec {m1}\times \vec {r}|\) is \(\sqrt{21}\). The norm \(|\vec {m1}|\) of the directional vector of L1, is \(\sqrt{6}\).

Final Computation

Substitute the values in the formula from Step 2, so the distance \(D\) between the lines would be \(D=\frac{\sqrt{21}}{\sqrt{6}} =\sqrt{3.5}\).