Krdytkyu

Given $2$ lines
$$L_1 : r = (1, 2, 3) + lambda (1, 2, 1.5)\
L_2 : r = (4, 0, -1) + mu (2, 4, 3)$$

Are parallel to each other with points $A(1, 2, 3)$ and $B(4, 0, -1)$.
Determine the distance between the two lines.

Method 1: distance from a point B to line L1

$AB(to) = OB(to) - OA(to) = (3, -2, -4)$

the distance is
$$frac{|(3, -2, -4) times (1, 2, 1.5) |}{sqrt{1 + 4 + 2.25}}= 4.716$$

Method 2: foot of perpendicular

Let foot of perpendicular of $A$ on $L2$ be $F$

$$OF(→) = (4+2mu, 4mu, -1+3mu) text{ for some }muinmathbb R\
AF(→) = (3+2mu, 4mu-2, -4+3mu) $$
Since $AF$ perpendicular to $L2$ :
$$(3+2mu, 4mu-2, -4+3mu) cdot (2, 4, 3) = 0\
μ = frac{14}{29}\
AF(to) = left(frac{115}{29}, frac{56}{29}, -frac{74}{29}right)\
|AF(to)| = 5.095$$

Method 3: Length of projection and Pythagoras Theorem

Length of projection of vector $vec{AB}$ onto $L_1$:

$$frac{|(3, -2, -4)cdot (1, 2, 1.5)|}{sqrt{1 + 4 + 2.25}} = 2.90065391$$

Length of $vec{AB} = | (3, -2, -4) |$.

Distance between two lines by Pythagoras theorem:

$$sqrt{| (3, -2, -4) |^2 - 2.90065391^2 } = 4.537237 . $$

Why do I get different answers for methods? I believe the three methods are correct.

The vector $AF$ should be $$left(frac{115}{29}, -frac{2}{29}, -frac{74}{29}right)$$

This vector has a norm of approximately $4.716$ which is what you hot with method 1.

Your steps were correct up to the point where you inserted $mu$ into the formula

$$AF = (3+2mu, 4mu - 2, -4+3mu)$$

since

$$3+2cdot frac{14}{29} = frac{87 + 28}{29} = frac{115}{29}\
4cdot frac{14}{29} - 2 = frac{56 - 58}{29} = -frac{2}{29}\
-4+3cdotfrac{14}{29} = frac{-116 + 42}{29} = -frac{74}{29}$$