8. Let L₁ : (x − 1)/2 = (y − 2)/3 = (z − 3)/4 and L₂ : (x − 2)/3 = (y − 4)/4 = (z − 5)/5 be two lines. Which of the following points lies on the line of the shortest distance between L₁ and L₂? (1) (−5/3, −7, 1) (2) (2, 3, 1/3) (3) (8/3, −1, 1/3) (4) (14/3, −3, 22/3)

1. Direction vectors of the two lines

For line L₁ $\vec a = \langle 2,\,3,\,4 \rangle$ For line L₂ $\vec b = \langle 3,\,4,\,5 \rangle$

2. Direction of the shortest–distance line (common perpendicular)

It must be perpendicular to both $\vec a$ and $\vec b$. The vector that satisfies this automatically is their cross product:

$\vec n = \vec a \times \vec b = (-1,\,2,\,-1)$

Any point on the required line can be written as $\vec r = \vec r_0 + \lambda\,\vec n$ where $\vec r_0$ is one particular point on that line and $\lambda$ is any real number.

3. Finding one such particular point

Let general points on L₁ and L₂ be $\vec r_1 = (1,2,3) + t\,\vec a$ $\vec r_2 = (2,4,5) + s\,\vec b$

The segment $\vec r_2-\vec r_1$ should be parallel to $\vec n$, which we enforce by setting it perpendicular to both $\vec a$ and $\vec b$:

What’s the question?

We have two sticks (lines) floating in space that do not touch and are not parallel. We want the tiny bridge (the common perpendicular) that connects them with the shortest possible length. The question simply asks: Which of the four given points is sitting somewhere on that little bridge?

How to think about it

1. Each stick has a direction.
   L₁ points along $(2,3,4)$ and L₂ along $(3,4,5)$.
2. The bridge must stand straight up from both sticks.
   That means the bridge’s direction is at right–angles (perpendicular) to both stick directions.
3. A vector that is perpendicular to two given vectors at once is their cross-product.
   Do the cross-product and you get $(-1,2,-1)$ (or any multiple of it).
4. Pick the right point.
   Take one end of the bridge (a point we calculate on L₁). Any other point on the same straight line will differ from it by some multiple of $(-1,2,-1)$.
5. Check the options.
   Only choice (4) gives a difference that is exactly a multiple of $(-1,2,-1)$, so that’s the one.

Step-by-step Calculation

1. Write parametric form of the lines

   L_1 &: \; \vec r_1 = (1,2,3) + t\,(2,3,4) \\ L_2 &: \; \vec r_2 = (2,4,5) + s\,(3,4,5) \end{aligned}$$

2. Vector connecting any two points on the lines:
   $\vec c = \vec r_2 - \vec r_1 = \bigl(-1 + 2t - 3s,\; -2 + 3t - 4s,\; -2 + 4t - 5s\bigr)$

3. Perpendicularity conditions
   $\vec a \cdot \vec c = 0,\qquad \vec b \cdot \vec c = 0$ where $\vec a=(2,3,4)$ and $\vec b=(3,4,5)$.

   Working them out:

   29t - 38s &= 16 \quad (1)\\[2pt] 38t - 50s &= 21 \quad (2) \end{aligned}$$

4. Solve (1) & (2) $s = -\frac16, \qquad t = \frac13$

5. Get one point on the common perpendicular (take the one on $L_1$): $\vec r_0 = (1,2,3) + \frac13(2,3,4) = \left(\frac53,\,3,\,\frac{13}{3}\right)$

6. Direction of common perpendicular $\vec n = \vec a \times \vec b = (-1,2,-1)$

7. Check each option $P$ by computing $P-\vec r_0$ and verifying if $P-\vec r_0 = k\,\vec n$ Only for option (4) do we get $\bigl(3,-6,3\bigr) = -3\,(-1,2,-1)$ confirming collinearity.

[ \boxed{\text{Option (4) }\left(\tfrac{14}{3},-3,\tfrac{22}{3}\right)} ]