Question

Two lines \(L_1\) and \(L_2\) are given as \(L_1: \vec{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(2\hat{i}+3\hat{j}+4\hat{k})\) and \(L_2: \vec{r} = (2\hat{i}+4\hat{j}+5\hat{k}) + \mu(4\hat{i}+6\hat{j}+8\hat{k})\), where \(\lambda\) and \(\mu\) are real parameters. Then which of the following statements are correct ?
A. Shortest distance between line \(L_1\) and \(L_2\) is \(\sqrt{\dfrac{5}{29}}\).
B. line \(L_1\) and \(L_2\) are parallel.
C. line \(L_1\) and \(L_2\) are concurrent lines.
D. line \(L_1\) and \(L_2\) are perpendicular.
Choose the correct answer from the options given below:

• A. B and C only
• B. C and D only
• C. A and C only
• D. A and B only

Hint:

Note \(\vec b_2=2\vec b_1\), so the lines are parallel. Use \(|(\vec a_2-\vec a_1)\times\vec b_1|/|\vec b_1|\) for the distance.

Answer 1

The Correct Option is D

Step 1: Compare directions first.
Divide $\vec b_2=(4,6,8)$ by $2$ to get $(2,3,4)=\vec b_1$. The lines run in the same direction, so they are parallel and B holds.

Step 2: Rule out D and C.
Parallel lines cannot be perpendicular, so D fails (the dot product $58$ is not zero).
Two parallel lines meet only if they are the same line. Check if $\vec a_2-\vec a_1=(1,2,2)$ is a multiple of $(2,3,4)$. The ratios $1/2, 2/3, 2/4$ are not equal. So the lines are distinct and never meet. C fails.

Step 3: Find the distance by a projection method.
Let $\hat b=\dfrac{(2,3,4)}{\sqrt{29}}$. The vector between the lines is $\vec v=(1,2,2)$. Its component along the lines is $\vec v\cdot\hat b = \dfrac{2+6+8}{\sqrt{29}}=\dfrac{16}{\sqrt{29}}$.

Step 4: Remove the parallel part.
The distance is the perpendicular part: $d^2=|\vec v|^2-(\vec v\cdot\hat b)^2 = 9-\dfrac{256}{29}=\dfrac{261-256}{29}=\dfrac{5}{29}$.

Step 5: Read off the answer.
$d=\sqrt{5/29}$, so A holds. The true statements are A and B, which is option 4.

Final Answer:
A and B are correct. \[ \boxed{\text{Option 4}} \]