Question

If the shortest distance between the lines.
L1: $\vec{r} = (2 + \lambda)\hat{i} + (1 - 3\lambda)\hat{j} + (3 + 4\lambda)\hat{k}$, $\lambda \in \mathbb{R}$.
L2: $\vec{r} = 2(1 + \mu)\hat{i} + 3(1 + \mu)\hat{j} + (5 + \mu)\hat{k}$, $\mu \in \mathbb{R}$ is $\frac{m}{\sqrt{n}}$, where gcd(m, n) = 1, then the value of m + n equals.

• A. 384
• B. 387
• C. 377
• D. 390

Answer 1

The Correct Option is B

The shortest distance between the skew lines \( L_1 \) and \( L_2 \) is calculated using the vector formula. The lines are defined as:

• \( L_1: \vec{r} = (2 + \lambda)\hat{i} + (1 - 3\lambda)\hat{j} + (3 + 4\lambda)\hat{k} \)
• \( L_2: \vec{r} = 2(1 + \mu)\hat{i} + 3(1 + \mu)\hat{j} + (5 + \mu)\hat{k} \)

From these equations, we extract the direction vectors and points on each line:

• Direction vector of \( L_1 \): \( \vec{a_1} = \hat{i} - 3\hat{j} + 4\hat{k} \)
• Direction vector of \( L_2 \): \( \vec{a_2} = 2\hat{i} + 3\hat{j} + \hat{k} \)

A point on \( L_1 \) (for \(\lambda = 0\)) is \(\vec{b_1} = 2\hat{i} + \hat{j} + 3\hat{k} \). A point on \( L_2 \) (for \(\mu = 0\)) is \(\vec{b_2} = 2\hat{i} + 3\hat{j} + 5\hat{k} \).

The vector connecting these points is \(\vec{b_2} - \vec{b_1} = 0\hat{i} + 2\hat{j} + 2\hat{k} \).

The formula for the shortest distance \(d\) between two skew lines is:

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

First, compute the cross product \(\vec{a_1} \times \vec{a_2}\):

\[ \vec{a_1} \times \vec{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{vmatrix} \]

The result of the determinant calculation is:

\(= \hat{i}((-3) \cdot 1 - 4 \cdot 3) - \hat{j}(1 \cdot 1 - 4 \cdot 2) + \hat{k}(1 \cdot 3 + 3 \cdot 2)\)

\(= \hat{i}(-3 - 12) - \hat{j}(1 - 8) + \hat{k}(3 + 6)\)

\(= -15\hat{i} + 7\hat{j} + 9\hat{k}\)

Next, calculate the dot product of \(\vec{b_2} - \vec{b_1}\) with the cross product:

\((\vec{b_2} - \vec{b_1}) \cdot (\vec{a_1} \times \vec{a_2}) = (0\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-15\hat{i} + 7\hat{j} + 9\hat{k})\)

\(= 0 \cdot (-15) + 2 \cdot 7 + 2 \cdot 9\)

\(= 0 + 14 + 18 = 32\)

Now, determine the magnitude of the cross product:

\(|\vec{a_1} \times \vec{a_2}| = \sqrt{(-15)^2 + 7^2 + 9^2}\)

\(= \sqrt{225 + 49 + 81} = \sqrt{355}\)

The shortest distance is thus:

\(d = \frac{|32|}{\sqrt{355}}\)

\(= \frac{32}{\sqrt{355}}\)

Given the form \(\frac{m}{\sqrt{n}}\), we have \(m = 32\) and \(n = 355\). The greatest common divisor of 32 and 355 is 1, confirming the simplest form.

The sum \(m + n = 32 + 355 = 387\).