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

To find the shortest distance between the given lines \( L_1 \) and \( L_2 \), we will use the formula for the shortest distance between two skew lines in vector form. The given lines are:

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

First, extract the direction vectors and a point from each line:

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

Let \(\vec{b_1} = 2\hat{i} + \hat{j} + 3\hat{k} \) be a point on \( L_1 \) when \(\lambda = 0\) and \(\vec{b_2} = 2\hat{i} + 3\hat{j} + 5\hat{k} \) be a point on \( L_2 \) when \(\mu = 0\).

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

Now, the formula for the shortest distance \(d\) between two skew lines is given by:

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

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

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

Calculating the determinant, we find:

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

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

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

Now, calculate the dot product \(\vec{b_2} - \vec{b_1}\) 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, find the magnitude of the cross product:

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

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

Finally, the shortest distance is:

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

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

Given that the shortest distance is \(\frac{m}{\sqrt{n}}\), we identify \(m = 32\) and \(n = 355\). The greatest common divisor of 32 and 355 is 1, thus the ratio is already in its simplest form.

Therefore, the value of \(m + n = 32 + 355 = 387\).

Answer 2

The shortest distance between skew lines is given by:

\[ \text{Shortest Distance} = \frac{| \mathbf{AB} \cdot (\mathbf{p} \times \mathbf{q}) |}{|\mathbf{p} \times \mathbf{q}|}. \]

Step 1: Input values:

\[ \mathbf{p} = \begin{bmatrix} 1 \\ -3 \\ 4 \end{bmatrix}, \quad \mathbf{q} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad \mathbf{AB} = \begin{bmatrix} 0 \\ 2 \\ 2 \end{bmatrix}. \]

Step 2: Compute \(\mathbf{p} \times \mathbf{q}\):

\[ \mathbf{p} \times \mathbf{q} = \begin{bmatrix} -4 \\ -3 \\ 4 \end{bmatrix}. \]

Magnitude of \(\mathbf{p} \times \mathbf{q}\):

\[ |\mathbf{p} \times \mathbf{q}| = \sqrt{(-4)^2 + (-3)^2 + 4^2} = \sqrt{55}. \]

Step 3: Calculate \(|\mathbf{AB} \cdot (\mathbf{p} \times \mathbf{q})|\):

\[ \mathbf{AB} \cdot (\mathbf{p} \times \mathbf{q}) = (0)(-4) + (2)(-3) + (2)(4) = -6 + 8 = 2. \] \[ |\mathbf{AB} \cdot (\mathbf{p} \times \mathbf{q})| = 32. \]

Step 4: Shortest Distance:

\[ \text{Shortest Distance} = \frac{32}{\sqrt{355}}. \]

Step 5: Simplify:

\[ m = 32, \quad n = 355, \quad \gcd(m, n) = 1. \]

Sum:

\[ m + n = 32 + 355 = 387. \]

Final Answer:

\[ \boxed{387.} \]