Question

The shortest distance between lines \( L_1 \) and \( L_2 \), where \( L_1 : \frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 4}{2} \) and \( L_2 \) is the line passing through the points \( A(-4, 4, 3) \), \( B(-1, 6, 3) \) and perpendicular to the line \( \frac{x - 3}{-2} = \frac{y}{3} = \frac{z - 1}{1} \), is

• A. \( \frac{121}{\sqrt{221}} \)
• B. \( \frac{24}{\sqrt{117}} \)
• C. \( \frac{141}{\sqrt{221}} \)
• D. \( \frac{42}{\sqrt{117}} \)

Answer 1

The Correct Option is C

To find the shortest distance between the two given lines \( L_1 \) and \( L_2 \), we first need to understand the properties and representations of the lines.

Step 1: Define the equations of the lines.

The line \( L_1 \) is given in symmetric form:

\(\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 4}{2}\)

This can be rewritten as:

• Direction ratios for \( L_1 \) are \( \langle 2, -3, 2 \rangle \).
• A point on \( L_1 \) is \( P(1, -1, -4) \).

The line \( L_2 \) passes through points \( A(-4, 4, 3) \) and \( B(-1, 6, 3) \).

• Direction ratios for line \( AB \) are \( \langle 3, 2, 0 \rangle \), calculated as \( B - A = (-1 + 4, 6 - 4, 3 - 3) \).

Since \( L_2 \) is perpendicular to the line:

\(\frac{x - 3}{-2} = \frac{y}{3} = \frac{z - 1}{1}\)

which has direction ratios \( \langle -2, 3, 1 \rangle \), the scalar product of these direction ratios with \( L_2 \) should be zero to be perpendicular.

Let \( \langle a, b, c \rangle \) be the direction ratios of \( L_2 \), then \( -2a + 3b + c = 0 \). We assume the ratio direction compatible with \( \langle 3, 2, 0 \rangle \).

Step 2: Calculate the shortest distance.

The shortest distance between skew lines is given by:

\(d = \frac{|(\mathbf{b_1} \times \mathbf{b_2}) \cdot \mathbf{PQ}|}{|\mathbf{b_1} \times \mathbf{b_2}|}\)

Where:

• \(\mathbf{b_1} = \langle 2, -3, 2 \rangle\) (direction ratios of \( L_1 \))
• \(\mathbf{b_2} = \langle 3, 2, 0 \rangle\) (direction ratios of \( L_2 \))
• \(\mathbf{PQ} = \langle -4 - 1, 4 + 1, 3 + 4 \rangle = \langle -5, 5, 7 \rangle\)

Now, calculate \(\mathbf{b_1} \times \mathbf{b_2}\):

\(\mathbf{b_1} \times \mathbf{b_2} = |\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}|\)

\(= \mathbf{i}(0 - 4) - \mathbf{j}(0 - 6) + \mathbf{k}(4 + 9)\)

\(= -4 \mathbf{i} + 6 \mathbf{j} + 13 \mathbf{k} = \langle -4, 6, 13 \rangle\)

Then calculate the dot product with \(\mathbf{PQ}\):

\((\mathbf{b_1} \times \mathbf{b_2}) \cdot \mathbf{PQ} = \langle -4, 6, 13 \rangle \cdot \langle -5, 5, 7 \rangle\)

\(= (20 + 30 + 91) = 141\)

Magnitude of \(\mathbf{b_1} \times \mathbf{b_2}\):

\(|\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-4)^2 + 6^2 + 13^2} = \sqrt{16 + 36 + 169} = \sqrt{221}\)

Therefore, the shortest distance is:

\(d = \frac{141}{\sqrt{221}}\)

This matches the correct answer: \( \frac{141}{\sqrt{221}} \).

Answer 2

Identify direction ratios and vector between points on the lines. The direction ratios of \( L_1 \) are \(\langle 2, -3, 2 \rangle\), and the direction ratios of \( L_2 \) are \(\langle 3, 2, 0 \rangle\) (since \( z \) is constant, indicating parallel planes along the \( z \)-axis).

Compute vector \(\overrightarrow{AB}\) between points on \( L_1 \) and \( L_2 \). Select points \( A(1, -1, -4) \) on \( L_1 \) and \( B(-4, 4, 3) \) on \( L_2 \). Calculate \(\overrightarrow{AB}\):

\[ \overrightarrow{AB} = \langle -4 - 1, 4 - (-1), 3 - (-4) \rangle = \langle -5, 5, 7 \rangle. \]

Use the shortest distance formula. The shortest distance (S.D.) between two skew lines with direction vectors \(\overrightarrow{d_1} = \langle 2, -3, 2 \rangle\) and \(\overrightarrow{d_2} = \langle 3, 2, 0 \rangle\), and a vector \(\overrightarrow{AB}\) between points on each line, is given by:

\[ \text{S.D} = \frac{|\overrightarrow{AB} \cdot (\overrightarrow{d_1} \times \overrightarrow{d_2})|}{|\overrightarrow{d_1} \times \overrightarrow{d_2}|}. \]

Calculate \(\overrightarrow{d_1} \times \overrightarrow{d_2}\):

\[ \overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix}. \]

Expanding the determinant:

\[ = \mathbf{i}((-3)(0) - (2)(2)) - \mathbf{j}((2)(0) - (2)(3)) + \mathbf{k}((2)(2) - (-3)(3)), \] \[ = \mathbf{i}(-4) - \mathbf{j}(-6) + \mathbf{k}(4 + 9), \] \[ = \langle -4, 6, 13 \rangle. \]

Compute \(\overrightarrow{AB} \cdot (\overrightarrow{d_1} \times \overrightarrow{d_2})\)

\[ \overrightarrow{AB} \cdot (\overrightarrow{d_1} \times \overrightarrow{d_2}) = \langle -5, 5, 7 \rangle \cdot \langle -4, 6, 13 \rangle, \] \[ = (-5)(-4) + (5)(6) + (7)(13), \] \[ = 20 + 30 + 91 = 141. \]

Calculate \(|\overrightarrow{d_1} \times \overrightarrow{d_2}|\)

\[ |\overrightarrow{d_1} \times \overrightarrow{d_2}| = \sqrt{(-4)^2 + 6^2 + 13^2} = \sqrt{16 + 36 + 169} = \sqrt{221}. \]

Substitute values into the shortest distance formula:

\[ \text{S.D} = \frac{|141|}{\sqrt{221}} = \frac{141}{\sqrt{221}}. \]

Therefore, the answer is:

\[ \frac{141}{\sqrt{221}}. \]