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

Answer: (C)

$\frac{141}{\sqrt{221}}$

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:

$\frac{141}{\sqrt{221}}.$