Question

The distance between parallel lines $\bar{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(2\hat{i} + \hat{j} - 2\hat{k})$ and $\bar{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + \hat{j} - 2\hat{k})$ is

• A. $2\text{ units}$
• B. $\frac{1}{3}\text{ units}$
• C. $\frac{1}{\sqrt{3}}\text{ units}$
• D. $\frac{\sqrt{2}}{3}\text{ units}$

Hint:

Don't confuse the parallel line distance formula with the skew line distance formula! Parallel lines use a cross product in the numerator $|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|$, while skew lines use a dot product scalar triple product $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the shortest perpendicular distance between two lines in 3D space. The lines have the exact same direction vector, confirming they are parallel.

Step 2: Key Formula or Approach:
The distance $d$ between two parallel lines defined by vector equations $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$ is given by:
$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$$ Where $\vec{a}_1$ and $\vec{a}_2$ are position vectors of points on the respective lines, and $\vec{b}$ is their common direction vector.

Step 3: Detailed Explanation:
Identify the vectors from the given equations:
$\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{a}_2 = \hat{i} - \hat{j} + 2\hat{k}$
$\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$
First, find the difference vector $(\vec{a}_2 - \vec{a}_1)$:
$$\vec{a}_2 - \vec{a}_1 = (1 - 2)\hat{i} + (-1 - (-1))\hat{j} + (2 - 1)\hat{k}$$ $$\vec{a}_2 - \vec{a}_1 = -\hat{i} + 0\hat{j} + \hat{k}$$ Next, compute the cross product $(\vec{a}_2 - \vec{a}_1) \times \vec{b}$ using a determinant:
$$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 0 & 1 \\ 2 & 1 & -2 \end{vmatrix}$$ $$= \hat{i}(0 - 1) - \hat{j}(2 - 2) + \hat{k}(-1 - 0)$$ $$= -\hat{i} - \hat{k}$$ Now, calculate the magnitude of this cross product vector:
$$|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{(-1)^2 + (0)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$$ Calculate the magnitude of the direction vector $\vec{b}$:
$$|\vec{b}| = \sqrt{(2)^2 + (1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ Finally, plug these magnitudes into the distance formula:
$$d = \frac{\sqrt{2}}{3}$$

Step 4: Final Answer:
The distance between the parallel lines is $\frac{\sqrt{2}}{3}\text{ units}$, corresponding to option (D).