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: Set up the parallel line distance.
$$d = \frac{|(\vec a_2 - \vec a_1)\times \vec b|}{|\vec b|}$$ with $\vec b = 2\hat i + \hat j - 2\hat k$.

Step 2: Find the joining vector and cross product.
$\vec a_2 - \vec a_1 = -\hat i + 0\hat j + \hat k$. Its cross product with $\vec b$ is $-\hat i - \hat k$, of length $\sqrt2$.

Step 3: Divide by $|\vec b|$.
$|\vec b| = \sqrt{4 + 1 + 4} = 3$, so $d = \dfrac{\sqrt2}{3}$.
\[ \boxed{\dfrac{\sqrt2}{3}\ \text{units}} \]