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: Note that the lines are parallel.
Both lines share the same direction $\bar{b}=2\hat{i}+\hat{j}-2\hat{k}$, so they are parallel and we use the parallel-line distance formula.
Step 2: Write down the two starting points.
Point on first line $\bar{a}_1=2\hat{i}-\hat{j}+\hat{k}$, point on second $\bar{a}_2=\hat{i}-\hat{j}+2\hat{k}$.
Step 3: Find the joining vector.
$\bar{a}_2-\bar{a}_1=(1-2)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}=-\hat{i}+0\hat{j}+\hat{k}$.
Step 4: Cross it with the direction.
$(-\hat{i}+\hat{k})\times(2\hat{i}+\hat{j}-2\hat{k})=\hat{i}(0\cdot(-2)-1\cdot1)-\hat{j}((-1)(-2)-1\cdot2)+\hat{k}((-1)(1)-0\cdot2)=-\hat{i}-0\hat{j}-\hat{k}$.
Step 5: Take the two magnitudes.
Length of the cross product is $\sqrt{(-1)^2+(-1)^2}=\sqrt{2}$. Length of $\bar{b}$ is $\sqrt{4+1+4}=3$.
Step 6: Divide to get the distance.
$d=\dfrac{\sqrt{2}}{3}$ units, which is option (4). \[ \boxed{d=\frac{\sqrt{2}}{3}\text{ units}} \]