Question

The shortest distance between lines $\vec{r} = (2\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{j} - 3\hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + \hat{j} - 5\hat{k})$ is

• A. $\frac{1}{\sqrt{5}}$ units
• B. $3$ units
• C. $\sqrt{5}$ units
• D. $2$ units

Hint:

Always glance at the direction vectors first! Notice that the first two components of $\vec{b}_1$ and $\vec{b}_2$ are identical ($2\hat{i} + \hat{j}$). This indicates that the lines share a common directional plane alignment, which simplifies the cross product greatly since the $\hat{k}$ component drops straight to zero ($2 - 2 = 0$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two skew lines in vector form, $\vec{r}=\vec{a}_1+\lambda\vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu\vec{b}_2$. We need to compute the shortest distance between them.

Step 2: Key Formula or Approach:
The shortest distance between two skew lines is given by $$d=\frac{|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)|}{|\vec{b}_1\times\vec{b}_2|}.$$

Step 3: Detailed Explanation:
From the given lines, $$\vec{a}_1=2\hat{i}-\hat{j},\qquad \vec{b}_1=2\hat{i}+\hat{j}-3\hat{k},$$ $$\vec{a}_2=\hat{i}-\hat{j}+2\hat{k},\qquad \vec{b}_2=2\hat{i}+\hat{j}-5\hat{k}.$$ Hence, $$\vec{a}_2-\vec{a}_1=(\hat{i}-2\hat{i})+(-\hat{j}+\hat{j})+(2\hat{k}-0\hat{k})=-\hat{i}+2\hat{k}.$$ Now, $$\vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-3\\2&1&-5\end{vmatrix}=-2\hat{i}+4\hat{j}.$$ Therefore, $$|\vec{b}_1\times\vec{b}_2|=\sqrt{(-2)^2+4^2}=\sqrt{20}=2\sqrt{5}.$$ Also, $$(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)=(-\hat{i}+2\hat{k})\cdot(-2\hat{i}+4\hat{j})=2.$$ Substituting into the formula, $$d=\frac{|2|}{2\sqrt{5}}=\frac{1}{\sqrt{5}}\ \text{units}.$$

Step 4: Final Answer:
The shortest distance is $\frac{1}{\sqrt{5}}$ units, which corresponds to option (A).