Question

The shortest distance between the lines \(\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})\) and \(\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})\) is

• A. \(\frac{1}{\sqrt{5}}\) units
• B. \(\frac{6}{\sqrt{5}}\) units
• C. \(\frac{2}{\sqrt{5}}\) units
• D. \(\frac{3}{\sqrt{5}}\) units

Hint:

Use \(d=\dfrac{|(\bar a_2-\bar a_1)\cdot(\bar b_1\times\bar b_2)|}{|\bar b_1\times\bar b_2|}\).

Answer 1

The Correct Option is B

Step 1: Compute the Normal Direction
The common perpendicular is along $\bar b_1\times\bar b_2=(2,-1,0)$, so the unit normal is $\dfrac{1}{\sqrt5}(2,-1,0)$.

Step 2: Project the Connecting Vector
The vector joining the points is $(-3,0,2)$. Its component along the unit normal is $\dfrac{-6}{\sqrt5}$.

Step 3: Result
The distance is the magnitude, $\dfrac{6}{\sqrt5}$ units, option (B).

Final Answer:
Projecting the joining vector on the common normal gives 6/root 5. \[ \boxed{\text{(B) }\dfrac{6}{\sqrt{5}}\text{ units}} \]