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

Show Hint

Use \(d=\dfrac{|(\bar a_2-\bar a_1)\cdot(\bar b_1\times\bar b_2)|}{|\bar b_1\times\bar b_2|}\).
Updated On: Nov 13, 2026
  • \(\frac{1}{\sqrt{5}}\) units
  • \(\frac{6}{\sqrt{5}}\) units
  • \(\frac{2}{\sqrt{5}}\) units
  • \(\frac{3}{\sqrt{5}}\) units
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept
For skew lines \(\bar r=\bar a_1+\lambda\bar b_1\) and \(\bar r=\bar a_2+\mu\bar b_2\), the shortest distance is \(\dfrac{|(\bar a_2-\bar a_1)\cdot(\bar b_1\times\bar b_2)|}{|\bar b_1\times\bar b_2|}\).

Step 2: Cross Product
\(\bar b_1\times\bar b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&-3\\2&4&-5\end{vmatrix}=2\hat i-\hat j\), with magnitude \(\sqrt5\).

Step 3: Dot Product
\(\bar a_2-\bar a_1=-3\hat i+2\hat k\), so \((\bar a_2-\bar a_1)\cdot(2\hat i-\hat j)=-6\).

Step 4: Distance
\(d=\dfrac{|-6|}{\sqrt5}=\dfrac{6}{\sqrt5}\text{ units}\)

Final Answer:
The shortest distance is 6 over root 5 units. \[ \boxed{\text{(B) }\dfrac{6}{\sqrt{5}}\text{ units}} \]
Was this answer helpful?
0
0

Top MHT CET Distance between Two Lines Questions

View More Questions