ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 11. three dimension geometry

If the shortest distance between the straight lines \(3(x-1)=6(y-2)=2(z-1)\) and \(4(\mathrm{x}-2)=2(\mathrm{y}-\lambda)=(\mathrm{z}-3), \lambda \in \mathrm{R}\) is \(\frac{1}{\sqrt{38}}\), then the integral value of \(\lambda\) is equal to :

  1. A \(-1\)
  2. B \(2\)
  3. C \(3\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

\(L_{1}: \frac{(x-1)}{2}=\frac{(y-2)}{1}=\frac{(z-1)}{3} \quad \vec{r}_{1}=2 \hat{i}+\hat{j}+3 \hat{k}\) \(L_{2}: \frac{(x-2)}{1}=\frac{y-\lambda}{2}=\frac{z-3}{4} \quad \vec{r}_{2}=\hat{i}+2 \hat{j}+4 \hat{k}\) Shortest distance \(=\) Projection of \(\vec{a}\) on…
From JEE Mains
Explore more questions on app