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 :
- A \(-1\)
- B \(2\)
- C \(3\)
- D \(5\)
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…
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