JEE Mains · Maths · STD 12 - 11. three dimension geometry
The shortest distance between the lines \(\vec{r}=\left(\dfrac{1}{3}\hat{i}+2\hat{j}+\dfrac{8}{3}\hat{k}\right)+\lambda(2\hat{i}-5\hat{j}+6\hat{k})\) and \(\vec{r}=\left(-\dfrac{2}{3}\hat{i}-\dfrac{1}{3}\hat{k}\right)+\mu(\hat{j}-\hat{k})\), \(\lambda,\mu \in \mathbb{R}\), is:
- A \(\sqrt{5}\)
- B \(3\)
- C \(2\sqrt{3}\)
- D \(\sqrt{15}\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
The given lines are \(\vec{r} = \vec{a}_1 + \lambda \vec{b}_1\) and \(\vec{r} = \vec{a}_2 + \mu \vec{b}_2\), where: \(\vec{a}_1 = \dfrac{1}{3}\hat{i} + 2\hat{j} + \dfrac{8}{3}\hat{k}\) \(\vec{b}_1 = 2\hat{i} - 5\hat{j} + 6\hat{k}\)…
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