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

यदि रेखाओं \(\bar{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R\), \(\alpha>0\) तथा \(\overrightarrow{ r _{2}}=-4 \hat{ i }-\hat{ k }+\mu(3 \hat{ i }-2 \hat{ j }-2 \hat{ k }), \mu \in R\) के मध्य न्यूनतम दूरी \(9\) है, तो \(\alpha\) बराबर है .......... |

  1. A \(21\)
  2. B \(4\)
  3. C \(66\)
  4. D \(6\)
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Correct Answer

(D) \(6\)

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Detailed explanation

If \(\vec{r}=\bar{a}+\lambda \vec{b}\) and \(\vec{r}=\vec{c}+\lambda \vec{d}\) then shortest distance between two lines is \(L=\frac{(\bar{a}-\vec{c}) \cdot(\bar{b} \times \bar{d})}{|b \times d|}\) \(\therefore \vec{a}-\vec{c}=((\alpha+4) \hat{i}+2 \hat{j}+3 k)\)…
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