JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \(\vec{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R, \alpha>0\) and \(\vec{r}_{2}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R\) is \(9\), then \(\alpha\) is equal to \(.....\)
- A \(21\)
- B \(4\)
- C \(66\)
- D \(6\)
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
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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