JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the line of the shortest distance between the lines \(L_1: \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and \(L_2: \vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})\) intersect \(\mathrm{L}_1\) and \(\mathrm{L}_2\) at \(\mathrm{P}\) and \(\mathrm{Q}\) respectively. If \((\alpha, \beta, \gamma)\) is the midpoint of the line segment \(PQ\), then \(2(\alpha+\beta+\gamma)\) is equal to ...........
- A \(21\)
- B \(25\)
- C \(30\)
- D \(35\)
Answer & Solution
Correct Answer
(A) \(21\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\left(D R^{\prime} \text { s of } L_1\right)\) \(\overrightarrow{\mathrm{d}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\left(\mathrm{DR}\right.\) 's of \(\left.L_2\right)\)…
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