JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the shortest distance between the lines \(\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}\) and \(\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}\) is \(\frac{44}{\sqrt{30}}\), then the largest possible value of \(|\lambda|\) is equal to ..........
- A \(45\)
- B \(49\)
- C \(43\)
- D \(40\)
Answer & Solution
Correct Answer
(C) \(43\)
Step-by-step Solution
Detailed explanation
\( \overline{\mathrm{a}}_1=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \) \( \overline{\mathrm{a}}_2=-2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \) \( \overrightarrow{\mathrm{p}}-=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \)…
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