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

Let the shortest distance between the lines \(\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}\) and \(\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}\) be \(3 \sqrt{30}\). Then the positive value of \(5 \alpha+\beta\) is

  1. A 42
  2. B 46
  3. C 48
  4. D 40
Verified Solution

Answer & Solution

Correct Answer

(B) 46

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & {\mathrm{A}(3, \alpha, 3) \& B(-3,-7, \beta)} \\ & \overrightarrow{\mathrm{BA}}=6 \hat{\mathrm{i}}+(\alpha+7) \hat{\mathrm{j}}+(3-\beta) \hat{\mathrm{k}} \\ \end{aligned}\)…
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