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
- A 42
- B 46
- C 48
- D 40
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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