COMEDK · Maths · 34. Three Dimensional Geometry
The shortest distance between the lines \(\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\) and \(\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}\) is
\(\begin{array}{ll}\text { a. } \sqrt{30} \quad \text { b. } 2 \sqrt{30} & \text { c. } 5 \\ \lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right) & =\end{array}\)
- A \(\frac{b+1}{2 a}\)
- B \(\frac{b}{a}\)
- C 0
- D \(\frac{2 b}{a}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 b}{a}\)
Step-by-step Solution
Detailed explanation
\(L_{1}: \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}\) \(L_{2}: \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}\) Shortest distance between two lines…
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