JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the point \((-1, \alpha, \beta)\) lie on the line of the shortest distance between the lines \(\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}\) and \(\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}\) Then \((\alpha-\beta)^2\) is equal to ....................
- A \(65\)
- B \(45\)
- C \(32\)
- D \(25\)
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
\( \mathrm{P}(-3 \lambda-2,4 \lambda+2,2 \lambda+5) \) \( \mathrm{Q}(-\mu-2,2 \mu-6,1) \) \( \mathrm{DRS} \text { of } \mathrm{PQ}=(3 \lambda-\mu, 2 \mu-4 \lambda-8,-2 \lambda-4)\) \(D R S\) of…
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