JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\) be two lines. Then which of the following points lies on the line of the shortest distance between \(L_1\) and \(L_2\) ?
- A \(\left(\frac{14}{3},-3, \frac{22}{3}\right)\)
- B \(\left(-\frac{5}{3},-7,1\right)\)
- C \(\left(2,3, \frac{1}{3}\right)\)
- D \(\left(\frac{8}{3},-1, \frac{1}{3}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{14}{3},-3, \frac{22}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\ & L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\end{aligned}\) \(\begin{aligned} & P(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & Q(3 \mu+2,4 \mu+4,5 \mu+5) \end{aligned}\) Dr's of…
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