JEE Mains · Maths · STD 12 - 11. three dimension geometry
The line \(l_1\) passes through the point \((2,6,2)\) and is perpendicular to the plane \(2 x+y-2 z=10\). Then the shortest distance between the line \(l_1\) and the line \(\frac{ x +1}{2}=\frac{ y +4}{-3}=\frac{ z }{2}\) is :
- A \(7\)
- B \(\frac{19}{3}\)
- C \(\frac{19}{3}\)
- D \(9\)
Answer & Solution
Correct Answer
(D) \(9\)
Step-by-step Solution
Detailed explanation
Line \(\ell\), is given by \(L_1: \frac{x-2}{2}=\frac{y-6}{1}=\frac{z-2}{-2}\) Given, \(L _2: \frac{ x +1}{2}=\frac{ y +4}{-3}=\frac{ z }{2}\) \(\text { Shortest distance }=\left|\frac{\overline{ AB } \cdot \overline{ MN }}{ MN }\right|\)…
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