CUET · MATHS · PYQ PAPER 2023
The shortest distance between the lines\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\)and\(\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}\) is :
- A \(\frac{\sqrt{293}}{7}\)
- B 0
- C \(7 \sqrt{293}\)
- D \(\frac{7}{\sqrt{293}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{293}}{7}\)
Step-by-step Solution
Detailed explanation
\(\vec{d_1} = (2, 3, 6)\), \(\vec{d_2} = (4, 6, 12)\). Since \(\vec{d_2} = 2\vec{d_1}\), the lines are parallel. \(P_1 = (1, 2, -4)\), \(P_2 = (3, 3, -5)\). \(\vec{P_1P_2} = (3-1, 3-2, -5-(-4)) = (2, 1, -1)\). Distance \(D = \frac{|\vec{P_1P_2} \times \vec{d_1}|}{|\vec{d_1}|}\).…
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