Question
Download Solution PDFनिम्न सरल रेखाओं के मध्य कोण है:
r̅ = 4î - ĵ + λ(î + 2ĵ - 2k̂) and
r̅ = î - ĵ + 2k̂ + μ(2î + 4ĵ - 4k̂)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
रेखाओं का युग्म है:
r̅ = 4î - ĵ + λ(î + 2ĵ - 2k̂)
r̅ = î - ĵ + 2k̂ + μ(2î + 4ĵ - 4k̂)
अवधारणा:
दो रेखाओं के बीच का कोण जब उनका समीकरण दिया गया हो।
यदि θ रेखाओं के बीच न्यून कोण है
\(\vec r = \vec a_1 \ + \ \lambda \vec b_1 \) और \(\vec r = \vec a_2 \ + \ \mu \vec b_2\)
\(cos θ = \left| \frac{ \vec b_1 . \vec b_2}{|\vec b_1| |\vec b_2|} \right|\)
प्रयुक्त सूत्र:
cos0° = 1
गणना:
हमारे पास है,
r̅ = 4î - ĵ + λ(î + 2ĵ - 2k̂)
r̅ = î - ĵ + 2k̂ + μ(2î + 4ĵ - 4k̂)
यहाँ, \(\vec b_1 = \hat i + 2\hat j - 2\hat k \) और \(\vec b_2 = 2\hat i \ + \ 4\hat j\ -\ 4 \hat k\)
दो रेखाओं के बीच का कोण निम्न द्वारा दिया जाता है
⇒ \(cos θ = \left| \frac{ \vec b_1 . \vec b_2}{|\vec b_1| |\vec b_2|} \right|\)
⇒ cosθ = \( \left| \frac{ (\hat i \ + \ 2\hat j \ -\ 2\hat k) .(2\hat i \ + \ 4\hat j \ - \ 4 \hat k)}{\sqrt{ 1 \ + \ 4 \ + 4} \sqrt{4 \ + \ 16 \ +\ 16}} \right|\)
⇒ cosθ = \(\left|\frac{ 2 \ + \ 8 \ + 8 }{3 \ \times \ 6} \right|\)
⇒ cosθ = \(\frac{18}{18}\)
⇒ cosθ = 1
⇒ θ = \(cos^{-1}1\)
⇒ θ = 0
∴ रेखाओं के युग्म के बीच का कोण 0° है।
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