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RBSE Solutions for Class 12 Maths Chapter 14 त्रि – विमीयज्यामिति Ex 14.5

May 24, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 12 Maths Chapter 14 त्रि – विमीयज्यामिति Ex 14.5

प्रश्न 1.
रेखाओं
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
और
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
के मध्य की न्यूनतम दूरी ज्ञात कीजिए।
हल :
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5

प्रश्न 2.
रेखाओं
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
के मध्य की न्यूनतम दूरी ज्ञात कीजिए।
हल :
रेखाओं
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
के बीच की न्यूनतम दूरी
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5

प्रश्न 3.
रेखाएँ, जिनके सदिश समीकरण निम्नलिखित है, के मध्य की न्यूनतम दूरी ज्ञात कीजिए
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
और
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
हल :
रेखाएँओं
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5

प्रश्न 4.
रेखाएँ, जिसकी सदिश समीकरण निम्नलिखित हैं, के मध्य की न्यूनतम दूरी ज्ञात कीजिए
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
और
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
हल :
रेखा
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5

प्रश्न 5.
निम्न रेखाओं के मध्य लघुत्तम दूरी ज्ञात कीजिए
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
तथा लघुतम दूरी वाली रेखा का समीकरण भी ज्ञात कीजिए।
हल :
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
अतः रेखा (1) पर कोई बिंदु P(2r1 + 1, 3r1 – 1, r1) तथा
रेखा (2) पर कोई बिंदु Q(3r2 – 1, r2 + 2, r2 + 2)
तब रेखा PQ के दिक्-अनुपात
= 3r2 – 2r1 – 2, r2 – 3r1 + 3, 2 – r1
PQ रेखा (1) के लम्बवत् हैं, इसलिए
a1a2 + b1b2 + c1c2 = 0
2(3r2 – 2r1 – 2) + 3(r2 – 3r1 + 3) + 1(2 – r1) = 0
9r2 – 14r1 = 7 …(3)
PQ रेखा (2) के लम्बवत् है।।
इसलिए a1a2 + b1b2 + c1c2 = 0
3(3r2 – 2r1 – 2) + 1(r2 – 3r1 + 3) + 0(2 – r1) = 0
10r2 – 9r1 – 3 = 0
समी. (3) व (4) को हल करने पर
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5
r1 व r2 के ये मान बिन्दु P व Q में रखने पर
RBSE Solutions for Class 12 Maths Chapter 14 त्रि - विमीयज्यामिति Ex 14.5

RBSE Solutions for Class 12 Maths

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