Ex 11.2

Ex 11.2, 1

Ex 11.2, 2

Ex 11.2, 3 Important

Ex 11.2, 4

Ex 11.2, 5 Important

Ex 11.2, 6

Ex 11.2, 7 Important

Ex 11.2, 8 (i) Important

Ex 11.2, 8 (ii)

Ex 11.2, 9 (i) Important

Ex 11.2, 9 (ii)

Ex 11.2, 10 Important

Ex 11.2, 11

Ex 11.2, 12 Important

Ex 11.2, 13 Important

Ex 11.2, 14

Ex 11.2, 15 Important You are here

Question 1 Important Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at April 16, 2024 by Teachoo

Ex 11.2, 15 Find the shortest distance between the lines whose vector equations are š ā = (1 ā t) š Ģ + (t ā 2) š Ģ + (3 ā 2t) š Ģ and š ā = (s + 1) š Ģ + (2s ā 1) š Ģ ā (2s + 1) š Ģ Shortest distance between lines with vector equations š ā = (š1) ā + š (š1) ā and š ā = (š2) ā + š(š2) ā is |("(" (šš) āĆ (šš) ā")" ."(" (šš) ā ā (šš) ā")" )/|(šš) ā Ć (šš) ā | | š ā = (š ā t) š Ģ + (šāš)š Ģ + (3 ā 2t) š Ģ = 1š Ģ ā tš Ģ + tš Ģ ā 2š Ģ + 3š Ģ ā 2tš Ģ = (1š Ģ ā 2š Ģ + 3š Ģ) + t(ā1š Ģ + 1š Ģ ā 2š Ģ) Comparing with š ā = (š1) ā + t (š1) ā, (š1) ā = 1š Ģ ā 2š Ģ + 3š Ģ & (š1) ā = ā 1š Ģ + 1š Ģ ā 2š Ģ š ā = (š + 1) š Ģ + (šš" ā " š)š Ģ ā (2s + 1) š Ģ = sš Ģ + 1š Ģ + 2sš Ģ ā 1š Ģ ā 2sš Ģ ā 1š Ģ = (1š Ģ ā 1š Ģ ā 1š Ģ) + s(1š Ģ + 2š Ģ ā 2š Ģ) Comparing with š ā = (š2) ā + s (š2) ā, (š2) ā = 1š Ģ ā 1š Ģ ā 1š Ģ & (š2) ā = 1š Ģ + 2š Ģ ā 2š Ģ Now, ((šš) ā ā (š_š ) ā) = (1š Ģ ā 1š Ģ ā 1š Ģ) ā (1š Ģ ā 2š + 3š Ģ) = (1 ā 1) š Ģ + ( ā 1 + 2)š Ģ + ( ā 1 ā 3)š Ģ = 0š Ģ + 1š Ģ ā 4š Ģ ( (š_š ) āĆ (š_š ) ā ) = |ā 8(š Ģ&š Ģ&š Ģ@ ā1&1& ā2@1&2& ā2)| = š Ģ[(1Ćā 2)ā(2Ćā 2)] ā š Ģ[(ā1Ćā2)ā(1Ćā 2)] + š Ģ[(ā 1Ć2)ā(1Ć1)] = š Ģ[ā2+4] ā š Ģ[2+2] A + š Ģ[ā2ā1] = 2š Ģ ā 4š Ģ ā 3š Ģ Magnitude of ((š1) āĆ(š2) ā) = ā(22+(ā 4)2+(ā 3)2) |(šš) āĆ(šš) ā | = ā(4+16+9) = āšš Also, ((šš) ā Ć (šš) ā) . ((šš) ā ā (šš) ā) = (2š Ģ ā 4š Ģ ā 3š Ģ) . (0š Ģ + 1š Ģ ā 4š Ģ) = (2 Ć 0) + (ā4 Ć 1) + (ā3 Ć ā4) = ā0 + (ā4) + 12 = 8 So, shortest distance = |(((š_1 ) ā Ć (š_2 ) ā ) . ((š_2 ) ā Ć (š_1 ) ā ).)/((š_1 ) ā Ć (š_2 ) ā )| = |8/ā29| = š/āšš Therefore, the shortest distance between the given two lines is 8/ā29 .