Ex 11.2

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

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 = (š1) ā + š (š1) ā and š ā = (š2) ā + š(š2) ā is |("(" (šš) āĆ (šš) ā")" ."(" (šš) ā ā (šš) ā")" )/|(šš) ā Ć (šš) ā | | š ā = (š ā t) š Ģ + (šāš)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(ā1š Ģ + 1š Ģ ā 2š Ģ) Comparing with š ā = (š1) ā + t (š1) ā, (š1) ā = 1š Ģ ā 2š Ģ + 3š Ģ & (š1) ā = ā 1š Ģ + 1š Ģ ā 2š Ģ š ā = (š + 1) š Ģ + (šš" 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 = (š2) ā + s (š2) ā, (š2) ā = 1š Ģ ā 1š Ģ ā 1š Ģ & (š2) ā = 1š Ģ + 2š Ģ ā 2š Ģ Now, ((šš) 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 ) āĆ (š_š ) ā ) = |ā 8(š Ģ&š Ģ&š Ģ@ ā1&1& ā2@1&2& ā2)| = š Ģ[(1Ćā 2)ā(2Ćā 2)] ā š Ģ[(ā1Ćā2)ā(1Ćā 2)] + š Ģ[(ā 1Ć2)ā(1Ć1)] = š Ģ[ā2+4] ā š Ģ[2+2] A + š Ģ[ā2ā1] = 2š Ģ ā 4š Ģ ā 3š Ģ Magnitude of ((š1) āĆ(š2) ā) = ā(22+(ā 4)2+(ā 3)2) |(šš) āĆ(šš) ā | = ā(4+16+9) = āšš Also, ((šš) ā Ć (šš) ā) . ((šš) ā ā (šš) ā) = (2š Ģ ā 4š Ģ ā 3š Ģ) . (0š Ģ + 1š Ģ ā 4š Ģ) = (2 Ć 0) + (ā4 Ć 1) + (ā3 Ć ā4) = ā0 + (ā4) + 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/ā29| = š/āšš Therefore, the shortest distance between the given two lines is 8/ā29 .