Ex 11.2

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Ex 11.2, 10 Find the values of 𝑝 so that the lines (1 − 𝑥)/3 = (7𝑦 − 14)/2𝑝 =(𝑧 − 3)/2 and (7 − 7𝑥)/3𝑝 = (𝑦 − 5)/1 = (6 − 𝑧)/5 are at right angles. Two lines (𝑥 − 𝑥1)/𝑎1 = (𝑦 − 𝑦1)/𝑏1 = (𝑧 − 𝑧1)/𝑐1 and (𝑥 − 𝑥2)/𝑎2 = (𝑦 − 𝑦2)/𝑏2 = (𝑧 − 𝑧2)/𝑐2 are at right angles to each other if 𝒂𝟏 𝒂𝟐 + 𝒃𝟏 𝒃𝟐 + 𝒄𝟏 𝒄𝟐 = 0 (𝟏 − 𝒙)/𝟑 = (𝟕𝒚 − 𝟏𝟒)/𝟐𝒑 = (𝒛 − 𝟑)/𝟐 ( −(𝑥 − 1))/3 = (7(𝑦 − 2))/2𝑝 = (𝑧 − 3)/2 (𝑥 − 1)/( −3) = (𝑦 − 2)/(2𝑝/7) = (𝑧 − 3)/2 Comparing with (𝑥 − 𝑥1)/𝑎1 = (𝑦 − 𝑦1)/𝑏1 = (𝑧 − 𝑧1)/𝑐1 𝑥1 = 1, y1 = 2, z1 = 3 & a1 = −3, b1 = 𝟐𝒑/𝟕 , c1 = 2 (𝟕 − 𝟕𝒙)/𝟑𝒑 = (𝒚 − 𝟓)/𝟏 = (𝟔 − 𝒛)/𝟓 ( −7(𝑥 − 1))/3𝑝 = (𝑦 − 5)/1 = ( − (𝑧 − 6))/5 (𝒙 − 𝟏)/( (−𝟑𝒑)/𝟕) = (𝒚 − 𝟓)/𝟏 = (𝒛 − 𝟔)/( −𝟓) Comparing with (𝑥 − 𝑥2)/𝑎2 = (𝑦 − 𝑦2)/𝑏2 = (𝑧 − 𝑧2)/𝑐2, x2 = 1, y2 = 5, z2 = 6 & 𝒂2 = ( − 𝟑𝒑)/𝟕, b2 = 1, c2 = −5 Since the lines are perpendicular 𝒂1𝒂𝟐+𝒃𝟏𝒃𝟐+𝒄𝟏𝒄𝟐 = 0 (−3×( − 3𝑝)/7) + (2𝑝/7×1 ) + (2 × −5) = 0 𝟗𝒑/𝟕 + 𝟐𝒑/𝟕 − 10 = 0 11𝑝/7 = 10 p = 10 × 7/11 ∴ p = 𝟕𝟎/𝟏𝟏

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.