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Ex 11.2, 12 - Class 12 - Find p so that lines are right angles

Ex 11.2, 12 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Ex 11.2, 12 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

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Ex 11.2, 12 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.