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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

Ex 11.2, 9 Important

Ex 11.2, 10 (i) Important

Ex 11.2, 10 (ii)

Ex 11.2, 11 (i) Important

Ex 11.2, 11 (ii)

Ex 11.2, 12 Important You are here

Ex 11.2, 13

Ex 11.2, 14 Important

Ex 11.2, 15 Important

Ex 11.2, 16

Ex 11.2, 17 Important

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Feb. 1, 2020 by Teachoo

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