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Ex 11.2
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
Last updated at March 22, 2023 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