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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 (i) Important

Ex 11.2, 8 (ii)

Ex 11.2, 9 (i) Important

Ex 11.2, 9 (ii)

Ex 11.2, 10 Important

Ex 11.2, 11 You are here

Ex 11.2, 12 Important

Ex 11.2, 13 Important

Ex 11.2, 14

Ex 11.2, 15 Important

Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Aug. 14, 2023 by Teachoo

Ex 11.2, 11 Show that the lines (𝑥 − 5)/7 = (𝑦 + 2)/( −5) = 𝑧/1 and 𝑥/1 = 𝑦/2 = 𝑧/3 are perpendicular to each other. Two lines (𝑥 − 𝑥1)/𝑎1 = (𝑦 − 𝑦1)/𝑏1 = (𝑧 − 𝑧1)/𝑐1 and (𝑥 − 𝑥2)/𝑎2 = (𝑦 − 𝑦2)/𝑏2 = (𝑧 − 𝑧2)/𝑐2 are perpendicular to each other if 𝒂𝟏 𝒂𝟐 + 𝒃𝟏 𝒃𝟐 + 𝒄𝟏 𝒄𝟐 = 0 (𝒙 − 𝟓)/𝟕 = (𝒚 + 𝟐)/( − 𝟓) = 𝒛/𝟏 (𝑥 − 5)/7 = (𝑦 − (−2))/( −5) = (𝑧 − 0)/1 Comparing with (𝑥 − 𝑥1)/𝑎1 = (𝑦 − 𝑦1)/𝑏1 = (𝑧 − 𝑧1)/𝑐1, So, 𝑥1 = 5, y1 = −2, 𝑧1 = 0 & 𝒂𝟏 = 7, 𝒃𝟏 = − 5, 𝒄𝟏 = 1, 𝒙/𝟏 = 𝒚/𝟐 = 𝒛/𝟑 (𝑥 − 0)/1 = (𝑦 − 0)/2 = (𝑧 − 0)/3 Comparing with (𝑥 − 𝑥2)/𝑎2 = (𝑦 − 𝑦2)/𝑏2 = (𝑧 − 𝑧2)/𝑐2, So, x2 = 0, y2 = 0, z2 = 0, & 𝒂𝟐 = 1, b2 = 2, c2 = 3 So, 𝒂𝟏 𝒂𝟐 + 𝒃𝟏 𝒃𝟐 + 𝒄𝟏 𝒄𝟐 = (7 × 1) + (−5 × 2) + (1 × 3) = 7 + (−10) + 3 = 0 Therefore, the two given lines are perpendicular to each other.