Chapter 11 Class 12 Three Dimensional Geometry
Ex 11.1, 2
Example, 6 Important
Example, 7
Example 10 Important
Ex 11.2, 5 Important
Ex 11.2, 9 (i) Important
Ex 11.2, 10 Important You are here
Ex 11.2, 12 Important
Ex 11.2, 13 Important
Ex 11.2, 15 Important
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 4 (a) Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Important Deleted for CBSE Board 2024 Exams
Question 14 (a) Important Deleted for CBSE Board 2024 Exams
Question 17 Important Deleted for CBSE Board 2024 Exams
Question 19 Important Deleted for CBSE Board 2024 Exams
Question 20 Important Deleted for CBSE Board 2024 Exams
Misc 3 Important
Misc 4 Important
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 14 Important Deleted for CBSE Board 2024 Exams
Misc 5 Important
Question 16 Important Deleted for CBSE Board 2024 Exams
Chapter 11 Class 12 Three Dimensional Geometry
Last updated at April 16, 2024 by Teachoo
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