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
Ex 11.2, 12 Important
Ex 11.2, 13 Important
Ex 11.2, 15 Important
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Question 11 Important Deleted for CBSE Board 2024 Exams
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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
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Question 20 Important Deleted for CBSE Board 2024 Exams
Misc 3 Important
Misc 4 Important
Question 10 Important Deleted for CBSE Board 2024 Exams You are here
Question 14 Important Deleted for CBSE Board 2024 Exams
Misc 5 Important
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Chapter 11 Class 12 Three Dimensional Geometry
Last updated at April 16, 2024 by Teachoo
Question 10 If the points (1, 1 , p) and (–3 , 0, 1) be equidistant from the plane 𝑟 ⃗. (3𝑖 ̂ + 4𝑗 ̂ − 12𝑘 ̂) + 13 = 0, then find the value of p. The distance of a point with position vector 𝑎 ⃗ from the plane 𝑟 ⃗.𝑛 ⃗ = d is |(𝒂 ⃗.𝒏 ⃗ − 𝒅)/|𝒏 ⃗ | | Given, the points are The equation of plane is 𝑟 ⃗. (3𝑖 ̂ + 4𝑗 ̂ − 12𝑘 ̂) + 13 = 0 𝑟 ⃗.(3𝑖 ̂ + 4𝑗 ̂ − 12𝑘 ̂) = −13 (1, 1, p) So, (𝑎_1 ) ⃗ = 1𝑖 ̂ + 1𝑗 ̂ + p𝑘 ̂ (−3, 0, 1) So, (𝑎_2 ) ⃗ = −3𝑖 ̂ + 0𝑗 ̂ + 1𝑘 ̂ –𝑟 ⃗.(3𝑖 ̂ + 4𝑗 ̂ − 12𝑘 ̂) = 13 𝒓 ⃗.(–3𝒊 ̂ – 4𝒋 ̂ + 12𝒌 ̂) = 13 Comparing with 𝑟 ⃗.𝑛 ⃗ = d, 𝑛 ⃗ = −3𝑖 ̂ − 4𝑗 ̂ + 12𝑘 ̂ d = 13 Magnitude of 𝑛 ⃗ = √((−3)^2+(−4)^2+〖12〗^2 ) |𝑛 ⃗ | = √(9+16+144) = √169 = 13 Distance of point (𝒂𝟏) ⃗ from plane |((𝑎1) ⃗"." 𝑛 ⃗" " − 𝑑)/|𝑛 ⃗ | | = |((1𝑖 ̂ + 1𝑗 ̂ + 𝑝𝑘 ̂ ).(−3𝑖 ̂−4𝑗 ̂+12𝑘 ̂ )−13)/13| = |((1×−3)+(1×−4) +(𝑝×12)−13)/13| = |(−3−4+12𝑝−13)/13| = |(12𝑝 − 20)/13| Distance of point (𝒂𝟐) ⃗ from plane |((𝑎2) ⃗"." 𝑛 ⃗ − 𝑑)/|𝑛 ⃗ | | = |((−3𝑖 ̂ +0𝑗 ̂ +1𝑘 ̂ ).(−3𝑖 ̂−4𝑗 ̂+12𝑘 ̂ )−13)/13| = |((−3×−3)+(0×−4) +(1×12)−13)/13| = |(9 + 0 +12−13)/13| = |8/13| = 8/13 Since the plane is equidistance from both the points, |(𝟏𝟐𝒑 − 𝟐𝟎)/𝟏𝟑| = 𝟖/𝟏𝟑 |12𝑝−20| = 8 (12p – 20) = ± 8 12p − 20 = 8 12p = 8 + 20 12p = 28 p = 28/12 p = 7/3 12p − 20 = −8 12p = −8 + 20 12p = 12 p = 12/12 p = 1 Answer does not match at end. If mistake, please comment