Question 12 - Plane - Chapter 11 Class 12 Three Dimensional Geometry

Last updated at May 29, 2023 by Teachoo

Ex 11.3, 12 - Find angle between planes r.(2i + 2j - 3k) = 5

Ex 11.3, 12 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Book a free demo

Transcript

Question 12 Find the angle between the planes whose vector equations are 𝑟 ⃗ . (2𝑖 ̂ + 2𝑗 ̂ – 3𝑘 ̂) = 5 and 𝑟 ⃗ . (3𝑖 ̂ – 3𝑗 ̂ + 5𝑘 ̂) = 3 .Angle between two planes 𝑟 ⃗ . (𝑛_1 ) ⃗ = d1 and 𝑟 ⃗.(𝑛2) ⃗ = d2 is given by cos 𝜃 = |((𝒏𝟏) ⃗. (𝒏𝟐) ⃗)/|(𝒏𝟏) ⃗ ||(𝒏𝟐) ⃗ | | 𝒓 ⃗.(2𝒊 ̂ + 2𝒋 ̂ − 3𝒌 ̂) = 5 Comparing with 𝑟 ⃗.(𝑛1) ⃗ = (𝑑1) ⃗, (𝑛1) ⃗ = 2𝑖 ̂ + 2𝑗 ̂−3𝑘 ̂ Magnitude of (𝑛1) ⃗ = √(2^2+2^2+〖(−3)〗^2 ) |(𝑛1) ⃗ | = √(4+4+9) = √17 So, cos θ = |((2𝑖 ̂ + 2𝑗 ̂ − 3𝑘 ̂ ) . (3𝑖 ̂ − 3𝑗 ̂ + 5𝑘 ̂ ))/(√17 × √43)| = |((2 × 3) + (2 × −3) + (−3 × 5))/√(17 × 43)| = |(6 − 6 − 15)/√731| = |(−15)/√731| = 15/√731 So, cos θ = 15/√731 ∴ θ = cos−1(𝟏𝟓/√𝟕𝟑𝟏) Therefore, the angle between the planes is cos−1(15/√731).

Next: Question 13 (a) Important → Ask a doubt

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular