1. Chapter 11 Class 12 Three Dimensional Geometry
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  3. Plane
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Question 12 - Plane - Chapter 11 Class 12 Three Dimensional Geometry

Last updated at Dec. 16, 2024 by Teachoo

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Question 12 Find the angle between the planes whose vector equations are ๐‘Ÿ โƒ— . (2๐‘– ฬ‚ + 2๐‘— ฬ‚ โ€“ 3๐‘˜ ฬ‚) = 5 and ๐‘Ÿ โƒ— . (3๐‘– ฬ‚ โ€“ 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚) = 3 .Angle between two planes ๐‘Ÿ โƒ— . (๐‘›_1 ) โƒ— = d1 and ๐‘Ÿ โƒ—.(๐‘›2) โƒ— = d2 is given by cos ๐œƒ = |((๐’๐Ÿ) โƒ—. (๐’๐Ÿ) โƒ—)/|(๐’๐Ÿ) โƒ— ||(๐’๐Ÿ) โƒ— | | ๐’“ โƒ—.(2๐’Š ฬ‚ + 2๐’‹ ฬ‚ โˆ’ 3๐’Œ ฬ‚) = 5 Comparing with ๐‘Ÿ โƒ—.(๐‘›1) โƒ— = (๐‘‘1) โƒ—, (๐‘›1) โƒ— = 2๐‘– ฬ‚ + 2๐‘— ฬ‚โˆ’3๐‘˜ ฬ‚ Magnitude of (๐‘›1) โƒ— = โˆš(2^2+2^2+ใ€–(โˆ’3)ใ€—^2 ) |(๐‘›1) โƒ— | = โˆš(4+4+9) = โˆš17 So, cos ฮธ = |((2๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ ) . (3๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚ ))/(โˆš17 ร— โˆš43)| = |((2 ร— 3) + (2 ร— โˆ’3) + (โˆ’3 ร— 5))/โˆš(17 ร— 43)| = |(6 โˆ’ 6 โˆ’ 15)/โˆš731| = |(โˆ’15)/โˆš731| = 15/โˆš731 So, cos ฮธ = 15/โˆš731 โˆด ฮธ = cosโˆ’1(๐Ÿ๐Ÿ“/โˆš๐Ÿ•๐Ÿ‘๐Ÿ) Therefore, the angle between the planes is cosโˆ’1(15/โˆš731).
  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

    3. Plane

    Direction cosines and ratios →
Direction cosines and ratios →
Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo