Example 23 - Find a unit vector perpendicular to a + b, a - b

Example 23 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 23 - Chapter 10 Class 12 Vector Algebra - Part 3

Example 23 - Chapter 10 Class 12 Vector Algebra - Part 4

  1. Chapter 10 Class 12 Vector Algebra (Term 2)
  2. Serial order wise

Transcript

Example 23 Find a unit vector perpendicular to each of the vectors ๐‘Ž โƒ— + ๐‘ โƒ— and ๐‘Ž โƒ— โˆ’ ๐‘ โƒ— where ๐‘Ž โƒ— = ๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚, b = ๐‘– ฬ‚ + 2 ๐‘— ฬ‚ + 3๐‘˜ ฬ‚ . Finding (๐’‚ โƒ— + ๐’ƒ โƒ—) and (๐’‚ โƒ— โˆ’ ๐’ƒ โƒ—) (๐’‚ โƒ— + ๐’ƒ โƒ—) = (1 + 1) ๐‘– ฬ‚ + (1 + 2) ๐‘— ฬ‚ + (1 + 3) ๐‘˜ ฬ‚ = 2๐’Š ฬ‚ + 3๐’‹ ฬ‚ + 4๐’Œ ฬ‚ (๐’‚ โƒ— โˆ’ ๐’ƒ โƒ—) = (1 โˆ’ 1) ๐‘– ฬ‚ + (1 โˆ’ 2) ๐‘— ฬ‚ + (1 โˆ’ 3) ๐‘˜ ฬ‚ = 0๐’Š ฬ‚ โˆ’ 1๐’‹ ฬ‚ โˆ’ 2๐’Œ ฬ‚ Now, we need to find a vector perpendicular to both ๐‘Ž โƒ— + ๐‘ โƒ— and ๐‘Ž โƒ— โˆ’ ๐‘ โƒ—, We know that (๐‘Ž โƒ— ร— ๐‘ โƒ—) is perpendicular to ๐‘Ž โƒ— and ๐‘ โƒ— Replacing ๐‘Ž โƒ— by (๐‘Ž โƒ— + ๐‘ โƒ—) & ๐‘ โƒ— by (๐‘Ž โƒ— โˆ’ ๐‘ โƒ—) (๐’‚ โƒ— + ๐’ƒ โƒ—) ร— (๐’‚ โƒ— โˆ’ ๐’ƒ โƒ—) will be perpendicular to (๐‘Ž โƒ— + ๐‘ โƒ—) and (๐‘Ž โƒ— โˆ’ ๐‘ โƒ—) Let ๐‘ โƒ— = (๐‘Ž โƒ— + ๐‘ โƒ—) ร— (๐‘Ž โƒ— โˆ’ ๐‘ โƒ—) ๐’„ โƒ— = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@2&3&4@0&โˆ’1&โˆ’2)| = ๐‘– ฬ‚ [(3ร—โˆ’2)โˆ’(โˆ’1ร—4)] โˆ’๐‘— ฬ‚ [(2ร—โˆ’2)โˆ’(0ร—4)] + ๐‘˜ ฬ‚ [(2ร—โˆ’1)โˆ’(0ร—3)] = ๐‘– ฬ‚ [โˆ’6โˆ’(โˆ’4)] โˆ’๐‘— ฬ‚ [โˆ’4โˆ’0] + ๐‘˜ ฬ‚ [โˆ’2โˆ’0] = ๐‘– ฬ‚ (โˆ’6 + 4) โˆ’๐‘— ฬ‚ (โˆ’4) + ๐‘˜ ฬ‚(โˆ’2) = โˆ’2๐’Š ฬ‚ + 4๐’‹ ฬ‚ โˆ’ 2๐’Œ ฬ‚ Since we need to find unit vector perpendicular Unit vector of ๐‘ โƒ— = ๐Ÿ/(๐‘ด๐’‚๐’ˆ๐’๐’Š๐’•๐’–๐’…๐’† ๐’๐’‡๐’„ โƒ— ) ร— ๐’„ โƒ— = 1/โˆš((โˆ’2)2 + (4)^2 + (โˆ’2)2) ร— (โˆ’2๐‘– ฬ‚ + 4๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚) = 1/โˆš(4 + 16 + 4) ร— (โˆ’2๐‘– ฬ‚ + 4๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚) = 1/(2โˆš6) ร— (โˆ’2๐‘– ฬ‚ + 4๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚) = (โˆ’๐Ÿ)/โˆš๐Ÿ” ๐’Š ฬ‚ + ๐Ÿ/โˆš๐Ÿ” ๐’‹ ฬ‚ โˆ’ ๐Ÿ/โˆš๐Ÿ” ๐’Œ ฬ‚ Note: There are always two perpendicular vectors So, another vector would be = โˆ’((โˆ’1)/โˆš6 ๐‘– ฬ‚" + " 2/โˆš6 ๐‘— ฬ‚" โˆ’ " 1/โˆš6 ๐‘˜ ฬ‚ ) = ๐Ÿ/โˆš๐Ÿ” ๐’Š ฬ‚" โˆ’" ๐Ÿ/โˆš๐Ÿ” ๐’‹ ฬ‚" + " ๐Ÿ/โˆš๐Ÿ” ๐’Œ ฬ‚ Hence, Perpendicular vectors are (โˆ’1)/โˆš6 ๐‘– ฬ‚ + 2/โˆš6 ๐‘— ฬ‚ โˆ’ 1/โˆš6 ๐‘˜ ฬ‚ & 1/โˆš6 ๐‘– ฬ‚" โˆ’" 2/โˆš6 ๐‘— ฬ‚" + " 1/โˆš6 ๐‘˜ ฬ‚

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.