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Last updated at Jan. 31, 2020 by Teachoo
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Example 23 Find a unit vector perpendicular to each of the vectors ๐ โ + ๐ โ and ๐ โ โ ๐ โ where ๐ โ = ๐ ฬ + ๐ ฬ + ๐ ฬ, b = ๐ ฬ + 2 ๐ ฬ + 3๐ ฬ . ๐ โ = ๐ ฬ + ๐ ฬ + ๐ ฬ = 1๐ ฬ + 1๐ ฬ + 1๐ ฬ ๐ โ = ๐ ฬ + 2๐ ฬ + 3๐ ฬ = 1๐ ฬ + 2๐ ฬ + 3๐ ฬ (๐ โ + ๐ โ) = (1 + 1) ๐ ฬ + (1 + 2) ๐ ฬ + (1 + 3) ๐ ฬ = 2๐ ฬ + 3๐ ฬ + 4๐ ฬ (๐ โ โ ๐ โ) = (1 โ 1) ๐ ฬ + (1 โ 2) ๐ ฬ + (1 โ 3) ๐ ฬ = 0๐ ฬ โ 1๐ ฬ โ 2๐ ฬ Now, we need to find a vector perpendicular to both ๐ โ + ๐ โ and ๐ โ โ ๐ โ, We know that (๐ โ ร ๐ โ) is perpendicular to ๐ โ and ๐ โ Replacing ๐ โ by (๐ โ + ๐ โ) & ๐ โ by (๐ โ โ ๐ โ) (๐ โ + ๐ โ) ร (๐ โ โ ๐ โ) will be perpendicular to (๐ โ + ๐ โ) and (๐ โ โ ๐ โ) Let (๐ โ + ๐ โ) ร (๐ โ โ ๐ โ) = ๐ โ ๐ โ = |โ 8(๐ ฬ&๐ ฬ&๐ ฬ@2&3&4@0&โ1&โ2)| = ๐ ฬ [(3รโ2)โ(โ1ร4)] โ๐ ฬ [(2รโ2)โ(0ร4)] + ๐ ฬ [(2รโ1)โ(0ร3)] = ๐ ฬ [โ6โ(โ4)] โ๐ ฬ [โ4โ0] + ๐ ฬ [โ2โ0] = ๐ ฬ (โ6 + 4) โ๐ ฬ (โ4) + ๐ ฬ(โ2) = โ2๐ ฬ + 4๐ ฬ โ 2๐ ฬ โด ๐ โ = -2๐ ฬ + 4๐ ฬ โ 2๐ ฬ Now, Unit vector of ๐ โ = 1/(๐๐๐๐๐๐ก๐ข๐๐ ๐๐๐ โ ) ร ๐ โ Magnitude of ๐ โ = โ((โ2)2+(4)^2+(โ2)2) |๐ โ | = โ(4+16+4) = โ24 = โ(2. ร 2 ร 6) = 2โ6 Unit vector of ๐ โ = 1/|๐ โ | ร ๐ โ = 1/(2โ6) [โ2๐ ฬ+4๐ ฬโ2๐ ฬ ] = 1/(2โ6) ร 2 [โ๐ ฬ+2๐ ฬโ๐ ฬ ] = (โ1)/โ6 ๐ ฬ + 2/โ6 ๐ ฬ โ 1/โ6 ๐ ฬ Therefore, required the unit vector is = (โ๐)/โ๐ ๐ ฬ + ๐/โ๐ ๐ ฬ โ ๐/โ๐ ๐ ฬ Note: There are always two perpendicular vectors So, another vector would be = โ((โ1)/โ6 ๐ ฬ" + " 2/โ6 ๐ ฬ" โ " 1/โ6 ๐ ฬ ) = ๐/โ๐ ๐ ฬ" โ" ๐/โ๐ ๐ ฬ" + " ๐/โ๐ ๐ ฬ Hence, perpendicular vectors are (โ1)/โ6 ๐ ฬ + 2/โ6 ๐ ฬ โ 1/โ6 ๐ ฬ & 1/โ6 ๐ ฬ" โ" 2/โ6 ๐ ฬ" + " 1/โ6 ๐ ฬ
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Example 26 (Supplementary NCERT) Not in Syllabus - CBSE Exams 2021
Example 27 (Supplementary NCERT) Not in Syllabus - CBSE Exams 2021
Example 28 (Supplementary NCERT) Important Not in Syllabus - CBSE Exams 2021
Example 29 (Supplementary NCERT) Important Not in Syllabus - CBSE Exams 2021
Example 30 (Supplementary NCERT) Important Not in Syllabus - CBSE Exams 2021
Example 31 (Supplementary NCERT) Important Not in Syllabus - CBSE Exams 2021
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