Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Example 28 - Chapter 10 Class 12 Vector Algebra
Last updated at May 29, 2023 by Teachoo
Examples
Example 1
Important
Example 2 Important
Example 3 Important
Example 4
Example 5
Example 6 Important
Example 7
Example 8 Important
Example 9
Example 10
Example 11 (i) Important
Example 11 (ii) Important
Example 12 Important
Example 13
Example 14 Important
Example 15
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20
Example 21 Important
Example 22
Example 23 Important
Example 24
Example 25 Important
Example 26
Example 27 Important
Example 28 Important You are here
Example 29 Important
Example 30 Important
Example 26 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams
Example 27 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams
Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams
Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams
Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams
Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams
Transcript
Example 28 Let 𝑎 ⃗, 𝑏 ⃗ and 𝑐 ⃗ be three vectors such that |𝑎 ⃗|= 3, |𝑏 ⃗|= 4, |𝑐 ⃗|= 5 and each one of them being perpendicular to the sum of the other two, find |𝑎 ⃗" + " 𝑏 ⃗" + " 𝑐 ⃗ |.Given, |𝑎 ⃗| = 3 , |𝑏 ⃗| = 4 , |𝑐 ⃗|= 5 Also, Each one of them being perpendicular to the sum of the other two 𝒂 ⃗ is perpendicular to 𝒃 ⃗ + 𝒄 ⃗ 𝑎 ⃗. (𝑏 ⃗ + 𝑐 ⃗) = 0 𝒂 ⃗. 𝒃 ⃗ + 𝒂 ⃗. 𝒄 ⃗ = 0 𝒃 ⃗ is perpendicular to 𝒂 ⃗ + 𝒄 ⃗ 𝑏 ⃗. (𝑎 ⃗ + 𝑐 ⃗) = 0 𝒃 ⃗. 𝒂 ⃗ + 𝒃 ⃗. 𝒄 ⃗ = 0 𝒄 ⃗ is perpendicular to 𝒂 ⃗ + 𝒃 ⃗ 𝑐 ⃗. (𝑎 ⃗ + 𝑏 ⃗) = 0 𝒄 ⃗. 𝒂 ⃗ + 𝒄 ⃗. 𝒃 ⃗ = 0 Now, |𝒂 ⃗+𝒃 ⃗+𝒄 ⃗ |2 = (𝒂 ⃗ + 𝒃 ⃗ + 𝒄 ⃗) . (𝒂 ⃗ + 𝒃 ⃗ + 𝒄 ⃗) = 𝑎 ⃗. 𝑎 ⃗ + 𝑎 ⃗ . 𝑏 ⃗ + 𝑎 ⃗ . 𝑐 ⃗ + 𝑏 ⃗ . 𝑎 ⃗ + 𝑏 ⃗ . 𝑏 ⃗ + 𝑏 ⃗ . 𝑐 ⃗ + 𝑐 ⃗ . 𝑎 ⃗ + 𝑐 ⃗ . 𝑏 ⃗ + 𝑐 ⃗ . 𝑐 ⃗ = 𝑎 ⃗. 𝑎 ⃗ + (𝒂 ⃗ . 𝒃 ⃗ + 𝒂 ⃗ . 𝒄 ⃗) + 𝑏 ⃗ . 𝑏 ⃗ + (𝒃 ⃗ . 𝒂 ⃗ + 𝒃 ⃗ . 𝒄 ⃗) + 𝑐 ⃗ . 𝑐 ⃗ + (𝒄 ⃗ . 𝒂 ⃗ + 𝒄 ⃗ . 𝒃 ⃗ = 𝑎 ⃗. 𝑎 ⃗ + (0) + 𝑏 ⃗ . 𝑏 ⃗ + (0) + 𝑐 ⃗ . 𝑐 ⃗ + (0) = 𝒂 ⃗. 𝒂 ⃗ + 𝑏 ⃗. 𝑏 ⃗ + 𝑐 ⃗. 𝑐 ⃗ = "|" 𝒂 ⃗"|"2 + "|" 𝑏 ⃗"|"2 + "|" 𝑐 ⃗"|"2 = 32 + 42 + 52 = 9 + 16 + 25 = 50 So, |𝒂 ⃗+𝒃 ⃗+𝒄 ⃗ |2 = 50 Taking square root both sides, "|" 𝑎 ⃗ "+ " 𝑏 ⃗" + " 𝑐 ⃗"|" = √50 "|" 𝑎 ⃗ "+ " 𝑏 ⃗ "+ " 𝑐 ⃗"|" = √25 × √2 "|" 𝒂 ⃗ "+ " 𝒃 ⃗ "+ " 𝒄 ⃗"|" = 𝟓√𝟐 = 𝑎 ⃗. 𝑎 ⃗ + (0) + 𝑏 ⃗ . 𝑏 ⃗ + (0) + 𝑐 ⃗ . 𝑐 ⃗ + (0) = 𝒂 ⃗. 𝒂 ⃗ + 𝑏 ⃗. 𝑏 ⃗ + 𝑐 ⃗. 𝑐 ⃗ = "|" 𝒂 ⃗"|"2 + "|" 𝑏 ⃗"|"2 + "|" 𝑐 ⃗"|"2 = 32 + 42 + 52 = 9 + 16 + 25 = 50 So, |𝒂 ⃗+𝒃 ⃗+𝒄 ⃗ |2 = 50 Taking square root both sides, "|" 𝑎 ⃗ "+ " 𝑏 ⃗" + " 𝑐 ⃗"|" = √50 "|" 𝑎 ⃗ "+ " 𝑏 ⃗ "+ " 𝑐 ⃗"|" = √25 × √2 "|" 𝒂 ⃗ "+ " 𝒃 ⃗ "+ " 𝒄 ⃗"|" = 𝟓√𝟐
