Last updated at May 29, 2018 by Teachoo

Transcript

Ex 10.4, 2 Find a unit vector perpendicular to each of the vector 𝑎 + 𝑏 and 𝑎 − 𝑏, where 𝑎 = 3 𝑖 + 2 𝑗 + 2 𝑘 and 𝑏 = 𝑖 + 2 𝑗 − 2 𝑘 . 𝑎 = 3 𝑖 + 2 𝑗 + 2 𝑘 𝑏 = 1 𝑖 + 2 𝑗 − 2 𝑘 ( 𝑎 + 𝑏) = (3 + 1) 𝑖 + (2 + 2) 𝑗 + (2 − 2) 𝑘 = 4 𝑖 + 4 𝑗 + 0 𝑘 ( 𝑎 − 𝑏) = (3 − 1) 𝑖 + (2 − 2) 𝑗 + (2 − (−2)) 𝑘 = 2 𝑖 + 0 𝑗 + 4 𝑘 Now, we need to find a vector perpendicular to both 𝑎 + 𝑏 and 𝑎 − 𝑏, We know that ( 𝑎 × 𝑏) is perpendicular to 𝑎 and 𝑏 Replacing 𝑎 by ( 𝑎 + 𝑏) & 𝑏 by ( 𝑎 − 𝑏) ( 𝒂 + 𝒃) × ( 𝒂 − 𝒃) will be perpendicular to ( 𝒂 + 𝒃) and ( 𝒂 − 𝒃) Let 𝑐 = ( 𝑎 + 𝑏) × ( 𝑎 − 𝑏) ∴ 𝑐 = 𝑖 𝑗 𝑘 42 40 04 = 𝑖 4×4−(0×0) − 𝑗 4×4−(2×0) + 𝑘 4×0−(2×4) = 𝑖 (16 − 0) − 𝑗 (16 − 0) + 𝑘 (0 − 8) = 16 𝑖 − 16 𝑗 − 8 𝑘 ∴ 𝑐 = 16 𝑖 − 16 𝑗 − 8 𝑘 Now, Unit vector of 𝑐 = 1𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑐 × 𝑐 Magnitude of 𝑐 = 162+ −162+ −82 𝑐 = 256+256+64 = 576 = 24 Unit vector of 𝑐 = 1 𝑐 × 𝑐 = 124 × 16 𝑖 − 16 𝑗 − 8 𝑘 = 𝟐𝟑 𝒊 − 𝟐𝟑 𝒋 − 𝟏𝟑 𝒌 . Therefore the required unit vector is 23 𝑖 − 23 𝑗 − 13 𝑘 . Note: There are always two perpendicular vectors So, another vector would be = − 𝟐𝟑 𝒊 − 𝟐𝟑 𝒋 − 𝟏𝟑 𝒌 = −𝟐𝟑 𝒊 + 𝟐𝟑 𝒋 + 𝟏𝟑 𝒌 Hence, the perpendicular vectors are 23 𝑖 − 23 𝑗 − 13 𝑘 & −23 𝑖 + 23 𝑗 + 13 𝑘

Ex 10.2, 7
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Ex 10.2, 9 Important

Ex 10.2, 10 Important

Ex 10.2, 13 Important

Ex 10.2, 17 Important

Example 14 Important

Example 16 Important

Example 21 Important

Ex 10.3, 2 Important

Ex 10.3, 3 Important

Ex 10.3, 10 Important

Ex 10.3, 13 Important

Ex 10.3, 16 Important

Example 23 Important

Example 24 Important

Example 25 Important

Ex 10.4, 2 Important You are here

Ex 10.4, 5 Important

Ex 10.4, 9 Important

Ex 10.4, 10 Important

Ex 10.4, 11 Important

Example 28 Important

Example 29 Important

Example 30 Important

Misc 6 Important

Misc 12 Important

Misc 13 Important

Misc 15 Important

Misc 19 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.