Last updated at March 11, 2017 by Teachoo

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Misc 20 (Method 1) Find the vector equation of the line passing through the point (1, 2, –4) and perpendicular to the two lines: 𝑥 − 83 = 𝑦 + 19−16 = 𝑧 − 107 and 𝑥 − 153 = 𝑦 − 298 = 𝑧 − 5−5 The vector equation of a line passing through a point with position vector 𝑎 and parallel to a vector 𝑏 is 𝒓 = 𝒂 + 𝜆 𝒃 The line passes through (1,2, −4) So, 𝑎 = 1 𝑖 + 2 𝑗 − 4 𝑘 Given, line is perpendicular to both lines ∴ 𝑏 is perpendicular to both lines We know that 𝑎 × 𝑏 is perpendicular to both 𝑎 & 𝑏 So, 𝑏 is cross product of both lines 𝑥 − 83 = 𝑦 + 19−16 = 𝑧 − 107 and 𝑥 − 153 = 𝑦 − 298 = 𝑧 − 5−5 Required normal = 𝑖 𝑗 𝑘3−16738−5 = 𝑖 (–16(-5) – 8(7)) – 𝑗 (3(-5) – 3(7)) + 𝑘(3(8) – 3(–16)) = 𝑖 (80 – 56) – 𝑗 (–15 – 21) + 𝑘(24 + 48) = 24 𝑖 + 36 𝑗 + 72 𝑘 Thus, 𝑏 = 24 𝑖 + 36 𝑗 + 72 𝑘 Now, Putting value of 𝑎 & 𝑏 in formula 𝑟 = 𝑎 + 𝜆 𝑏 ∴ 𝑟 = (1 𝑖 + 2 𝑗 – 4 𝑘) + 𝜆 (24 𝑖 + 36 𝑗 + 72 𝑘) = ( 𝑖 + 2 𝑗 – 4 𝑘) + 𝜆12 (2 𝑖 + 3 𝑗 + 6 𝑘) = ( 𝑖 + 2 𝑗 – 4 𝑘) + 𝜆 (2 𝑖 + 3 𝑗 + 6 𝑘) Therefore, the equation of the line is ( 𝒊 + 2 𝒋 – 4 𝒌) + 𝜆 (2 𝒊 + 3 𝒋 + 6 𝒌). Misc 20 (Method 2) Find the vector equation of the line passing through the point (1, 2, –4) and perpendicular to the two lines: 𝑥 − 83 = 𝑦 + 19−16 = 𝑧 − 107 and 𝑥 − 153 = 𝑦 − 298 = 𝑧 − 5−5 The vector equation of a line passing through a point with position vector 𝑎 and parallel to a vector 𝑏 is 𝒓 = 𝒂 + 𝜆 𝒃 The line passes through (1,2, −4) So, 𝑎 = 1 𝑖 + 2 𝑗 − 4 𝑘 Let 𝑏 = x 𝑖 + y 𝑗 + z 𝑘 Two lines with direction ratios 𝑎1 , 𝑏1 , 𝑐1 and 𝑎2 , 𝑏2 , 𝑐2 are perpendicular if 𝒂𝟏 𝒂𝟐 + 𝒃𝟏𝒃𝟐 + 𝒄𝟏 𝒄𝟐 = 0 Given, line 𝑏 is perpendicular to 𝑥 − 83 = 𝑦 + 1916 = 𝑧 − 107 and 𝑥 − 153 = 𝑦 − 298 = 𝑧 − 5 − 5 So, 3x − 16y + 7z = 0 and 3x + 8y − 5z = 0 𝑥80 − 56 = 𝑦21 − ( − 15) = 𝑧24 − ( − 48) 𝑥24 = 𝑦36 = 𝑧72 𝑥2 = 𝑦3 = 𝑧6 = k Hence, x = 2k , y = 3k , & z = 6k Thus, 𝑏 = x 𝑖 + y 𝑗 + z 𝑘 = 2k 𝑖 + 3k 𝑗 + 6k 𝑘 Now, Putting value of 𝑎 & 𝑏 in formula 𝑟 = 𝑎 + 𝜆 𝑏 ∴ 𝑟 = ( 𝑖 + 2 𝑗 − 4 𝑘) + 𝜆 (2k 𝑖 + 3k 𝑗 + 6k 𝑘) = ( 𝑖 + 2 𝑗 − 4 𝑘) + 𝜆k (2 𝑖 + 3 𝑗 + 6 𝑘) = ( 𝑖 + 2 𝑗 − 4 𝑘) + 𝜆 (2 𝑖 + 3 𝑗 + 6 𝑘) Therefore, the equation of the line is ( 𝒊 + 2 𝒋 − 4 𝒌) + 𝜆(2 𝒊 + 3 𝒋 + 6 𝒌)

Example, 3
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Ex 11.1, 2 Important

Example, 6 Important

Example, 9 Important

Example 12 Important

Ex 11.2, 5 Important

Ex 11.2, 11 Important

Ex 11.2, 12 Important

Ex 11.2, 14 Important

Ex 11.2, 15 Important

Ex 11.2, 17 Important

Example 20 Important

Example 21 Important

Example 23 Important

Example 24 Important

Example, 25 Important

Ex 11.3, 4 Important

Ex 11.3, 11 Important

Ex 11.3, 12 Important

Ex 11.3, 14 Important

Example 27 Important

Example 29 Important

Example 30 Important

Misc 6 Important

Misc 9 Important

Misc 14 Important

Misc 18 Important

Misc 20 Important You are here

Misc 21 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.