Last updated at May 29, 2018 by Teachoo

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Example, 6Find the vector and the Cartesian equations of the line through the point (5, 2, โ 4) and which is parallel to the vector 3๐ย ฬ + 2๐ย ฬ โ 8๐ย ฬ . Vector equation Equation of a line passing through a point with position vector ๐ย โ , and parallel to a vector ๐ย โ is ๐ย โ = ๐ย โ + ๐๐ย โ Since line passes through (5, 2, โ 4) ๐ย โ = 5๐ย ฬ + 2๐ย ฬ โ 4๐ย ฬ Since line is parallel to 3๐ย ฬ + 2๐ย ฬ โ 8๐ย ฬ ๐ย โ = 3๐ย ฬ + 2๐ย ฬ โ 8๐ย ฬ Equation of line ๐ย โ = ๐ย โ + ๐๐ย โ ๐ย โ = (5๐ย ฬ + 2๐ย ฬ โ 4๐ย ฬ) + ๐ (3๐ย ฬ + 2๐ย ฬ โ 8๐ย ฬ) Therefore, equation of line in vector form is ๐ย โ = (5๐ย ฬ + 2๐ย ฬ โ 4๐ย ฬ) + ๐ (3๐ย ฬ + 2๐ย ฬ โ 8๐ย ฬ) Cartesian equation Equation of a line passing through a point (x, y, z) and parallel to a line with direction ratios a, b, c is (๐ฅ โ ๐ฅ1)/๐ = (๐ฆ โ ๐ฆ1)/๐ = (๐ง โ ๐ง1)/๐ Since line passes through (5, 2, โ4) ๐ฅ1 = 5, y1 = 2 , z1 = โ4 Also, line is parallel to 3๐ย ฬ + 2๐ย ฬ โ 8๐ย ฬ , ๐ = 3, b = 2, c = โ 8 Equation of line in Cartesian form is (๐ฅ โ ๐ฅ1)/๐ = (๐ฆ โ ๐ฆ1)/๐ = (๐ง โ ๐ง1)/๐ (๐ฅ โ 5)/3 = (๐ฆ โ 2)/2 = (๐ง โ ( โ 4))/( โ 8) (๐ โ ๐)/๐ = (๐ โ ๐)/๐ = (๐ + ๐)/(โ๐)

Example, 3
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Ex 11.1, 2 Important

Example, 6 Important You are here

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

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

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.