Last updated at March 11, 2017 by Teachoo

Transcript

Example 30 (Method 1) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane. The equation of a line passing through two points with position vectors 𝑎 & 𝑏 is 𝒓 = 𝒂 + 𝜆 ( 𝒃 − 𝒂) Given the line passes through the points ( 𝑏 − 𝑎) = (5 𝑖 + 1 𝑗 + 6 𝑘) − (3 𝑖 + 4 𝑗 + 1 𝑘) = 2 𝑖 − 3 𝑗 + 5 𝑘 ∴ 𝑟 = (3 𝑖 + 4 𝑗 + 1 𝑘) + 𝜆 (2 𝑖 − 3 𝑗 + 5 𝑘) Let the coordinates of the point where the line crosses the XY plane be (x, y, 0). So, 𝑟 = x 𝑖 + y 𝑗 + 0 𝑘 Since point crosses the plane, it will satisfy its equation Putting (2) in (1) x 𝑖 + y 𝑗 + 0 𝑘 = 3 𝑖 + 4 𝑗 + 1 𝑘 + 2𝜆 𝑖 − 3𝜆 𝑗 + 5𝜆 𝑘 = (3 + 2𝜆) 𝑖 + (4 − 3𝜆) 𝑗 + (1 + 5𝜆) 𝑘 Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 5𝜆 ∴ 𝜆 = −𝟏𝟓 So, x = 3 + 𝜆 = 3 + 2 × −15 = 3 − 25 = 135 & y = 4 − 3𝜆 = 4 − 3 × −15 = 4 + 35 = 235 Therefore, the required coordinates are 𝟏𝟑𝟓, 𝟐𝟑𝟓,𝟎 Example 30 (Method 2) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane. The equation of a line passing through two points A( 𝑥1, 𝑦1, 𝑧1) and B( 𝑥2, 𝑦2, 𝑧2) is 𝒙 − 𝒙𝟏 𝒙𝟐 − 𝒙𝟏 = 𝒚 − 𝒚𝟏 𝒚𝟐 − 𝒚𝟏 = 𝒛 − 𝒛𝟏 𝒛𝟐 − 𝒛𝟏 Given the line passes through the points So, the equation of line is 𝑥 − 35 − 3 = 𝑦 − 41 − 4 = 𝑧 − 16 − 1 𝑥 − 32 = 𝑦 − 4−3 = 𝑧 − 15 = k So, Since, the line crosses the XY plane at (x, y, 0) z = 0 5k + 1 = 0 5k = −1 ∴ k = −𝟏𝟓 So, x = 2k + 3 = 2 × −15 + 3 = 3 − 25 = 135 y = −3 × −15 + 4 = 4 + 35 = 235 Therefore, the coordinates of the required point are 𝟏𝟑𝟓, 𝟐𝟑𝟓, 𝟎.

Example, 3
Important

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 You are here

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.