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Question 7 - Equation of plane - Intercept Form - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at Dec. 16, 2024 by Teachoo
Equation of plane - Intercept Form
Chapter 11 Class 12 Three Dimensional Geometry
Concept wise
- Direction cosines and ratios
- Equation of line - given point and //vector
- Equation of line - given 2 points
- Angle between two lines - Vector
- Angle between two lines - Cartisian
- Angle between two lines - Direction ratios or cosines
- Shortest distance between two skew lines
- Shortest distance between two parallel lines
- Equation of plane - In Normal Form
- Equation of plane - Prependicular to Vector & Passing Through Point
- Equation of plane - Passing Through 3 Non Collinear Points
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Equation of plane - Intercept Form
- Equation of plane - Passing Through Intersection Of Planes
- Coplanarity of 2 lines
- Angle between two planes
- Distance of point from plane
- Angle between Line and Plane
- Equation of line under planes condition
- Point with Lines and Planes
Transcript
Question 7 Find the intercepts cut off by the plane 2x + y – z = 5. The equation of a plane with intercepts 𝑎, b, c on x, y, and z – axis respectively is 𝒙𝒂 + 𝒚𝒃 + 𝒛𝒄 = 1 Given, equation of plane is 2x + y – z = 5 Dividing by 5 2𝑥 + 𝑦 − 𝑧5 = 55 2𝑥5 + 𝑦5 − 𝑧5 = 1 𝑥 52 + 𝑦5 + 𝑧(−5) = 1 Comparing above equation with (1) a = 𝟓𝟐 , b = 5 , c = −5
