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Question 4 - Equation of plane - In Normal Form - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at Dec. 16, 2024 by Teachoo
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
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Equation of plane - In Normal Form
- Equation of plane - Prependicular to Vector & Passing Through Point
- Equation of plane - Passing Through 3 Non Collinear Points
- 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 4 Find the direction cosines of the unit vector perpendicular to the plane .(6 3 2 ) + 1 = 0 passing through the origin. Vector equation of a plane at a distance d from the origin and unit vector to normal from origin is . = d Unit vector of = = 1 ( ) Given, equation of plane is .(6 3 2 ) + 1 = 0 .(6 3 2 ) = 1 Multiplying with 1 on both sides, .(6 3 2 ) = 1 1 . ( 6 + 3 + 2 ) = 1 So; = 6 + 3 + 2 Magnitude of = 6 2+32+22 = 36+9+4 = 49 = 7 Now, = 1 ( ) = 1 7 ( 6 + 3 + 2 ) = + + Direction cosines of unit vector perpendicular to the given plane i.e. in are , , .
