You are here
Question 12 - Equation of plane - Prependicular to Vector & Passing Through Point - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at April 16, 2024 by Teachoo
Equation of plane - Prependicular to Vector & Passing Through Point
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
- 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 12 If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x − x1) + B(y − y1) + C(z − z1) = 0 The plane passes through P(1, 2, −3) So, x1 = 1, y1 = 2, z = −3 Normal vector to plane = (𝑂𝑃) ⃗ where O(0, 0, 0), P (1, 2, −3) Direction ratios of (𝑂𝑃) ⃗ = 1 − 0 , 2 − 0 , −3 − 0 = 1 , 2 , –3 ∴ A = 1, B = 2, C = −3 Equation of plane in Cartesian form is 1(x − 1) + 2 (y − 2) + (−3) (z − (−3)) = 0 x − 1 + 2y − 4 − 3 (z + 3) = 0 x − 1 + 2y − 4 − 3z − 9 = 0 x + 2y − 3z − 14 = 0
