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Question 4 (a) - Equation of plane - In Normal Form - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at August 13, 2026 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
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Transcript
Question 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (a) 2x + 3y + 4z 12 = 0 Assume a point P(x1, y1, z1) on the given plane Since perpendicular to plane is parallel to normal vector Vector is parallel to normal vector to the plane. Given equation of plane is 2x + 3y + 4z 12 = 0 2x + 3y + 4z = 12 Since, and are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional So, 1 2 = 1 2 = 1 2 = k 1 2 = 1 3 = 1 4 = k x1 = 2k , y1 = 3k , z1 = 4k Also, point P(x1, y1, z1) lies in the plane. Putting P (2k, 3k, 4k) in 2x + 3y + 4z = 12, 2(2k) + 3(3k) + 4(4k) = 12 4k + 9k + 16k = 12 29k = 12 k = 12 29 So, 1 = 2k = 2 12 29 = 24 29 1 = 3k = 3 12 29 = 36 29 & 1 = 4k = 4 12 29 = 48 29 Therefore, coordinate of foot of perpendicular are , , ,
