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Relations and Functions
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Determinants
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Linear Programming
Continuity and Differentiability
- Concept of Continuity
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Integrals
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Differential Equations
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Vectors
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Three - Dimensional Geometry
- Introduction of Three Dimensional Geometry
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- Equation of a Line in Space
- Angle Between Two Lines
- Shortest Distance Between Two Lines
- Three - Dimensional Geometry Examples and Solutions
- Equation of a Plane Passing Through Three Non Collinear Points
- Forms of the Equation of a Straight Line
- Coplanarity of Two Lines
- Distance of a Point from a Plane
- Angle Between Line and a Plane
- Angle Between Two Planes
- Vector and Cartesian Equation of a Plane
- Equation of a Plane in Normal Form
- Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point
- Distance of a Point from a Plane
- Plane Passing Through the Intersection of Two Given Planes
- Overview of Three Dimensional Geometry
Linear Programming
Probability
Notes
There can be many planes that are perpendicular to the given vector, but through a given point `P(x_1, y_1, z_1)`, only one such plane exists in following fig.
Let a plane pass through a point A with position vector `vec a` and perpendicular to the vector
`vec N `
Let `vec r` be the position vector of any point P(x,y,z) in the plane. Fig.
Then the point P lies in the plane if and only if `vec (AP)` is perpendicular to `vec N` . i.e., `vec (AP) . vec N = 0` . But `vec (AP) = vec r - vec a.` Therefore `(vec r - vec a) . vec N = 0` ...(1)
Cartesian form:
Let the given point A be `(x_1,y_1,z_1)` ,P be (x , y, z) and direction ratios of `vec N` are A ,B and C . Then,
`vec a = x_1 hat i + y_1 hat j + z_1 hat k , hat r = x hat i + y hat j +z hat k` and `vec N = A hat i + B hat j + C hat k`
Now `(vec r - vec a) . vec N = 0`
So `[(x - x_1) hat i + (y - y_1) hat j + (z - z_1) hat k] . (A hat i + B hat j +C hat k) = 0`
i.e. `A (x - x_1) + B (y - y_1) + C (z - z_1) = 0`
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