PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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Plane - Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point

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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`

Video link : https://youtu.be/q6ASCg_D_gY

Shaalaa.com | 3D Geometry Plane Basics 2 (Equation of Plane Passing Through The Points)

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