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Maths Exercices


Planes in the space

An equation of a plane in space can be obtained from a point in the plane and a vector normal (perpendicular) to the plane.

Consider the plane containing the point P(x1,y1,z1) having a nonzero normal vector n=<a,b,c>, as shown in the next figure:

This plane consists of all points Q(x,y,z) for which vector is orthogonal to n.

Using the dot product, you can write the following:


<a, b, c>·<x-x1, y-y1, z-z1>=0

a(x-x1) + b(y-y1) + c(z-z1) = 0

The third equation of the plane is said to be in standard form.


The plane containing the point (x1,y1,z1) and having normal vector n = <a,b,c> can be represented by the standard form of the equation of a plane:

a(x-x1) + b(y-y1) + c(z-z1) = 0

Standard equation of a plane

By regruping terms, you obtain the general form of the equation of a plane in space:

ax + by + cz + d = 0
General form of equation of plane

Given the general form of the equation of a plane, it is easy to find a normal vector to the plane. Simply use the coefficients of x, y and z and write n = <a,b,c>

Find the general equation of the plane containing the points (2,1,1), (0,4,1), and (-2,1,4)

To find the general equiation you need a point in the plane and a vector that is normal to the plane. There are three choices for the point, but no normal vector is given. To obtain a normal vector, use the cross product of vectors u and v extending from the point (2,1,1) to the points (0,4,1) and (-2,1,4) as shown in the next figure:

The component forms of u and v are:

u = <0 - 2, 4 - 1, 1 - 1> = <-2, 3, 0>

v = <-2 - 2, 1 - 1, 4 - 1> = <-4, 0, 3>

and it follows that

n = u x v =

i j k
-2 3 0
-4 0 3

=9i + 6j +12k

= <a,b,c>

is normal to the given plane. Using the direction numbers for n and the point (x1,y1,z1)=(2,1,1), you can determine an equation of the plane to be

a(x-x1) + b(y-y1) + c(z-z1) = 0  
9(x-2) + 6(y-1) + 12(z-1) = 0 Standard form
9x+6y+12z-36=0 General form
3x+2y+4z-12=0 Simplified Standard form



Supose that u = <u1,u2,u3> and v = <v1,v2,v3> are nonparallel vectors that lie in a plane V. Let P0=(x0,y0.z0) be any fixed point that lies on V. We can think of P0 as an "origin" in the plane. Any point P=(x,y,z) on the plane can be reached from P0 by adding a scalar multiple of u and then a scalar multiple of v, as illustrate in the following picture:

The position vector can be obtained by vector addition as:

= + u + v

If we write out this vector equation in coordinates, then we get three scalar parametric equations for the plane:


Parametric equations of a plane

The parameters and may be thought of as coordinates in the plane V: Each point of V is determined by specifying the values of and .

Find the general equation of the plane containing the point A=(-1,-4,-2) with u=(-2,1,-3) and v=(3,1,5)

x+ y+ z+ =0