Since, the point belongs to the plane, its coordinates satisfy the general equation of the plane, and

z M₃

M₂ y

M₁

х .

Therefore, ; .

By analogy, we obtain

; ; ,

; ; .

Substituting the obtained values of the coefficients А,B, and C into the general equation of the plane, we obtain

,

which gives, after the reduction by D, the three-intercept equation of the plane:

.

The equation of a plane passing through three points. Suppose given, three points , , and . As is known form elementary geometry, there exists a unique plane passing through these points. It is required to write its equation.

Following the general scheme, of we take an arbitrary point M(x;y;z) in the plane.

z M₂

M₁ M(x;y;z)

M₃

y

0

x

The characteristic feature of a plane is that if a point М belongs to the plane, then the three vectors

={x–x₁;y–y₁; z–z₁},

= {x₂–x₁;y₂–y₁;z₂–z₁},

={x₃–x₁;y₃–y₁;z₃–z₁},

are coplanar.

Therefore, the triple product of these vectors must be zero:

.

Expressing the triple product in terns of coordinates, we obtain an equation of the plane, passing through the three given points:

. (24)

Example. Write an equation of the plane passing through the

; ; .

By formula (24), we have

or .

The Angle between Planes

Consider the planes given by the equations

; ,

which have normal vectors and . Using inner product, we find the cosine of the angle:

.

The condition for two planes to be parallel coincides with the condition for the normal vectors and to be collinear:

.

The perpendicularity conditioin coincides with the perpendicularity condition for the vectors and :

( · )=0, i.e., A₁A₂+B₁B₂+C₁C₂=0.

Example 1. Show that the following planes are parallel or perpendicular:

.

The planes are perpendicular, because

.

Example 2. Write an equation of the plane passing through the point М₀(–1;2; 4) and parallel to the plane 6x-7y+5z+11=0.

The normal vector is normal to the required plane also. We have

A(x–x₀)+B(y–y₀)+C(z–z₀)=0,

6(x+1)–7(y–2)+5(z–4)=0; 6x–7y+5z=0.

The normal equation of a plane. Consider a plane. Let us draw the perpendicular ОР from the origin to this plane. Let a,b, and g be the angels between this perpendicular and the coordinate axes x,y, and z, and let . It is required to write an equation of this plane.

z

P

M(x,y,z)

0 y

x

Take an arbitrary point M(x;y;z) in the plane and consider the radius-vector . The unit vector on the perpendicular ОР has the coordinates

={cos α; cos b; cos g}.

For any point М in the plane, the projection of the vector on the unit vector equals р:

.

Consider the scalar product

,

or, in coordinate form, .

Thus, we have obtained the normal equation of the plane:

. (25)

The normalizing factor. Consider the plane given by the general equation Ax+By+Cz+D=0.

It is required to reduce this equation to the normal form (25).

Definition. Number m is called the normalizing factor if the equation multiplied by it is normal.

To find the normalizing factor, we multiply the general equation of the plane by a number m term by term:

mAx+mBy+mCz+mD=0.

This equation is normal if the two normality conditions hold, i.e.,

1. (mA)²+(mB)²+(mC)²=1,

2. mD<0.

From the first condition, taking out m² and extracting the square root, we find the normalizing factor.

.

The sign opposite to that of the free term D must be taken.

Date: 2015-01-02; view: 630