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

z M3

M2 y

M1

.

Therefore, ; .

By analogy, we obtain

; ; ,

; ; .

Substituting the obtained values of the coefficients ,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 belongs to the plane, then the three vectors

={xx1;yy1; zz1},

= {x2x1;y2y1;z2z1},

={x3x1;y3y1;z3z1},

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 0(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(xx0)+B(yy0)+C(zz0)=0,

6(x+1)7(y2)+5(z4)=0; 6x7y+5z=0.

The normal equation of a plane. Consider a plane. Let us draw the perpendicular 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 has the coordinates

={cos α; cos b; cos g}.

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

.

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: 703


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