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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

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: 703

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