Analytic geometry. Textbook

3.2.3Distance of a point from a plane

Theorem 6. The distance of the point M(x1; y1; z1) from the plane de-ned by the equation Ax + By + Cz + D = 0 may be found by the formula

jAx1 + By1 + Cz1 + Dj d = p :

A2 + B2 + C2

Proof. Let us rst assume that the point M1 does not lie on the plane. We denote by ~n the normal vector of this plane, ~n = fA; B; Cg. Let M1M0 be the perpendicular dropped from the point M1 on the plane S. Then

!

d = jM0M1j.

!

Assume that M0(x0; y0; z0), then M0M1 = fx1 x0; y1 y0; z1 z0g.

! !

The vectors ~n and M0M1 are collinear. The nonzero vectors ~n and M0M1 have the same direction or opposite directions. Let ' be the angle between these vectors. If these vectors have the same direction then ' = 0, if these vectors have the opposite directions then ' = . Therefore cos ' = 1 or cos ' = 1.

M0

O

y

x~n

Fig. 12. Distance of a point from a plane

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We nd the scalar product ~n M0M1 using two ways.

The rst way: we use coordinates of these vectors,

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~n M0M1 = A(x1 x0) + B(y1 y0) + C(z1 z0) = Ax1 + By1 + Cz1 (Ax0 + By0 + Cz0):

The point M0 lies on the plane, therefore

Ax0 + By0 + Cz0 + D = 0; Ax0 + By0 + Cz0 = D;

!

~n M0M1 = Ax1 + By1 + Cz1 + D:

The second way: we use the de nition of the scalar product:

|{z}

| {z }

d1

p

d A2 + B2 + C2 = Ax1 + By1 + Cz1 + D:

We have not yet considered the case when the point M1 lies on the plane. In this case d = 0, Ax1 + By1 + Cz1 + D = 0 and the formula under consideration remains true.

Corollary 1. The plane divides the space into two half-spaces. If a plane is de ned by the equation Ax + By + Cz + D = 0 then one of these half-spaces is de ned by the inequality Ax + By + Cz + D > 0 and the other one by the inequality Ax + By + Cz + D < 0.

Remark. As in the case of half-planes introduced in chapter 1 (page 23) the \positive" and \negative" half-spaces are not de ned by the plane itself, they depend on the choice of the equation of the the plane. The normal vector of the plane de ned by its equation is directed into the \positive" half-space. Here are considered open half-spaces. It means that they do not contain the dividing plane. The closed half-spaces contain the given plane. To de ne these half-spaces it is necessary to replace the strict inequalities by non-strict.

Corollary 2. The distance between parallel planes de ned by the equations Ax + By + Cz + D1 = 0 and Ax + By + Cz + D2 = 0 may be found by the formula

The proof of this corollary is completely similar to the one given for the case of parallel straight lines (p. 23).

3.2.4Equation of the plane in the intercept form

Definition. The abscissas of the points in which the surface cuts the x-axis are called the x-intercepts of the surface. Similarly there are de ned its y-intercepts and z-intercepts.

We get the following rules.

Assume that a surface is de ned by some equation of the form f(x; y; z) = 0 (for instance, x + 2y + 6z 12 = 0 or x2 + y2 + z2 1 = 0).

•To nd the x-intercepts of the surface de ned by some equation, we put y = 0 and z = 0 in the equation and solve for x.

•To nd the y-intercepts of the surface de ned by some equation, we put x = 0 and z = 0 in the equation and solve for y.

•To nd the z-intercepts of the surface de ned by some equation, we put x = 0 and y = 0 in the equation and solve for z.

Assume that in the equation Ax + By + Cz + D = 0 of the plane all the values A, B, C and D do not vanish. We can nd the intercepts using the previous rules. We put y = 0 and z = 0 into the equation of the plane, and get the following equation: Ax + D = 0, x = DA .

Now we use another approach. We transfer the value D to the right side of the equation and divide the both sides of the equation by D:

Denoting a = DA , b = DB , c = DC we get the nal form of the equation

xa + yb + zc = 1:

Fig. 14. Pencil of planes

Theorem 7. The planes from the pencil with the axis being the intersection of two non-parallel planes A1x + B1y + C1z + D1 = 0 and

A2x + B2y + C2z + D2 = 0 may be de ned by the equations

1(A1x + B1y + C1z + D1) + 2(A2x + B2y + C2z + D2) = 0 (6) where j 1j + j 2j 6= 0.

Proof. Let us denote the intersection line of the given planes by `.

We assume that j 1j + j 2j 6= 0 and prove that (6) is the equation of the plane through `.

We rewrite (6) as follows:

( 1A1 + 2A2)x + ( 1B1 + 2B2)y

+( 1C1 + 2C2)z + ( 1D1 + 2D2) = 0:

We are to prove that the coe cients

1A1 + 2A2; 1B1 + 2B2; 1C1 + 2C2

do not vanish at the same time. Assuming the contrary we get that

1A1 + 2A2 = 0; 1B1 + 2B2 = 0; 1C1 + 2C2 = 0:

These relations mean that the normal vectors fA1; B1; C1g and fA2; B2; C2g of the given planes are collinear and the planes are parallel. We have obtained a contradiction. Similar arguments lead to a contradiction in the case 2 6= 0.

Let M0(x0; y0; z0) be an arbitrary point on the axis `. Then

A1x0 + B1y0 + C1z0 + D1 = 0; A2x0 + B2y0 + C2z0 + D2 = 0;

therefore

1(A1x0 + B1y0 + C1z0 + D1) + 2(A2x0 + B2y0 + C2z0 + D2) = 0;

hence the point M0 is on the plane under consideration. Due to the arbitrariness of this point we get that the straight line ` is on this plane.

Now we are going to prove that any plane passing through the straight line ` is de ned by the equation (6) with appropriate values 1 and 2. Such a plane is uniquely de ned by a point M1(x1; y1; z1) on this plane such that M1 62`. We put the coordinates of this point into (6) with unknown values 1 and 2:

1(A1x1 + B1y1 + C1z1 + D1) + 2(A2x1 + B2y1 + C2z1 + D2) = 0:

The values

A1x1 + B1y1 + C1z1 + D1 and A2x1 + B2y1 + C2z1 + D2

do not vanish at the same time since otherwise M1 would be the point on the both planes and therefore M1 2 `. Let

1 = A2x1 + B2y1 + C2z1 + D1; 2 = (A1x1 + B1y1 + C1z1 + D2):

To reduce the general equation of the plane Ax+By+Cz +D = 0 to the p

normal form, we divide the both sides of this equation by A2 + B2 + C2 where the sign is chosen in order to satisfy the inequality

In the case D = 0 any sign is suitable, in the case D 6= 0 we take the sign opposite to the sign of D. We denote

D

p = p

A2 + B2 + C2

with the chosen sign. Let ~n be the normal vector de ned by the transformed

equation, then j~nj = 1. We denote by , , the angles between ~n and ~{,

~ f g

~|, k respectively. Then ~n = cos ; cos ; cos and the normal equation takes the following most used form

x cos + y cos + z cos p = 0:

We emphasize that cos2 + cos2 + cos2 = 1. The value p is the distance of the origin O(0; 0; 0) from the plane, since

3.3Straight line in the space

A straight line in the space is uniquely de ned if we know a point lying on this line and a nonzero vector that is parallel to this line. Let ` be a straight line passing through a point M0(x0; y0; z0) and parallel to a nonzero vector ~e = fk; l; mg. A point M(x; y; z) is on the line ` if and only if the

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vectors M0M and ~e are collinear (see Fig. 15).

These relations are called canonical or standard equations of the straight line. Vector ~e is called the direction vector of this line. The direction vector is de ned up to a nonzero factor.

Denoting the value of the fractions in (8) by t, we get that x = x0 + kt, y = y0 + lt, z = z0 + mt. Conversely, if the values x, y, z are obtained from these equalities for an arbitrary value t, then they satisfy the equations (8). The equations

8

> x = x0 + kt;

>

>

>

< y = y0 + lt;

> z = z0 + mt;

>

>

> t 2 R

:

are called parametric equations of the straight line.

Sometimes straight lines in the space are de ned as intersection of two non-parallel planes. There is usually used the following notation

(

A1x + B1y + C1z + D1 = 0;

A2x + B2y + C2z + D2 = 0: