Analytic geometry. Textbook

vector product and the properties given above we get the following equalities

Now for two arbitrary vectors

Taking into account relations (2) and collecting the similar terms we get:

It is possible to rewrite the obtained formula as follows:

~

~{ ~| k

~

~a b = a1 a2 a3 :

b1 b2 b3

Remark. Here and below we use third order determinants that are

somewhat di erent from \standard" determinants with numerical entries.

The rst row of these determinants consists of vectors. Such determinants

are de ned by the similar relation as in the \purely numerical" case. Each

term in the sum de ning the determinant is a product of a vector and

two scalars, so this term is a vector and the determinant is also a vector.

Many properties of such determinants are similar to the ones of numerical

determinants, e.g. expanding the determinant written above along the rst

triple product is also called the mixed product.

Proof. We have the following relations:

At the last step, we used the expansion of the third-order determinant along the rst row.

Now let us turn to the properties of the scalar triple product. In the

The other cases are investigated similarly.

~ ~ ~

2) ~a b~c = b~c~a = ~c~a b.

Swapping the rows of the determinant we get:

The next equality follows from the previous.

~ ~ ~ ~

3) ~a b~c = ~a~c b = ~c b~a = b~a~c.

~ ~a b

S

~a

~c

~

b

~

Fig. 8. The case of non-collinear vectors ~a and b

~ ~ ~

The vector product ~a b is orthogonal to ~a and b, therefore ~a b is orthogonal

to the plane S. Taking into account that ~c lies on the plane S, we get that

~ ~ ~

the vectors ~a b and ~c are orthogonal, ~a b~c = (~a b) ~c = 0.

~

Theorem 3. Assume that the vectors ~a, b and ~c are not coplanar. Let V

be the volume of the parallelepiped de ned by these vectors reduced to the

these vectors form the right-handed triple and the sign \ " is taken if these

vectors form the left-handed triple.

Proof.

~a

Fig. 9. The parallelepiped de ned by three vectors

86 Chapter 3. Analytic geometry in the space

Let us consider the case of right-handed triple. We take the parallelo-

~

gram with the sides ~a and b as the base of this parallelepiped, the area of

the base is denoted by A, let h be the height of this parallelepiped. The volume of the parallelepiped equals the product of the area of the base and

3.2Plane in the space

3.2.1Equation of the plane

The plane is uniquely determined if we know some its point and a nonzero vector which is orthogonal to this plane.

Let us assume that the plane S is determined by a point M0 and a vector

6 ~ ~n = 0.

The point M belongs to the plane S if and only if the vectors ~n and

!

M0M are orthogonal. Therefore the equation of the plane may be written

!

in the following vector form: ~n M0M = 0.

Let us denote M0(x0; y0; z0), ~n = fA; B; Cg. For a point M(x; y; z) we

!

get that M0M = fx x0; y y0; z z0g. Rewriting equation of the plane

The vector ~n is called the normal vector of the plane. It is de ned up to a nonzero scalar factor.

Theorem 4. Let S1 and S2 be the planes de ned by the equations

A1x + B1y + C1 + D1 = 0; A2x + B2y + C2 + D2 = 0:

The planes S1 and S2 are parallel if and only if

A1 = B1 = C1 :

A2 B2 C2

The planes are perpendicular if and only if

A1A2 + B1B2 + C1C2 = 0:

Proof. The planes are parallel if and only if their normal vectors n1 = fA1; B1; C1g; n2 = fA2; B2; C2g

88 Chapter 3. Analytic geometry in the space

are collinear. The condition given in the statement of this theorem is exactly the collinearity condition for ~n1 and ~n2.

The planes are perpendicular if and only if their normal vectors

n1 = fA1; B1; C1g; n2 = fA2; B2; C2g

are orthogonal. In the second part of the statement of the theorem there is

Proof. Assume that the equality (4) is valid. Then

A1 = A2; B1 = B2; C1 = C2; D1 = D2

for some 6= 0. We consequently get that the equivalent equations

A1x + B1y + C1z + D1 = 0;A2x + B2y + C2z + D2 = 0; A2x + B2y + C2z + D2 = 0:

Therefore S1 = S2.

Now we assume that the equations under consideration de ne equal

for some 6= 0. Let M0(x0; y0; z0) be a point on this plane. Then

A1x0 + B1y0 + C1z0 + D1 = 0;

A2x0 + B2y0 + C2z0 + D2 = 0;

D2 = (A2x0 + B2y0 + C2z0);

D1 = (A1x0 + B1y0 + C1z0) = (A2x0 + B2y0 + C2z0) = D2:

Adding condition D1 = D2 to the relation (5) we get relations equivalent to (4).

3.2.2Special cases of equations of planes

Now we analyze special cases of equations of planes.

First we consider equations of planes Ax + By + Cx + D = 0 under assumptions that some of the coe cients A, B, C, D vanish.

In the case C = 0, the equation takes the form Ax + By + D = 0. For a point M0(x0; y0; z0) on this plane, there holds the equality Ax0+By0+D = 0. Since this equality does not contain the third coordinate we get that the points M1(x0; y0; z) are on this plane for all z 2 R. The plane contains \vertical" straight line passing through the point M0. It means that the plane is parallel to the z-axis.

z

M1

O

y

M0

x

Fig. 11. Equation Ax + By + D = 0, the plane is parallel to the z-axis

Let us prove the converse assertion: if the plane de ned by the equation

Ax + By + Cz + D = 0 is parallel to the z-axis then C = 0.

We take two di erent points of this plane that lie on one vertical straight line, say M0(x0; y0; z0) and M1(x0; y0; z1), where z0 6= z1. Then

Ax0 + By0 + Cz0 + D = 0; Ax0 + By0 + Cz1 + D = 0:

Taking di erence of these equalities we get that C(z1 z0) = 0. Since z1 z0 6= 0 we obtain that C = 0. Similarly there are investigated the case

A = 0 and the case B = 0.

In the case A = 0, B = 0 the equation of the plane takes the form Cz + D = 0. Since C 6= 0 we get that z = DC , or z = z0 where z0 = DC . The plane is parallel to the plane xOy.

It is obvious that there is valid the converse assertion: if a plane de ned by the equation Ax + By + Cz + D = 0 is parallel to the plane xOy then

A = 0 and B = 0. Such a plane may be de ned by the equation z = z0 or z z0 = 0 for some z 2 R. This equation and the equation given above di er only by a nonzero factor. Therefore A = 0, B = 0. Similarly there are investigated the cases A = 0, C = 0 and B = 0, C = 0.

If D = 0, then the equation of the plane takes the form Ax+By+Cz = 0. It means that the plane passes through the origin: A 0 + B 0 + C 0 = 0.

There is valid the converse: if the plane de ned by the equation Ax +

By + Cz + D = 0 passes through the origin then D = 0. In this case

A 0 + B 0 + C 0 + D = 0, therefore D = 0.

Now we assume that the plane is de ned by the equation Ax + By +

Cz + D = 0 and C = 0, D = 0. Such a plane is parallel to the z-axis and passes through the origin. It means that the plane passes through the z-axis. The converse assertion is also obviously valid.

The following table contains all the cases analyzed above.