APPLICATIONS TO ANALYTIC
GEOMETRY

We can use vectors to derive equations of lines and planes in three-space. Also vectors make the solution of some geometric problems fairly easy.

If we think of a desk top as representing a plane and stand a pencil (unsharpened) on it (see Fig. 2.4.1), the pencil points in a direction perpendicular to any line on the desk top. Such a vector is called a normal vector.

Definition 2.4.1   A vector N which is perpendicular to all vectors in a plane $P$ is called a normal vector for the plane. A normal vector for a plane is said to be perpendicular to the plane.

A plane can be specified by giving a normal to the plane and a point in the plane. This is similar to the case for lines in the plane for which two quantities specify a line: the slope and a point. In three-space, the normal vector serves the function of the slope in two-space. Note that two planes are parallel if their normals are parallel.

To find an equation for a plane $P$, let the point in $P$ be $(x_0,y_0,z_0)$ and the normal be ${\bf N} = (a,b,c)$. If $(x,y,z)$ is any other point in $P$ (see Fig. 2.4.2), then the vector $(x-x_0, y-y_0, z-z_0)$ lies in $P$ and ${\bf N} \cdot (x-x_0, y-y_0, z-z_0)=0$ since N is perpendicular to all vectors in $P$. Writing the dot product, we have

$$0 = {\bf N}\cdot (x-x_0, y-y_0, z-z_0) = (a,b,c) \cdot (x-x_0, y-y_0, z-z_0)$$

$$= a(x-x_0) + b(y-y_0) + c(z-z_0)$$

FIGURE

Therefore

An equation of the plane containing $(x_0,y_0,z_0)$ with normal $(a,b,c)$ is

$$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$$

This form of the equation is called the point-normal form since the coordinates of the point and components of the normal appear explicitly. The equation

$$(a,b,c)\cdot (x-x_0, y-y_0, z-z_0) = 0$$

is called the vector form of equation for the plane.

Example 1   Find the point-normal form of equation for the plane passing through $(1,-2,4)$, having normal vector $(2,3,-1)$.

Solution The vector form is

$$(2,3,-1) \cdot (x-1, y-(-2), z-4)=0$$

Writing out the dot product, we find

$$2(x-1) + 3(y+2) - 1(z-4) = 0$$

Example 2   A plane $P$ has equation

$$2(x-2) - 7(x-3) + 4(x+2) = 0$$

Find a normal vector to $P$ and a point in $P$.

Solution Since the equation is in point-normal form, the desired information can be read off:

$${\bf N} = (2,-7,4)$$

$${\text{Point }} (2,3,-2)$$

The equation found in Example 1 could be simplified to

$$2x+3y - z+8 = 0$$

This equation fits the general form

$$ax+by + cz+d=0$$

From the general form the normal vector can still be read. Because the equation is linear, a plane is called a linear structure in three-space.

Example 3   A plane $P$ has equation

$$2x-3y + 4z+12 = 0$$

Find a normal vector and two points in the plane.

Solution A normal is $(2,-3,4)$. Any nonzero multiple of this vector is also a normal. To find points in the plane, we must find values of $x,y$, and $z$ which satisfy the equation. To do this, we can give values to any two of the variables and solve for the third. If we let $x=2$ and $y=0$, we find $z=-4$, so $(2,0,-4)$ is in the plane. If we let $y=0$ and $z=0$, we find $x=-6$ and obtain $(-6,0,0)$ as another point in the plane.

The cross product can sometimes be used to find a normal.

Example 4   A plane $P$ contains the points $p_1\colon (1,1,2)$, $p_2\colon (0,-3,4)$ and $p_3\colon (2,0,3)$. Find a normal vector for $P$ and an equation for $P$.

Solution By subtracting points we find as vectors in $P\colon \ (0,-3,4) - (1,1,2) = (-1,-4,2) = {\bf A}$ and $(2,0,3)-(1,1,2) = (1,-1,1) = {\bf B}$. Now ${\bf A}\times {\bf B}$ is perpendicular to both A and B and hence is normal to $P$ (see Fig. 2.4.3). Now

$${\bf A\times B} = (-2,3,5)$$

A vector equation is

$$(-2,3,5)\cdot (x-1,y-1,z-2) = 0$$

which is equivalent to

$$-2(x-1) + 3(y-1) + 5(z-2) = 2$$

or

$$-2x+3y + 5z-11=0$$

A straight line through the point $(x_0,y_0,z)$ is the set of all points $(z,y,z)$ such that the vector from $(x_0,y_0,z_0)$ to $(x,y,z)$ is a multiple of a given vector

FIGURE

V (see Fig. 2.4.4). From Fig. 2.4.4 we see that

$$(x-x_0, y-y_0, z-z_0) = k{\bf V} = k(a,b,c)$$

where $k$ is allowed to run through the real numbers, which means

$$x-x_0 = ka \qquad y-y_0 =kb\qquad z-z_0=kc\qquad {\text{(parametric equations)}}$$

or

$$\frac{x-x_0}a = k = \frac{y-y_0}b = k = \frac{z-z_0}c$$

(unless $a,b$, or $c$ is zero).

Example 5   Find equations of the line passing through $(1,-1,2)$ and (7,0,5). Is $(-5,-2,-1)$ on the line? What about (13,1,14)?

Solution We have a point (in fact, we have two) and need a vector in the direction of the line. For this purpose a vector in the line will do. So we subtract the points to get

$${\bf V} = (6,1,3)$$

Using $(1,-1,2)$ as a point in the line, we see that

$$(x-1,y+1,z-2) = k(6,1,3)$$

is a vector equation for the line. Parametric equations are

$$\begin{array}{l} x-1=6k\ y+1=k\ z-2=3k\end{array}\quad {\... ...y}{l} x=6k+1\ y=k-1\ z=3k+2\end{array}\qquad k\in\mathbb{R}$$

To see whether $(-5,-2,-1)$ is on the line, we ask if there is a value for $k$ (the parameter) which gives $x=-5$, $y=-2$, $z=-1$. Using the last equations, we try to solve

$$-5 = 6k+1$$

$$-2 =k-1$$

$$-1 =3k+2$$

The first equation gives $k=-1$. This value for $k$ also works in the second and third equations so $(-5,-2,-1)$ is on the line.

Finally we consider the point (13,1,14). We have this time

$$13 = 6k+1$$

$$1 =k-1$$

$$14 =3k+2$$

and the first equation implies that $k=2$. This value for $k$ satisfies the second equation but not the third. Thus (13,1,14) is not on the line.

Lines can be generated as the intersection of planes. In general, two planes can intersect in one line, not intersect at all, or be the same plane.

Example 6   Determine the line of intersection of $x+2y - z=3$ and $x+y-3z=5$.

Solution We solve the equations simultaneously:

$$\begin{array}{r} x+2y-\phantom{3}z=3\ x+\phantom{2}y - 3z=5... ...rray}{r} x+2y-\phantom{2}z =3\ \phantom{x2}-y-2z=2\end{array}$$

We have two equations in three unknowns. Let $z=t$, so that

$$y = -2t-2$$

$$x = 5t+7$$

Thus, parametric equations for the line of intersection are

$$x=5t+7\qquad y=-2t-2\qquad z=t\qquad t\in\mathbb{R}$$

Example 7   Show that the planes $x-y+z=3$ and $3x-3y+3z=11$ do not intersect.

Solution Again we consider the equations simultaneously:

$$\begin{array}{r} \phantom{3}x-\phantom{3}y+\phantom{3}z =3\ph... ...ray} \longrightarrow \begin{array}{r} x-y+z=3\ 0=2\end{array}$$

Since the reduced system is inconsistent, no points lie on both planes (they are parallel, since their normals are in the same direction) and so there is no line of intersection for these planes.

Note in Example 7 that if the second plane had had equation

$$3x-3y+3z=9$$

then the planes would be not only parallel but also coincident, and there would be no unique line of intersection.

The use of vectors in the study of analytic geometry is a very effective method. Interested readers can find further information in texts on vector geometry.