Throughout calculus I and II, we have only dealt with functions of two variables or two-dimensional graphs. Unfortunately, there are not many real world applications that fit into a two-dimensional space. Therefore, we will have to define a three-dimensional space. Here is the coordinate axis for a three-dimensional space.

The x-, y-, and z-axis cut space up into 8 octants. (Remember, the x- and y-axis cut the plane into four quadrants.) Most of the time we will show the first octant. (This is the octant in which x-, y-, and z-axis is positive.) The way the frame is set up is called the right-handed orientation in which the z-axis is straight up and the x-axis is straight out. Any point in space is given in an ordered triple of the form (x, y, z). Any points that lie on the x-axis have y- and z-coordinates of zero, and the point is of the form (x, 0, 0). Any points that lie on the y-axis have x- and z-coordinates of zero, and the point is of the form (0, y, 0). Finally, any points that lie on the z-axis have x- and y-coordinates of zero, and the point is of the form (0, 0, z).

EXAMPLE 1:

Sketch the graph of x = 4.

SOLUTION:

This is a plane through the point (4, 0, 0) parallel to the yz-plane.

FACT:

The xy-plane has the equation z = 0.

The yz-plane has the equation x = 0.

The xz-plane has the equation y = 0.

EXAMPLE 2:

Sketch the graph of x = 2 and z = 4.

SOLUTION:

This is the graph of two intersecting planes that form a line in which the points have the form (2, y, 4).

EXAMPLE 3:

Sketch the graphs of x ² + y ² = 9 and z = 3.

SOLUTION:

The graph of this region will be the circle with radius 3 centered at (0, 0, 3) in the plane z = 3 when the cylinder x ² + y ² = 9 is cut by the plane z = 3.

EXAMPLE 4:

Sketch the graphs of x ² + y ² + z ²= 25 and x = 3.

SOLUTION:

The region that we have when the plane x = 3 cuts the sphere x ² + y ² + z² = 25 is the circle y ² + z ² = 16. This is a circle in the x = 3 plane with center at (3, 0, 0) and radius 4.

EXAMPLE 5:

A) Graph the inequality 0 £ x £ 1.

SOLUTION:

The graph of this inequality is a slab of planes parallel to the yz-plane and perpendicular to the x-axis.

EXAMPLE 5:

B) Graph the inequalities 0 £ x £ 1, 0 £ y £ 2.

SOLUTION:

This is a rectangular slab that has infinite height.

EXAMPLE 6:

A) Write an equation of the plane through the point (2, -1, 5) perpendicular to the x-axis.

SOLUTION:

Since the plane is to be perpendicular to the x-axis, the form of the equation of this plane is x = k and the equation of this plane is x = 2.

EXAMPLE 6:

B) Write an equation of the plane through the point (2, -1, 5) perpendicular to the y-axis.

SOLUTION:

Since the plane is to be perpendicular to the y-axis, the form of the equation of this plane is y = k and the equation of this plane is y = -1.

EXAMPLE 7:

Write and equation or a set of equations for the circle of radius 4 centered at (2, -1, 5) and lying in a plane parallel to the yz-plane.

SOLUTION:

Since the circle has to be in a plane parallel to the yz-plane, then it is a plane perpendicular to the x-axis. Therefore, the plane is x = 2. The equation of the circle will have the form (y - k) ² + (z - h) ² = r ². Therefore, the equations for this circle are x = 2 and (y + 1) ² + (z - 5) ² = 16.

Since we are working in three dimensions, we need to talk about the equation of a sphere. A circle is a two-dimensional figure, and by adding another dimension, we can have the equation of a sphere. The equation of a circle centered at (h, k) with radius r is (x - h) ² + (y - k) ² = r ². The equation for a sphere centered at (x ₀, y ₀, z ₀) and radius a is (x - x ₀) ² + (y - y ₀) ² + (z - z ₀) ² = a ².

EXAMPLE 8:

Find the equation of the sphere centered at (2, -1, 4) with radius of 3.

SOLUTION:

(x - 2) ² + (y + 1) ² + (z - 4) ² = 9

EXAMPLE 9:

Find the center and radius for x ² + y ² + z ² - 4x + 2y = 16.

SOLUTION:

We will have to complete the square before we can determine the center and radius of this sphere.

Vectors in space are no different than vectors in two dimensions. All the facts discussed in supplemental notes 1 still apply. The basic vectors in three dimensions are i: (0, 0, 0) ® (1, 0, 0), j: (0, 0, 0) ® (0, 1, 0), and k: (0, 0, 0) ® (0, 0, 1).

Suppose we are given the point P (x, y, z) in space. The position vector r = OP = xi +yj + zk where O is the origin.

FACT:

For A = a₁i + a₂j + a₃k and B = b₁i + b₂j + b₃k,

A + B = (a₁ + b₁)i + (a₂ + b₂)j + (a₃ + b₃)k

A - B = (a₁ - b₁)i + (a₂ - b₂)j + (a₃ - b₃)k.

EXAMPLE 10:

Given A = 3i + 2j - k and B = i - 3j + 4k, find A + B and A - 3B.

SOLUTION:

A + B = (3 + 1)i + (2 - 3)j + (-1 + 4)k = 4i - j + 3k

To find A - 3B, I will first find 3B.

3B = (3)( i - 3j + 4k) = 3i - 9j + 12k

A - 3B = (3 - 3)i + (2 - (-9))j + (-1 -12)k = 11j - 13k

FACT:

The vector from P₁ (x₁, y₁, z₁) to P₂ (x₂, y₂, z₂) is P₁P₂ = (x₂ - x₁)i + (y₂ - y₁)j + (z₂ - z₁)k.

EXAMPLE 11:

Find the vector from (3, 2, -1) to (-4, 6, 5).

SOLUTION:

Let P₁ be (3, 2, -1) and P₂ be (-4, 6, 5).

P₁P₂ = (-4 - 3)i + (6 - 2)j + (5 - (-1))k = -7i + 4j + 6k

FACT:

The magnitude (length) of A = a₁i + a₂j + a₃k is

FACT:

The zero vector in space is 0 = 0i + 0j + 0k.

FACT:

Given A ¹ 0, the direction of A is A/ | A |.

EXAMPLE 12:

Find the vector 3 units long in the direction of A = i + j + k.

SOLUTION:

First, we must find the length of A.

Now to find the directions of A.

Now to find the new vector with length of 3 units in the directions of A.

FACT:

The distance between 2 point P₁ and P₂ in space is the length of the vector P₁P₂.

FACT:

The midpoint M of the line segment joining P₁ (x₁, y₁, z₁) and P₂ (x₂, y₂, z₂) is the point

EXAMPLE 13:

Given P (2, 4, 5) and Q (3, -1, 3), find (a) PQ, (b) the distance between P and Q, (c) the direction of PQ, and (d) the midpoint of the line segment PQ.

SOLUTION:

1. PQ = (3 - 2)i + (-1 - 4)j + (3 - 5)k = i - 5j - 2k
2. The distance between P and Q is the length of the vector PQ.



3. The direction of PQ.



4. The midpoint of the line segment PQ is