12.1. Cartesian Space

In most of your previous math classes, we worked with functions on the xy-plane only meaning we

were working only in 2D. Now we will be working in space, or rather 3D. Now we will have 3 axes,

x, y, and z.

Figure 12.1.1

We will now have 8 octants instead of 4 quadrants.

Figure 12.1.2

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Example 1: Plot the point (−1, 3, 2).

Solution:

Example 2: Give the equation of a plane that is parallel to the xz-plane that passes through the

point (−1, 3, 2).

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Important Formulas:

(1) Distance formula: d(P1, P2) =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

(2) Midpoint formula:

(x1 + x2

2,y1 + y2

2,z1 + z2

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)(3) Equation of a sphere: (x− a)2 + (y − b)2 + (z − c)2 = r2

Figure 12.1.3

Example 3: Give the equation of the sphere that has A(2, 0,−1) and B(2, 1, 3) as the endpoints of

a diameter.

Solution:

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12.3. Vectors

A vector is an ordered triple (in space) where addition and multiplication by scalars holds. Vectors

have a direction and a length (magnitude or norm).

Figure 12.3.1

Example 1: If the point P has coordinates of (0, 2,−3) and the point Q has coordinates (2,−1, 4),

find the vector−−→PQ.

Solution:

The commutative and associative properties of addition hold for vectors:

(1) Commutative property: a + b = b + a

(2) Associative property: (a + b) + c = a + (b + c)

The zero vector is: 0 = (0, 0, 0).

Vectors can be multiplied by a scalar value.

Example 2: Set a = (−5, 2, 0) and b = (1,−3, 2) and find 2a− 3b.

Solution:

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Two vectors are parallel if a = αb for some real number α. If α > 0, then a and b have the same

direction. If α < 0, then a and b have opposite directions.

Example 3: Determine if any of the following vectors are parallel:

a = (1,−1, 2),b = (2,−1, 2), c = (−3, 3,−6)

Solution:

The norm (or magnitude) of a vector a = (a1, a2, a3) is found by

‖a‖ =√a21 + a22 + a23

Unit Vectors are vectors of norm 1. To find a unit vector, we divide the vector by its norm:

ua =a

‖a‖

There are three special unit vectors:

i = (1, 0, 0), j = (0, 1, 0),k = (0, 0, 1)

.

Figure 12.3.2

All vectors can be represented by a linear combination of these:

(a1, a2, a3) = a1i + a2j + a3k

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12.4. The Dot Product

The dot product of two vectors a = (a1, a2, a3) and b = (b1, b2, b3) is found by

a · b = (a1)(b1) + (a2)(b2) + (a3)(b3)

Notice that the dot product of two vectors is commutative. Also note the fact that the dot product

gives a scalar answer.

The dot product is used in many formulas:

(1) Cosine of the angle between two vectors: cos θ =a · b‖a‖‖b‖

(2) The component of a in the b direction: compba = a · ub

(3) The projection of a in the b direction: projba = (a · ub)ub

Figure 12.4.1

Example 1: Givent a = i− 2j + 3k and b = −2i + j + k find the angle between a and b as well as

the component and projection of a in the b direction.

Solution:

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Example 2: Find all numbers x for which the angle beetween (x, 1, 1) and (1, x, 1) isπ

3.

Solution:

The angles α, β, γ that a vector makes with unit vectors i, j, and k are called direction angles of a.

A unit vector with these direction angles is: cosαi + cosβj + cos γk.

Figure 12.4.2

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12.5. The Cross Product

The cross product of two vectors a = (a1, a2, a3) and b = (b1, b2, b3) is found by

a× b = (a2b3 − a3b2)i + (a3b1 − a1b3)j + (a1b2 − a2b1)k

Notice that the cross product of two vectors is NOT commutative (in fact, it is anticommutative).

Also note the fact that the cross product gives a vector answer.

The cross product is really the determinant of a 3x3 matrix:

a× b =

∣∣∣∣∣∣∣i j k

a1 a2 a3b1 b2 b3

∣∣∣∣∣∣∣ .Example 1: Verify the formula for the cross product using the formula for determinant.

Solution:

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Example 2: Calculate (2i− j + 2k)× (i− 3k).

Solution:

If vectors a and b are NOT parallel, they form two adjacent sides of a parallelogram:

Figure 12.5.1

The vector a×b is perpendicular to the plane containing this parallelogram. If a ‖ b then a×b = 0.

The area of the parallelogram formed by vectors a and b is A = ‖a× b‖.

Example 3: Find the area of triangle PQR given P (2, 0,−3), Q(3,−1, 0), R(2, 3,−2).

Solution:

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If vectors a, b and c are noncoplanar, then they form sides of a parallelepiped. The volume of a

parallelepiped is found by V =| (a× b) · c |. Note that (a× b) · c is called the scalar triple product

of vectors a, b and c.

Figure 12.5.2

Example 4: Find the volume of the parallelepiped who sides are the vectors a = (2, 0, 1),b =

(1,−1, 2), and c = (−1,−1, 2).

Solution:

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12.6. Lines

The vector equation r(t) = r0 + td (t is a real number) parameterizes the line l . The tip of r0 is

the point (x0, y0, z0) and d (the direction vector) is (d1, d2, d3) so we can write l as:

r(t) = (x0 + t d1)i + (y0 + t d2)j + (z0 + t d3)k

The form of the line given above is in vector form. We can also write lines in scalar parametric form

and symmetric form.

Scalar parametric form:

x(t) = x0 + td1, y(t) = y0 + td2, z(t) = z0 + td3

Symmetric form:x− x0d1

=y − y0d2

=z − z0d3

Example 1: Find a vector parameterization for the line that passes through the point P (1,−2, 0)

and is parallel to the line r(t) = (i− j + 2k) + t(i + k)

Solution:

Example 2: Re-write the answer to example 1 in scalar parametric form then in symmetric form.

Solution:

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Two lines are parallel if their direction vectors are scalar multiples of each other and they have no

points in common. If they have direction vectors that are scalar multiples of each other and they

have points in common, we say the lines coincide.

If two lines l1 : r(t) = r0 + td and l2 : R(u) = R0 + uD intersect then there exists numbers t and u

such that r(t) = R(u).

Lines that are not parallel, coincident or intersecting are called skew.

Example 3: Determine if the following lines are parallel, coincident, intersecting or skew:

l1 : r(t) = (3i + j + 5k) + t(i− j + 2k)

l2 : R(u) = (i + 4j + 2k) + u(j + k)

Solution:

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Other important formulas involving lines l1 : r(t) = r0 + td and l2 : R(u) = R0 + uD:

(1) Cosine of the angle between two lines: cos θ =| ud · uD |

(2) The distance from point P to line l1: d(P, l1) =‖−−→Pr0 × d‖‖d‖

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12.7. Planes

We have worked with planes that are perpendicular to the x, y or z-axis such as the plane z = 2

graphed below:

Figure 12.7.1

What about planes that are not perpendicular to an axis? Let N = Ai + Bj + Ck be the nonzero

vector perpendicular to a plane through point P (x0, y0, z0) and let Q(x, y, z) be another point on

the plane as shown:

Figure 12.7.2

Notice that the vector−−→PQ = (x − x0, y − y0, z − z0) and because N is perpendicular to our plane,

N ·−−→PQ = 0. Using this information, we find the equation of the plane to be:

(A,B,C) · (x− x0, y − y0, z − z0) = 0

A(x− x0) +B(y − y0) + C(z − z0) = 0

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Example 1: Find an equation for the plane which passes through the point P (1, 2,−3) and is

perpendicular to the vector i− j + 2k

Solution:

Example 2: Find an equation in x, y, z for the plane that passes through the given points.

P1(3, 4, 1), P2(3, 2, 1), P3(1, 1, 2)

Solution:

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Given two planes P1 and P2, then exactly one of the following holds:

(1) P1 and P2 are parallel.

(2) P1 and P2 coincide.

(3) P1 and P2 intersect in a straight line.

If the planes intersect, the direction vector of their line of intersection is found by the cross product

of their normal vectors.

Example 3: Given P1 : x + y + z + 1 = 0 and P2 : x − y + z + 2 = 0, determine if the planes are

parallel, coincident or intersect. If they intersect, determine the line of intersection.

Solution:

Other important formulas involving planes:

(1) Cosine of the angle between two lines: cos θ =| uN1 · uN2 |

(2) The distance from point P (x1, y1, z1) to plane P : Ax+By + Cz +D = 0:

d(P,P) =| Ax1 +By1 + Cz1 +D |√

A2 +B2 + C2

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