MechForge · Calculus · Chapter 12: Vectors and the Geometry of Space

12.1

Three-Dimensional Coordinate Systems

A point in the plane needs two numbers; a point in space needs three. We fix an origin and three mutually perpendicular axes, then locate every point by an ordered triple of coordinates.

Coordinate Axes and Octants

Three mutually perpendicular directed lines through the origin form the x-, y-, and z-axes. Their orientation follows the right-hand rule. The three coordinate planes (xy, yz, xz) divide space into eight regions called octants; the first octant is where all three coordinates are positive.

Ordered Triple

Any point P in space corresponds to a unique triple (a, b, c), where a, b, c are its signed distances from the yz-, xz-, and xy-planes respectively. This one-to-one correspondence between points and triples is the three-dimensional rectangular coordinate system, the set ℝ³.

Surfaces vs. Curves

In ℝ² a single equation generally describes a curve; in ℝ³ a single equation generally describes a surface. For example, z = 3 is a horizontal plane in ℝ³, while the same equation would describe a line in ℝ².

Distance Between Two Points in Space

\[ |P_1P_2| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \]

Equation of a Sphere — center (h, k, l), radius r

\[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]

Midpoint of a Segment

\[ \left( \frac{x_1+x_2}{2},\; \frac{y_1+y_2}{2},\; \frac{z_1+z_2}{2} \right) \]

Figure 12.1 A point P(a, b, c) located by moving along the x-axis, then parallel to the y-axis, then parallel to the z-axis. (Original redrawn schematic.)

EXAMPLE 1Distance Between Two Points

Find the distance from P(2, −1, 7) to Q(1, −3, 5).

Solution

Apply the distance formula component by component:

\[ |PQ| = \sqrt{(1-2)^2 + (-3-(-1))^2 + (5-7)^2} \]

\[ = \sqrt{1 + 4 + 4} = 3 \]

EXAMPLE 2Identifying a Sphere by Completing the Square

Show that \(x^2 + y^2 + z^2 + 4x - 6y + 2z + 6 = 0\) is a sphere, and find its center and radius.

Solution

Group like variables and complete the square in each:

\[ (x^2+4x) + (y^2-6y) + (z^2+2z) = -6 \]

\[ (x+2)^2 + (y-3)^2 + (z+1)^2 = -6 + 4 + 9 + 1 = 8 \]

Compare with the standard form: the center is \((-2, 3, -1)\) and the radius is \(\sqrt{8} = 2\sqrt{2}\).

Instructor Note

When a region is described by inequalities (e.g. \(1 \le x^2+y^2+z^2 \le 4,\; z \le 0\)), translate each inequality into geometry separately, then intersect: two concentric spheres bounding a shell, cut by the half-space below the xy-plane. Sketching each constraint before combining avoids most mistakes.

12.2

Vectors

A vector carries both magnitude and direction — displacement, velocity, force. It can be drawn as an arrow, but for computation it is best handled through components in a coordinate system.

Vector

A quantity with both magnitude (length) and direction, drawn as a directed line segment. Two vectors are equal (equivalent) if they have the same length and direction, regardless of position. The zero vector 0 has length 0 and no direction.

Vector Addition — Triangle & Parallelogram Laws

To add u + v, place the tail of v at the tip of u; the sum runs from the tail of u to the tip of v (Triangle Law). Equivalently, drawn from a common starting point, the sum is the diagonal of the parallelogram they span (Parallelogram Law).

Scalar Multiplication

Multiplying a vector by a scalar c scales its length by |c|. The direction is preserved if c > 0 and reversed if c < 0. Two nonzero vectors are parallel exactly when one is a scalar multiple of the other.

Components, Standard Basis, Unit Vector

Placing a vector's tail at the origin, its tip's coordinates are its components: a = ⟨a₁, a₂, a₃⟩. The standard basis vectors are i = ⟨1,0,0⟩, j = ⟨0,1,0⟩, k = ⟨0,0,1⟩. A unit vector has length 1; dividing any nonzero vector by its length gives the unit vector in its direction.

Vector Between Two Points

\[ \overrightarrow{AB} = \langle x_2-x_1,\; y_2-y_1,\; z_2-z_1 \rangle \]

Magnitude (Length) of a Vector

\[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

Component Operations

\[ \mathbf{a}+\mathbf{b} = \langle a_1+b_1,\, a_2+b_2,\, a_3+b_3 \rangle, \qquad c\mathbf{a} = \langle ca_1,\, ca_2,\, ca_3 \rangle \]

Unit Vector in the Direction of a

\[ \mathbf{u} = \frac{1}{|\mathbf{a}|}\,\mathbf{a} = \frac{\mathbf{a}}{|\mathbf{a}|} \]

Properties

Properties of Vectors

For vectors a, b, c and scalars c, d: addition is commutative and associative; a + 0 = a; a + (−a) = 0; scalar multiplication distributes over both vector and scalar sums; (cd)a = c(da); and 1a = a.

EXAMPLE 3Vector Operations from Components

If a = ⟨4, 0, 3⟩ and b = ⟨−2, 1, 5⟩, find |a|, a + b, and 2a + 5b.

Solution

Length of a:

\[ |\mathbf{a}| = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{25} = 5 \]

Add componentwise:

\[ \mathbf{a}+\mathbf{b} = \langle 4+(-2),\, 0+1,\, 3+5 \rangle = \langle 2, 1, 8 \rangle \]

Scale, then add:

\[ 2\mathbf{a}+5\mathbf{b} = \langle 8,0,6 \rangle + \langle -10,5,25 \rangle = \langle -2, 5, 31 \rangle \]

EXAMPLE 4Finding a Unit Vector

Find the unit vector in the direction of 2i − j − 2k.

Solution

Compute the length:

\[ |2\mathbf{i}-\mathbf{j}-2\mathbf{k}| = \sqrt{2^2+(-1)^2+(-2)^2} = \sqrt{9} = 3 \]

Divide the vector by its length:

\[ \mathbf{u} = \tfrac{1}{3}\left(2\mathbf{i}-\mathbf{j}-2\mathbf{k}\right) = \tfrac{2}{3}\mathbf{i} - \tfrac{1}{3}\mathbf{j} - \tfrac{2}{3}\mathbf{k} \]

EXAMPLE 5Resolving Forces — A Hanging Weight

A 100-lb weight hangs from two wires making angles of 50° and 32° with the horizontal. Find the tensions in both wires.

Solution

Write each tension in horizontal and vertical components, then require the resultant to cancel the weight 100j. Horizontal balance: \(-|\mathbf{T_1}|\cos 50^\circ + |\mathbf{T_2}|\cos 32^\circ = 0\). Vertical balance: \(|\mathbf{T_1}|\sin 50^\circ + |\mathbf{T_2}|\sin 32^\circ = 100\).

Solve the system:

\[ |\mathbf{T_1}| \approx 85.6\ \text{lb}, \qquad |\mathbf{T_2}| \approx 64.9\ \text{lb} \]

This force-balance technique is the foundation of static equilibrium analysis in mechanics.

Figure 12.2 A weight suspended by two wires; the wire tensions resolve into horizontal and vertical components that balance the load. (Original redrawn schematic.)

Instructor Note

Force balance problems are everywhere in statics. Always (1) draw a free-body diagram, (2) resolve each force into components, (3) write one equilibrium equation per axis, and (4) solve the simultaneous system. The two-wire weight here is the prototype for trusses and cable systems you will meet later.

12.3

The Dot Product

The dot product multiplies two vectors to produce a scalar. It measures how much two vectors point in the same direction, and it is the tool of choice for angles, orthogonality, projections, and work.

Dot Product (Scalar Product)

The dot product of two vectors multiplies corresponding components and sums the results, yielding a single real number — not a vector. It is also called the scalar product or inner product.

Orthogonality

Two vectors are perpendicular (orthogonal) precisely when their dot product is zero. The zero vector is considered orthogonal to every vector.

Sign of the Dot Product

The dot product is positive when the vectors point in generally the same direction, zero when they are perpendicular, and negative when they point in generally opposite directions — it measures the extent of shared direction.

Direction Angles and Direction Cosines

The direction angles α, β, γ of a vector are the angles it makes with the positive x-, y-, and z-axes. Their cosines, the direction cosines, are exactly the components of the unit vector in that vector's direction, and satisfy cos²α + cos²β + cos²γ = 1.

Dot Product — Component Form

\[ \mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

Dot Product — Geometric Form

\[ \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta \]

Angle Between Two Vectors

\[ \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|} \]

Scalar and Vector Projections of b onto a

\[ \operatorname{comp}_{\mathbf{a}}\mathbf{b} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|}, \qquad \operatorname{proj}_{\mathbf{a}}\mathbf{b} = \left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|^2}\right)\mathbf{a} \]

Work Done by a Constant Force

\[ W = \mathbf{F}\cdot\mathbf{D} = |\mathbf{F}|\,|\mathbf{D}|\cos\theta \]

Theorem

Geometric Interpretation of the Dot Product

If θ is the angle between vectors a and b, then a·b = |a||b| cos θ. This is proved by applying the Law of Cosines to the triangle formed by a, b, and a − b.

Test

Orthogonality Criterion

Two vectors a and b are orthogonal if and only if a·b = 0.

EXAMPLE 6Angle Between Vectors

Find the angle between a = ⟨2, 2, −1⟩ and b = ⟨5, −3, 2⟩.

Solution

Compute lengths and the dot product:

\[ |\mathbf{a}| = 3,\quad |\mathbf{b}| = \sqrt{38},\quad \mathbf{a}\cdot\mathbf{b} = 10-6-2 = 2 \]

Apply the angle formula:

\[ \cos\theta = \frac{2}{3\sqrt{38}} \;\Rightarrow\; \theta = \cos^{-1}\!\left(\frac{2}{3\sqrt{38}}\right) \approx 84^\circ \]

EXAMPLE 7Projection

Find the scalar and vector projections of b = ⟨1, 1, 2⟩ onto a = ⟨−2, 3, 1⟩.

Solution

With |a| = √14, the scalar projection is:

\[ \operatorname{comp}_{\mathbf{a}}\mathbf{b} = \frac{-2+3+2}{\sqrt{14}} = \frac{3}{\sqrt{14}} \]

Multiply by the unit vector in the direction of a:

\[ \operatorname{proj}_{\mathbf{a}}\mathbf{b} = \frac{3}{14}\langle -2,3,1 \rangle = \left\langle -\tfrac{3}{7}, \tfrac{9}{14}, \tfrac{3}{14} \right\rangle \]

EXAMPLE 8Work as a Dot Product

A force F = 3i + 4j + 5k moves a particle from P(2, 1, 0) to Q(4, 6, 2). Find the work done.

Solution

The displacement is \(\mathbf{D} = \overrightarrow{PQ} = \langle 2, 5, 2 \rangle\).

Work is the dot product:

\[ W = \mathbf{F}\cdot\mathbf{D} = 6 + 20 + 10 = 36\ \text{J} \]

Instructor Note

Two practical reflexes: to test perpendicularity, just check whether the dot product is zero — no angle computation needed. And remember that work counts only the component of force along the motion; a force perpendicular to displacement does zero work. That single fact explains why carrying a load horizontally at constant height does no mechanical work.

12.4

The Cross Product

Unlike the dot product, the cross product of two vectors is itself a vector — one perpendicular to both inputs. It is defined only in three dimensions and underlies area, volume, and torque calculations.

Cross Product (Vector Product)

The cross product a × b of two 3-D vectors is a vector perpendicular to both a and b, with direction given by the right-hand rule and length equal to the area of the parallelogram they span. It is defined only for three-dimensional vectors.

Right-Hand Rule

Curl the fingers of the right hand from a toward b through the angle between them; the thumb points in the direction of a × b. Reversing the order reverses the result: a × b = −(b × a), so the cross product is not commutative.

Scalar Triple Product & Coplanarity

The scalar triple product a · (b × c) gives the signed volume of the parallelepiped spanned by the three vectors. If it equals zero, the three vectors lie in a common plane (they are coplanar).

Torque

The torque of a force F applied at position r is the cross product τ = r × F. It measures the tendency to produce rotation; its direction is the axis of rotation.

Cross Product — Determinant Form

\[ \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]

Cross Product — Component Form

\[ \mathbf{a}\times\mathbf{b} = \langle a_2b_3 - a_3b_2,\; a_3b_1 - a_1b_3,\; a_1b_2 - a_2b_1 \rangle \]

Magnitude of the Cross Product

\[ |\mathbf{a}\times\mathbf{b}| = |\mathbf{a}|\,|\mathbf{b}|\sin\theta \]

Scalar Triple Product (Volume of a Parallelepiped)

\[ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

Magnitude of Torque

\[ |\boldsymbol{\tau}| = |\mathbf{r}\times\mathbf{F}| = |\mathbf{r}|\,|\mathbf{F}|\sin\theta \]

Theorem

Orthogonality of the Cross Product

The vector a × b is orthogonal to both a and b. This is verified by showing both a·(a × b) = 0 and b·(a × b) = 0.

Corollary

Parallel Criterion

Two nonzero vectors a and b are parallel if and only if a × b = 0.

Key Idea The length |a × b| equals the area of the parallelogram with sides a and b; half of it gives the area of the triangle they form.

EXAMPLE 9Computing a Cross Product

If a = ⟨1, 3, 4⟩ and b = ⟨2, 7, −5⟩, find a × b.

Solution

Expand the determinant along the first row:

\[ \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & 4 \\ 2 & 7 & -5 \end{vmatrix} \]

\[ = (-15-28)\mathbf{i} - (-5-8)\mathbf{j} + (7-6)\mathbf{k} = -43\mathbf{i} + 13\mathbf{j} + \mathbf{k} \]

EXAMPLE 10Area of a Triangle

Find the area of the triangle with vertices P(1, 4, 6), Q(−2, 5, −1), and R(1, −1, 1).

Solution

Form two edge vectors and take their cross product:

\[ \overrightarrow{PQ}\times\overrightarrow{PR} = \langle -40, -15, 15 \rangle \]

The triangle's area is half the magnitude:

\[ A = \tfrac{1}{2}\sqrt{(-40)^2 + (-15)^2 + 15^2} = \tfrac{5}{2}\sqrt{82} \]

EXAMPLE 11Torque on a Wrench

A 40-N force is applied to the end of a 0.25-m wrench at 75° to the handle. Find the magnitude of the torque about the bolt.

Solution

Use the magnitude formula:

\[ |\boldsymbol{\tau}| = |\mathbf{r}|\,|\mathbf{F}|\sin\theta = (0.25)(40)\sin 75^\circ \approx 9.66\ \text{N·m} \]

Only the component of force perpendicular to the wrench produces rotation, which is exactly what the sine factor captures.

Figure 12.3 The cross product a × b is perpendicular to the plane of a and b; its length equals the area of the parallelogram they span. (Original redrawn schematic.)

Instructor Note

Torque is the rotational analogue of force, and τ = r × F makes the geometry explicit: a longer wrench or a more perpendicular pull both increase torque. This is why you choke up less on a long breaker bar and push at a right angle. Keep the right-hand rule consistent and the sign of every cross product takes care of itself.

12.5

Equations of Lines and Planes

A line in space is fixed by a point and a direction vector; a plane is fixed by a point and a normal vector. Vectors convert these geometric descriptions into clean parametric and scalar equations.

Line in Space

A line is determined by a point on it and a vector v parallel to it (the direction). The position vector of any point on the line is the base point's position vector plus a scalar multiple of v.

Direction Numbers & Skew Lines

The components of a line's direction vector are its direction numbers; any proportional set works equally well. Two lines in space are skew if they neither intersect nor are parallel — possible only in three dimensions.

Plane and Normal Vector

A plane is determined by a point in it and a normal vector n orthogonal to the plane. Since n is perpendicular to every vector lying in the plane, the plane is the set of points P with n · (r − r₀) = 0.

Angle Between Planes

Two planes are parallel when their normal vectors are parallel. If not, they meet in a line, and the angle between the planes equals the acute angle between their normal vectors.

Vector Equation of a Line

\[ \mathbf{r} = \mathbf{r_0} + t\mathbf{v} \]

Parametric Equations of a Line

\[ x = x_0 + at, \qquad y = y_0 + bt, \qquad z = z_0 + ct \]

Symmetric Equations of a Line

\[ \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} \]

Scalar Equation of a Plane — normal ⟨a, b, c⟩ through (x₀, y₀, z₀)

\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]

Linear (General) Equation of a Plane

\[ ax + by + cz + d = 0 \]

Distance from a Point to a Plane

\[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

EXAMPLE 12Equations of a Line Through Two Points

Find parametric and symmetric equations of the line through A(2, 4, −3) and B(3, −1, 1), and find where it crosses the xy-plane.

Solution

Direction vector from A to B:

\[ \mathbf{v} = \langle 1, -5, 4 \rangle \]

Parametric form using A:

\[ x = 2 + t,\quad y = 4 - 5t,\quad z = -3 + 4t \]

Symmetric form:

\[ \frac{x-2}{1} = \frac{y-4}{-5} = \frac{z+3}{4} \]

Set z = 0, solve t = 3/4, giving the crossing point \(\left(\tfrac{11}{4}, \tfrac{1}{4}, 0\right)\).

EXAMPLE 13Plane Through Three Points

Find an equation of the plane through P(1, 3, 2), Q(3, −1, 6), and R(5, 2, 0).

Solution

Form two in-plane vectors and cross them to get a normal:

\[ \mathbf{a} = \langle 2,-4,4 \rangle,\quad \mathbf{b} = \langle 4,-1,-2 \rangle,\quad \mathbf{n} = \mathbf{a}\times\mathbf{b} = \langle 12, 20, 14 \rangle \]

Use the scalar equation with point P:

\[ 12(x-1) + 20(y-3) + 14(z-2) = 0 \;\Rightarrow\; 6x + 10y + 7z = 50 \]

EXAMPLE 14Distance Between Parallel Planes

Find the distance between the parallel planes 10x + 2y − 2z = 5 and 5x + y − z = 1.

Solution

Pick any point on the first plane — setting y = z = 0 gives \(\left(\tfrac{1}{2}, 0, 0\right)\).

Apply the point-to-plane distance to the second plane:

\[ D = \frac{\left|5\left(\tfrac{1}{2}\right) + 0 - 0 - 1\right|}{\sqrt{5^2 + 1^2 + 1^2}} = \frac{3/2}{3\sqrt{3}} = \frac{\sqrt{3}}{6} \]

Instructor Note

Three reliable moves: (1) the direction vector of a line through two points is just B − A; (2) the normal of a plane through three points is the cross product of two edge vectors; (3) to find where a line pierces a plane, substitute the parametric x, y, z into the plane equation and solve for t. Master those three and most line-and-plane problems collapse to arithmetic.

12.6

Cylinders and Quadric Surfaces

Beyond planes and spheres lie two more families of surfaces: cylinders, generated by sliding a curve along a line, and quadric surfaces, the 3-D counterparts of the conic sections. Both are understood through their traces.

Trace

A trace (or cross-section) is the curve formed where a surface meets a plane parallel to a coordinate plane. Sketching several traces reveals the shape of an unfamiliar surface.

Cylinder

A cylinder is a surface made of all lines (rulings) parallel to a given line and passing through a given plane curve. A telltale sign: if one of the variables x, y, or z is missing from the equation, the surface is a cylinder whose rulings are parallel to that missing variable's axis.

Quadric Surface

A quadric surface is the graph of a second-degree equation in x, y, and z. By translation and rotation, every quadric reduces to one of two standard forms, producing six basic types: the ellipsoid, elliptic paraboloid, hyperbolic paraboloid, cone, hyperboloid of one sheet, and hyperboloid of two sheets.

General Second-Degree Equation

\[ Ax^2 + By^2 + Cz^2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0 \]

Standard Forms (after reduction)

\[ Ax^2 + By^2 + Cz^2 + J = 0 \qquad \text{or} \qquad Ax^2 + By^2 + Iz = 0 \]

The Six Basic Quadric Surfaces

Surface	Standard Equation	Trace Behavior
Ellipsoid	\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)	All traces are ellipses; a sphere when a = b = c.
Elliptic Paraboloid	\(\dfrac{z}{c}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\)	Horizontal traces ellipses, vertical traces parabolas; the first-power variable is the axis.
Hyperbolic Paraboloid	\(\dfrac{z}{c}=\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}\)	Horizontal traces hyperbolas, vertical traces parabolas; saddle-shaped.
Cone	\(\dfrac{z^2}{c^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\)	Horizontal traces ellipses; vertical traces are hyperbolas, or pairs of lines through the origin.
Hyperboloid of One Sheet	\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=1\)	Horizontal traces ellipses, vertical traces hyperbolas; axis is the negative-coefficient variable.
Hyperboloid of Two Sheets	\(-\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)	Two separated pieces; horizontal traces ellipses only beyond the gap, vertical traces hyperbolas.

EXAMPLE 15Sketching an Ellipsoid via Traces

Use traces to describe the surface \(x^2 + \dfrac{y^2}{9} + \dfrac{z^2}{4} = 1\).

Solution

Setting z = 0 gives \(x^2 + y^2/9 = 1\), an ellipse in the xy-plane.

The horizontal trace at z = k is an ellipse whenever k² < 4, i.e. −2 < k < 2; the vertical traces in x = k and y = k are likewise ellipses.

All traces being ellipses, the surface is an ellipsoid, symmetric about each coordinate plane.

EXAMPLE 16Classifying by Completing the Square

Classify the surface \(x^2 + 2z^2 - 6x - y + 10 = 0\).

Solution

Complete the square in x:

\[ y - 1 = (x-3)^2 + 2z^2 \]

This matches the elliptic paraboloid form with axis parallel to the y-axis, vertex at (3, 1, 0). Traces in y = k (k > 1) are ellipses; the trace in the xy-plane is a parabola.

Applications Quadric surfaces are everywhere in engineering: the earth is modeled as an ellipsoid; circular paraboloids focus light and radio signals in dishes and telescopes; and nuclear cooling towers take the shape of hyperboloids of one sheet for structural stability.

Figure 12.4 A hyperboloid of one sheet: horizontal traces are ellipses and vertical traces are hyperbolas — the shape used for cooling towers. (Original redrawn schematic.)

Instructor Note

To classify any quadric: first complete the square to eliminate linear terms and locate the center, then read off the signs of the squared terms. The pattern of plus and minus signs, plus which variable (if any) appears only to the first power, tells you exactly which of the six surfaces you have. Practice this sign-reading until it is automatic.