SECTION 9.1
Vectors in Two Dimensions
Scalar vs. Directed Quantity
A scalar is a quantity determined completely by its magnitude — for example length, mass, area, temperature, or energy; only one number is needed. A directed quantity (such as displacement, velocity, acceleration, or force) requires both a magnitude and a direction, and is represented mathematically by a vector.
Vector (Geometric Description)
A vector in the plane is a line segment with an assigned direction, sketched as an arrow. For the vector \( \vec{AB} \), point \( A \) is the initial point and \( B \) is the terminal point. The length of segment \( AB \) is the magnitude (or length), denoted \( |\vec{AB}| \). Vectors are written in boldface, e.g. \( \mathbf{u} = \vec{AB} \).
Equal Vectors
Two vectors are equal if they have equal magnitude and the same direction. A vector represents a magnitude and direction, not a fixed location, so the same vector has many representations depending on its initial point.
Vector Addition
If displacement \( \mathbf{u} = \vec{AB} \) is followed by \( \mathbf{v} = \vec{BC} \), the resulting displacement is \( \vec{AC} = \vec{AB} + \vec{BC} \). To add two vectors, place the initial point of one at the terminal point of the other (head-to-tail); the sum runs from the first initial point to the last terminal point. Equivalently, drawing both from the same point, \( \mathbf{u} + \mathbf{v} \) is the diagonal of the parallelogram they form. The zero vector \( \mathbf{0} \) represents no displacement.
Scalar Multiplication
For a real number \( c \) and vector \( \mathbf{v} \), the vector \( c\mathbf{v} \) has magnitude \( |c|\,|\mathbf{v}| \). It has the same direction as \( \mathbf{v} \) if \( c > 0 \) and the opposite direction if \( c < 0 \); if \( c = 0 \) then \( c\mathbf{v} = \mathbf{0} \). Scalar multiplication stretches or shrinks a vector. We write \( (-1)\mathbf{v} = -\mathbf{v} \), the vector of equal length and opposite direction.
Vector Subtraction
The difference of two vectors is defined by \( \mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}) \). Geometrically, \( \mathbf{u} - \mathbf{v} \) is the other diagonal of the parallelogram formed by \( \mathbf{u} \) and \( \mathbf{v} \).
Components of a Vector
Placing a vector in a coordinate plane lets us describe it analytically by components. A vector \( \mathbf{v} = \langle a_1, a_2 \rangle \) has horizontal component \( a_1 \) and vertical component \( a_2 \). Note the distinction between the vector \( \langle a_1, a_2 \rangle \) and the point \( (a_1, a_2) \).
Unit Vectors i and j
A unit vector has length 1. Two special unit vectors are \( \mathbf{i} = \langle 1, 0 \rangle \) and \( \mathbf{j} = \langle 0, 1 \rangle \). Any vector can be written in terms of them.
Direction of a Vector
The direction of a vector \( \mathbf{v} \) with initial point at the origin is the smallest positive angle \( \theta \) in standard position formed by the positive \( x \)-axis and \( \mathbf{v} \). Knowing magnitude and direction determines the horizontal and vertical components.
Velocity and Force as Vectors
The velocity of a moving object is a vector whose direction is the direction of motion and whose magnitude is the speed. Force describes a push or pull and is measured in pounds (or newtons). When several forces act on an object, the resultant force is the vector sum of those forces. The true velocity of an object (e.g. an airplane in wind) is its velocity relative to the ground, found by adding the velocity vectors involved.
Solution
(a) Subtract initial-point coordinates from terminal-point coordinates:
\[ \mathbf{u} = \langle 3 - (-2),\; 7 - 5 \rangle = \langle 5, 2 \rangle \]
(b) Let the terminal point be \( (x, y) \). Then \( \langle x - 2,\; y - 4 \rangle = \langle 3, 7 \rangle \), so \( x - 2 = 3 \) and \( y - 4 = 7 \), giving \( x = 5 \), \( y = 11 \). The terminal point is \( (5, 11) \).
Solution
(a) \( |\mathbf{u}| = \sqrt{2^{2} + (-3)^{2}} = \sqrt{13} \)
(b) \( |\mathbf{v}| = \sqrt{(-5)^{2} + 0^{2}} = \sqrt{25} = 5 \)
(c) \( |\mathbf{w}| = \sqrt{\left(\tfrac{3}{5}\right)^{2} + \left(\tfrac{4}{5}\right)^{2}} = \sqrt{\tfrac{9}{25} + \tfrac{16}{25}} = \sqrt{1} = 1 \) — a unit vector.
Solution
\[ \mathbf{u} + \mathbf{v} = \langle 2 + (-1),\; -3 + 2 \rangle = \langle 1, -1 \rangle \]
\[ \mathbf{u} - \mathbf{v} = \langle 2 - (-1),\; -3 - 2 \rangle = \langle 3, -5 \rangle \]
\[ 2\mathbf{u} = \langle 4, -6 \rangle, \qquad 3\mathbf{v} = \langle -3, 6 \rangle \]
\[ 2\mathbf{u} + 3\mathbf{v} = \langle 4, -6 \rangle + \langle -3, 6 \rangle = \langle 1, 0 \rangle \]
Solution
(a) \( \mathbf{u} = 5\mathbf{i} + (-8)\mathbf{j} = 5\mathbf{i} - 8\mathbf{j} \)
(b) Treat vectors like algebraic expressions:
\[ 2\mathbf{u} + 5\mathbf{v} = 2(3\mathbf{i} + 2\mathbf{j}) + 5(-\mathbf{i} + 6\mathbf{j}) = (6\mathbf{i} + 4\mathbf{j}) + (-5\mathbf{i} + 30\mathbf{j}) = \mathbf{i} + 34\mathbf{j} \]
Solution
(a) \( a = 8\cos\tfrac{\pi}{3} = 4 \) and \( b = 8\sin\tfrac{\pi}{3} = 4\sqrt{3} \), so
\[ \mathbf{v} = \langle 4,\; 4\sqrt{3} \rangle = 4\mathbf{i} + 4\sqrt{3}\,\mathbf{j} \]
(b) \( \tan\theta = \dfrac{1}{-\sqrt{3}} \), giving reference angle \( \pi/6 \). The terminal point lies in Quadrant II, so \( \theta = \dfrac{5\pi}{6} \).
Solution
(a) \( \mathbf{v} = 0\mathbf{i} + 300\mathbf{j} = 300\mathbf{j} \). The wind makes a 60° angle with the positive \( x \)-axis:
\[ \mathbf{u} = (40\cos 60^\circ)\mathbf{i} + (40\sin 60^\circ)\mathbf{j} = 20\mathbf{i} + 20\sqrt{3}\,\mathbf{j} \approx 20\mathbf{i} + 34.64\mathbf{j} \]
(b) True velocity \( \mathbf{w} = \mathbf{u} + \mathbf{v} = 20\mathbf{i} + (20\sqrt{3} + 300)\mathbf{j} \approx 20\mathbf{i} + 334.64\mathbf{j} \).
(c) \( |\mathbf{w}| = \sqrt{20^{2} + 334.64^{2}} \approx 335.2 \text{ mi/h} \). Since \( \tan\theta = 334.64/20 = 16.732 \), \( \theta \approx 86.6^\circ \), so the plane heads N 3.4° E.
Solution
Let \( \mathbf{u} = 5\mathbf{i} \) (river) and \( \mathbf{v} = (10\cos\theta)\mathbf{i} + (10\sin\theta)\mathbf{j} \) (boat). The true course is
\[ \mathbf{w} = \mathbf{u} + \mathbf{v} = (5 + 10\cos\theta)\mathbf{i} + (10\sin\theta)\mathbf{j} \]
To land directly across, the horizontal component must be 0:
\[ 5 + 10\cos\theta = 0 \;\Rightarrow\; \cos\theta = -\tfrac{1}{2} \;\Rightarrow\; \theta = 120^\circ \]
She should head in the direction \( \theta = 120^\circ \) (or N 30° W).
Solution
\[ \mathbf{F}_1 = (10\cos 45^\circ)\mathbf{i} + (10\sin 45^\circ)\mathbf{j} = 5\sqrt{2}\,\mathbf{i} + 5\sqrt{2}\,\mathbf{j} \]
\[ \mathbf{F}_2 = (20\cos 150^\circ)\mathbf{i} + (20\sin 150^\circ)\mathbf{j} = -10\sqrt{3}\,\mathbf{i} + 10\mathbf{j} \]
\[ \mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2 = (5\sqrt{2} - 10\sqrt{3})\mathbf{i} + (5\sqrt{2} + 10)\mathbf{j} \approx -10\mathbf{i} + 17\mathbf{j} \]
(image omitted to respect copyright)
Figure 9.1.1 — A vector \( \mathbf{u} = \vec{AB} \) drawn as an arrow from initial point \( A \) to terminal point \( B \).
Figure 9.1.1
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Figure 9.1.4 — Addition of vectors: (a) head-to-tail method and (b) the parallelogram method, both yielding \( \mathbf{u} + \mathbf{v} \).
Figure 9.1.4
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Figure 9.1.16–17 — Wind velocity \( \mathbf{u} \) (N 30° E) and airplane velocity \( \mathbf{v} \) combine to give the true ground velocity \( \mathbf{w} = \mathbf{u} + \mathbf{v} \).
Figure 9.1.16
Instructor Note
When solving force or velocity problems, always draw the vector diagram first and convert every bearing (like N 30° E) into a standard-position angle measured from the positive \( x \)-axis before computing components. Most student errors here come from mixing up the bearing with the standard angle, not from the algebra. A quick sketch saves you every time.
SECTION 9.2
The Dot Product
Dot Product
The dot product of \( \mathbf{u} = \langle a_1, a_2 \rangle \) and \( \mathbf{v} = \langle b_1, b_2 \rangle \) multiplies corresponding components and adds them. The result is a real number (a scalar), not a vector — which is why it is also called the scalar product.
Angle Between Vectors
Sketching \( \mathbf{u} \) and \( \mathbf{v} \) from the same initial point, the angle \( \theta \) between them is the smaller of the angles they form, with \( 0 \le \theta \le \pi \). The dot product relates directly to this angle.
Orthogonal (Perpendicular) Vectors
Two nonzero vectors are perpendicular (orthogonal) when the angle between them is \( \pi/2 \). This happens exactly when their dot product is zero.
Component of u Along v
The component of u along v (also the scalar projection of \( \mathbf{u} \) onto \( \mathbf{v} \)) is \( |\mathbf{u}|\cos\theta \) — the magnitude of the portion of \( \mathbf{u} \) pointing in the direction of \( \mathbf{v} \). It is a scalar, and is negative when \( \pi/2 < \theta \le \pi \). Resolving a force into perpendicular directions (e.g. a car on an incline, where weight splits into a component along the slope and a component into the surface) uses this idea.
Projection of u Onto v
The projection \( \operatorname{proj}_{\mathbf{v}}\mathbf{u} \) is the vector parallel to \( \mathbf{v} \) whose length is the component of \( \mathbf{u} \) along \( \mathbf{v} \). It is found by multiplying a unit vector in the direction of \( \mathbf{v} \) by that component. Any vector \( \mathbf{u} \) can be resolved as \( \mathbf{u} = \mathbf{u}_1 + \mathbf{u}_2 \) with \( \mathbf{u}_1 = \operatorname{proj}_{\mathbf{v}}\mathbf{u} \) parallel to \( \mathbf{v} \) and \( \mathbf{u}_2 = \mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u} \) orthogonal to \( \mathbf{v} \).
Work
If a constant force of magnitude \( F \) moves an object a distance \( d \) along the direction of motion, the work done is \( W = Fd \) (units: foot-pounds when \( F \) is in pounds and \( d \) in feet). In general, only the component of force along the direction of motion does work, so work is expressed as a dot product of force and displacement.
Solution
(a) \( \mathbf{u} \cdot \mathbf{v} = (3)(4) + (-2)(5) = 12 - 10 = 2 \)
(b) \( \mathbf{u} \cdot \mathbf{v} = (2)(5) + (1)(-6) = 10 - 6 = 4 \)
Solution
\[ \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|} = \frac{(2)(4) + (5)(-3)}{\sqrt{4 + 25}\,\sqrt{16 + 9}} = \frac{-7}{5\sqrt{29}} \]
\[ \theta = \cos^{-1}\!\left( \frac{-7}{5\sqrt{29}} \right) \approx 105.1^\circ \]
Solution
(a) \( \mathbf{u} \cdot \mathbf{v} = (3)(2) + (5)(-8) = -34 \neq 0 \) — not perpendicular.
(b) \( \mathbf{u} \cdot \mathbf{v} = (2)(-1) + (1)(2) = 0 \) — perpendicular.
Solution
Resolve the downward weight \( \mathbf{w} \) into components parallel and perpendicular to the driveway.
(a) \( |\mathbf{u}| = 3000\cos 75^\circ \approx 776 \) lb keeps the car from rolling.
(b) \( |\mathbf{v}| = 3000\cos 15^\circ \approx 2898 \) lb is exerted on the driveway.
Solution
\[ \operatorname{comp}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|} = \frac{(1)(-2) + (4)(1)}{\sqrt{4 + 1}} = \frac{2}{\sqrt{5}} \]
Solution
(a)
\[ \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \left( \frac{\langle -2, 9 \rangle \cdot \langle -1, 2 \rangle}{(-1)^{2} + 2^{2}} \right)\langle -1, 2 \rangle = \left( \frac{2 + 18}{5} \right)\langle -1, 2 \rangle = 4\langle -1, 2 \rangle = \langle -4, 8 \rangle \]
(b)
\[ \mathbf{u}_1 = \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \langle -4, 8 \rangle \]
\[ \mathbf{u}_2 = \mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \langle -2, 9 \rangle - \langle -4, 8 \rangle = \langle 2, 1 \rangle \]
Solution
Displacement \( \mathbf{D} = \langle 5 - 1,\; 9 - 3 \rangle = \langle 4, 6 \rangle \).
\[ W = \mathbf{F} \cdot \mathbf{D} = \langle 2, 3 \rangle \cdot \langle 4, 6 \rangle = 8 + 18 = 26 \text{ ft-lb} \]
Solution
Displacement \( \mathbf{D} = 100\mathbf{i} \). Force
\[ \mathbf{F} = (20\cos 60^\circ)\mathbf{i} + (20\sin 60^\circ)\mathbf{j} = 10\mathbf{i} + 10\sqrt{3}\,\mathbf{j} \]
\[ W = \mathbf{F} \cdot \mathbf{D} = (10\mathbf{i} + 10\sqrt{3}\,\mathbf{j}) \cdot (100\mathbf{i}) = 1000 \text{ ft-lb} \]
(image omitted to respect copyright)
Figure 9.2.3 — Geometric interpretation of the component of \( \mathbf{u} \) along \( \mathbf{v} \) as \( |\mathbf{u}|\cos\theta \), for acute and obtuse \( \theta \).
Figure 9.2.3
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Figure 9.2.5 — A car on a 15° incline: weight \( \mathbf{w} \) resolved into a component \( \mathbf{u} \) along the driveway and \( \mathbf{v} \) perpendicular to it.
Figure 9.2.5
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Figure 9.2.7 — Work done by force \( \mathbf{F} \) moving an object along \( \mathbf{D} = \vec{PQ} \); only \( |\mathbf{F}|\cos\theta \) acts along the motion.
Figure 9.2.7
Instructor Note
Keep the two projections straight — comp gives a number (scalar projection), while proj gives a vector. In engineering statics you almost always want proj when resolving a load into components along a member and perpendicular to it. The inclined-car example is the exact model you'll reuse for forces on ramps, roofs, and conveyor belts.
SECTION 9.3
Three-Dimensional Coordinate Geometry
The 3D Rectangular Coordinate System
Space is described using a fixed origin \( O \) and three mutually perpendicular coordinate axes — the \( x \)-, \( y \)-, and \( z \)-axes. Any point \( P \) in space is located by an ordered triple \( (a, b, c) \): the \( x \)-, \( y \)-, and \( z \)-coordinates. The set of all such triples forms the three-dimensional rectangular coordinate system.
Coordinate Planes and Octants
The three axes determine three coordinate planes: the \( xy \)-plane, the \( yz \)-plane, and the \( xz \)-plane. These planes divide space into eight parts called octants. (Visual aid: stand in a room corner — the left wall is the \( xz \)-plane, the right wall the \( yz \)-plane, and the floor the \( xy \)-plane.)
Surfaces in Space
In two dimensions an equation in \( x \) and \( y \) graphs as a curve. In three dimensions an equation in \( x \), \( y \), and \( z \) represents a surface. For example, \( z = 3 \) is a horizontal plane parallel to the \( xy \)-plane, and \( y = 5 \) is a vertical plane parallel to the \( xz \)-plane.
Sphere and Trace
A sphere is the set of all points at a fixed distance \( r \) (the radius) from a center \( C \). The trace of a sphere in a plane is the intersection (a circle) found by substituting the fixed coordinate of that plane into the sphere's equation.
Solution
For \( (2, 4, 7) \): move 2 along \( x \), 4 along \( y \), then 7 up along \( z \). For \( (-4, 3, -5) \): move \( -4 \) along \( x \), 3 along \( y \), then \( -5 \) down along \( z \). Each triple locates one unique point in space.
Solution
(a) All points with \( z \)-coordinate 3 — a horizontal plane parallel to the \( xy \)-plane, three units above it.
(b) All points with \( y \)-coordinate 5 — a vertical plane parallel to the \( xz \)-plane, five units to the right.
Solution
\[ d(P, Q) = \sqrt{(1 - 2)^{2} + (-3 - (-1))^{2} + (5 - 7)^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
Solution
\[ (x - 2)^{2} + (y + 1)^{2} + (z - 3)^{2} = 25 \]
Solution
Complete the square in each variable:
\[ (x^{2} - 4x + 4) + (y^{2} + 6y + 9) + (z^{2} + 2z + 1) = -6 + 4 + 9 + 1 \]
\[ (x - 2)^{2} + (y + 3)^{2} + (z + 1)^{2} = 8 \]
Center \( (2, -3, -1) \), radius \( \sqrt{8} = 2\sqrt{2} \).
Solution
(a) Set \( z = 0 \): \( (x - 2)^{2} + (y - 4)^{2} + 25 = 36 \), so \( (x - 2)^{2} + (y - 4)^{2} = 11 \) — a circle of radius \( \sqrt{11} \) in the \( xy \)-plane, centered at \( (2, 4, 0) \).
(b) Set \( z = 9 \): \( (x - 2)^{2} + (y - 4)^{2} + 16 = 36 \), so \( (x - 2)^{2} + (y - 4)^{2} = 20 \) — a circle of radius \( \sqrt{20} \), 9 units above the \( xy \)-plane, centered at \( (2, 4, 9) \).
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Figure 9.3.3 — A point \( P(a, b, c) \) located relative to the three coordinate axes.
Figure 9.3.3
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Figure 9.3.7 — Rectangular box construction proving the three-dimensional Distance Formula via the Pythagorean Theorem.
Figure 9.3.7
Instructor Note
Completing the square in three variables at once (Example 5) trips people up. Do them one variable at a time and remember to add the same constants to the right-hand side. This skill comes straight back when you study quadric surfaces and when checking whether a CAD-defined solid is actually a sphere.
SECTION 9.4
Vectors in Three Dimensions
Vectors in Space
A vector in space, placed with its initial point at the origin, is described by an ordered triple \( \mathbf{v} = \langle a_1, a_2, a_3 \rangle \), where \( a_1, a_2, a_3 \) are its components. All the properties that hold for plane vectors hold for space vectors.
Unit Vectors i, j, k
The standard unit vectors in space are \( \mathbf{i} = \langle 1, 0, 0 \rangle \), \( \mathbf{j} = \langle 0, 1, 0 \rangle \), and \( \mathbf{k} = \langle 0, 0, 1 \rangle \). Any space vector can be written in terms of them.
Dot Product in Space
The dot product extends naturally to three dimensions and remains a scalar. All dot-product properties, including the Dot Product Theorem and the perpendicularity test \( \mathbf{u} \cdot \mathbf{v} = 0 \), carry over.
Direction Angles and Direction Cosines
The direction angles \( \alpha, \beta, \gamma \) of a nonzero vector \( \mathbf{v} \) are the angles in \( [0, \pi] \) that \( \mathbf{v} \) makes with the positive \( x \)-, \( y \)-, and \( z \)-axes. Their cosines, \( \cos\alpha, \cos\beta, \cos\gamma \), are the direction cosines. If \( |\mathbf{v}| = 1 \), the direction cosines are simply the components of \( \mathbf{v} \). The direction angles fix a vector's direction but not its length. An angle is acute if \( 0 \le \theta < \pi/2 \) and obtuse if \( \pi/2 < \theta \le \pi \).
Solution
(a) \( \mathbf{v} = \langle 3 - 1,\; 1 - (-4),\; -1 - 5 \rangle = \langle 2, 5, -6 \rangle \)
(b) \( \langle x - 2,\; y - 1,\; z + 1 \rangle = \langle -2, 1, 3 \rangle \), so \( x = 0 \), \( y = 2 \), \( z = 2 \). Terminal point \( (0, 2, 2) \).
Solution
(a) \( |\mathbf{u}| = \sqrt{3^{2} + (-2)^{2} + 5^{2}} = \sqrt{38} \)
(b) \( |\mathbf{v}| = \sqrt{0^{2} + 3^{2} + (-1)^{2}} = \sqrt{10} \)
(c) \( |\mathbf{w}| = \sqrt{0 + 0 + 1} = 1 \)
Solution
\[ \mathbf{u} + \mathbf{v} = \langle 7, -1, 3 \rangle, \qquad \mathbf{u} - \mathbf{v} = \langle -5, -3, 5 \rangle \]
\[ 5\mathbf{u} - 3\mathbf{v} = \langle 5, -10, 20 \rangle - \langle 18, 3, -3 \rangle = \langle -13, -13, 23 \rangle \]
Solution
(a) \( \mathbf{u} = 5\mathbf{i} - 3\mathbf{j} + 6\mathbf{k} \)
(b) \( 2\mathbf{u} + 3\mathbf{v} = 2(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) + 3(4\mathbf{i} + 7\mathbf{k}) = (2\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}) + (12\mathbf{i} + 21\mathbf{k}) = 14\mathbf{i} + 4\mathbf{j} + 15\mathbf{k} \)
Solution
(a) \( (1)(6) + (2)(-5) + (3)(1) = 6 - 10 + 3 = -1 \)
(b) \( (2)(1) + (3)(-2) + (-1)(8) = 2 - 6 - 8 = -12 \)
Solution
\[ (2)(5) + (2)(-4) + (-1)(2) = 10 - 8 - 2 = 0 \]
Since the dot product is 0, the vectors are perpendicular.
Solution
\( |\mathbf{v}| = \sqrt{1 + 4 + 9} = \sqrt{14} \). Then
\[ \cos\alpha = \tfrac{1}{\sqrt{14}}, \quad \cos\beta = \tfrac{2}{\sqrt{14}}, \quad \cos\gamma = \tfrac{3}{\sqrt{14}} \]
\[ \alpha \approx 74^\circ, \quad \beta \approx 58^\circ, \quad \gamma \approx 37^\circ \]
Solution
\[ \cos^{2}\tfrac{\pi}{3} + \cos^{2}\tfrac{3\pi}{4} + \cos^{2}\gamma = 1 \]
\[ \left(\tfrac{1}{2}\right)^{2} + \left(-\tfrac{1}{\sqrt{2}}\right)^{2} + \cos^{2}\gamma = 1 \;\Rightarrow\; \cos^{2}\gamma = \tfrac{1}{4} \]
\[ \cos\gamma = \pm\tfrac{1}{2} \;\Rightarrow\; \gamma = \tfrac{\pi}{3} \text{ or } \tfrac{2\pi}{3} \]
Since \( \gamma \) is obtuse, \( \gamma = \dfrac{2\pi}{3} \).
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Figure 9.4.4 — Any space vector \( \mathbf{v} \) decomposed along the unit vectors as \( a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \).
Figure 9.4.4
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Figure 9.4.6 — The direction angles \( \alpha, \beta, \gamma \) that a vector \( \mathbf{v} \) makes with the positive \( x \)-, \( y \)-, and \( z \)-axes.
Figure 9.4.6
Instructor Note
Direction cosines are the backbone of 3D statics. When you write a force along a structural member as \( \mathbf{F} = |\mathbf{F}|\langle \cos\alpha, \cos\beta, \cos\gamma \rangle \), the identity \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \) is your built-in error check — if your three cosines don't square-sum to 1, you've made an arithmetic mistake. Use it every time.
SECTION 9.5
The Cross Product
Cross Product (Vector Product)
Given \( \mathbf{u} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{v} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \mathbf{u} \times \mathbf{v} \) is a vector (unlike the dot product, which is a scalar) — hence "vector product." It is defined only for three-dimensional vectors and is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \).
Determinant Notation
A determinant gives a convenient way to remember the cross product. A second-order determinant is \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \). A third-order determinant is expanded along the first row in terms of second-order determinants, with a minus sign on the middle term. Writing \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row makes the cross product easy to compute.
Cross Product Theorem and Right-Hand Rule
The vector \( \mathbf{u} \times \mathbf{v} \) is orthogonal to both \( \mathbf{u} \) and \( \mathbf{v} \). Its direction is given by the right-hand rule: if the fingers of your right hand curl from \( \mathbf{u} \) toward \( \mathbf{v} \) (through the angle less than 180°), your thumb points in the direction of \( \mathbf{u} \times \mathbf{v} \). Two nonzero vectors are parallel if and only if \( \mathbf{u} \times \mathbf{v} = \mathbf{0} \).
Area and Volume Interpretations
The length \( |\mathbf{u} \times \mathbf{v}| \) equals the area of the parallelogram determined by \( \mathbf{u} \) and \( \mathbf{v} \) (half of it is the area of the corresponding triangle). The scalar triple product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \) gives, in absolute value, the volume of the parallelepiped determined by \( \mathbf{u}, \mathbf{v}, \mathbf{w} \). If that volume is 0, the three vectors are coplanar.
Solution
\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 3 \\ 2 & 0 & -1 \end{vmatrix} = \begin{vmatrix} -1 & 3 \\ 0 & -1 \end{vmatrix}\mathbf{i} - \begin{vmatrix} 0 & 3 \\ 2 & -1 \end{vmatrix}\mathbf{j} + \begin{vmatrix} 0 & -1 \\ 2 & 0 \end{vmatrix}\mathbf{k} \]
\[ = (1 - 0)\mathbf{i} - (0 - 6)\mathbf{j} + (0 + 2)\mathbf{k} = \mathbf{i} + 6\mathbf{j} + 2\mathbf{k} \]
Solution
From Example 1, \( \mathbf{u} \times \mathbf{v} = \mathbf{i} + 6\mathbf{j} + 2\mathbf{k} \). Divide by its length \( \sqrt{1 + 36 + 4} = \sqrt{41} \):
\[ \frac{\mathbf{u} \times \mathbf{v}}{|\mathbf{u} \times \mathbf{v}|} = \frac{1}{\sqrt{41}}(\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}) \]
Solution
\[ \vec{PQ} = \langle -3, 1, -7 \rangle, \qquad \vec{PR} = \langle 0, -5, -5 \rangle \]
\[ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 1 & -7 \\ 0 & -5 & -5 \end{vmatrix} = (-5 - 35)\mathbf{i} - (15 - 0)\mathbf{j} + (15 - 0)\mathbf{k} = -40\mathbf{i} - 15\mathbf{j} + 15\mathbf{k} \]
So \( \langle -40, -15, 15 \rangle \) is perpendicular to the plane; any nonzero scalar multiple, such as \( \langle -8, -3, 3 \rangle \), works too.
Solution
From Example 3, \( \vec{PQ} \times \vec{PR} = \langle -40, -15, 15 \rangle \).
\[ |\vec{PQ} \times \vec{PR}| = \sqrt{(-40)^{2} + (-15)^{2} + 15^{2}} = \sqrt{2050} = 5\sqrt{82} \]
The triangle's area is half the parallelogram's: \( \dfrac{5}{2}\sqrt{82} \).
Solution
\[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} 1 & 4 & -7 \\ 2 & -1 & 4 \\ 0 & -9 & 18 \end{vmatrix} = 1(-18 + 36) - 4(36 - 0) + (-7)(-18 - 0) = 18 - 144 + 126 = 0 \]
The volume is 0, so the three vectors are coplanar.
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Figure 9.5.1 — \( \mathbf{u} \times \mathbf{v} \) is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \).
Figure 9.5.1
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Figure 9.5.2 — The right-hand rule fixing the direction of \( \mathbf{u} \times \mathbf{v} \).
Figure 9.5.2
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Figure 9.5.3 — The parallelogram determined by \( \mathbf{u} \) and \( \mathbf{v} \), with base \( |\mathbf{u}| \), altitude \( |\mathbf{v}|\sin\theta \), and area \( |\mathbf{u} \times \mathbf{v}| \).
Figure 9.5.3
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Figure 9.5.4 — The parallelepiped determined by \( \mathbf{u}, \mathbf{v}, \mathbf{w} \), with volume \( |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| \).
Figure 9.5.4
Instructor Note
The cross product is the tool for moments (torque) in mechanics: \( \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \). The order matters — \( \mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u}) \) — so never swap the operands. When you need a surface normal in CAD or a torque direction in dynamics, this is the operation you reach for. Verify your answer is perpendicular by dotting it back into \( \mathbf{u} \) and \( \mathbf{v} \); both should give 0.
SECTION 9.6
Equations of Lines and Planes
Determining a Line in Space
A line \( L \) in space is determined by a point \( P_0(x_0, y_0, z_0) \) on it and a direction vector \( \mathbf{v} = \langle a, b, c \rangle \) parallel to it. With \( \mathbf{r}_0 \) the position vector of \( P_0 \), the position vectors \( \mathbf{r} = \mathbf{r}_0 + t\mathbf{v} \) (for all real \( t \)) trace out the line — the vector equation of a line. Each value of the parameter \( t \) gives one point on \( L \). The position vector of a point \( (a_1, a_2, a_3) \) is the vector \( \langle a_1, a_2, a_3 \rangle \) from the origin to that point.
Determining a Plane in Space
A plane's direction cannot be captured by a single vector lying in it, but a vector perpendicular to the plane — the normal vector \( \mathbf{n} \) — completely specifies its orientation. A plane is determined by a point \( P_0 \) in it and a normal vector \( \mathbf{n} \). A point \( P \) lies in the plane exactly when \( \vec{P_0 P} \) is orthogonal to \( \mathbf{n} \), i.e. \( \mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0 \) — the vector equation of a plane.
Multiple Parametrizations
A line has infinitely many sets of parametric equations, since any point on the line and any vector parallel to it can be used. Different-looking equations may describe the same line.
Solution
\[ x = 5 + 3t, \qquad y = -2 - 4t, \qquad z = 3 + 2t, \qquad t \in \mathbb{R} \]
Solution
Direction vector \( \mathbf{v} = \langle 2 - (-1),\; -3 - 2,\; -7 - 6 \rangle = \langle 3, -5, -13 \rangle \). Using point \( (-1, 2, 6) \):
\[ x = -1 + 3t, \qquad y = 2 - 5t, \qquad z = 6 - 13t \]
Solution
(a)
\[ 4(x + 3) + 6(y - 1) - 3(z + 2) = 0 \;\Rightarrow\; 4x + 6y - 3z = -12 \]
(b) Set two variables to 0 at a time: \( x \)-intercept \( x = -3 \); \( y \)-intercept \( y = -2 \); \( z \)-intercept \( z = 4 \). The plane meets the axes at \( (-3, 0, 0) \), \( (0, -2, 0) \), \( (0, 0, 4) \).
Solution
A normal is \( \mathbf{n} = \vec{PQ} \times \vec{PR} = \langle -40, -15, 15 \rangle \) (from Section 9.5, Example 3). Using point \( P \):
\[ -40(x - 1) - 15(y - 4) + 15(z - 6) = 0 \]
\[ -40x - 15y + 15z = -10 \;\Rightarrow\; 8x + 3y - 3z = 2 \]
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Figure 9.6.1 — A line as the set of terminal points of \( \mathbf{r}_0 + t\mathbf{v} \), parallel to direction vector \( \mathbf{v} \).
Figure 9.6.1
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Figure 9.6.4 — A plane defined by a point \( P_0 \) and a normal vector \( \mathbf{n} \); point \( P \) lies in it when \( \mathbf{r} - \mathbf{r}_0 \perp \mathbf{n} \).
Figure 9.6.4
Instructor Note
Two takeaways you'll use constantly: a line needs a point + a parallel (direction) vector, while a plane needs a point + a perpendicular (normal) vector. When given three points and asked for a plane, the cross product of two edge vectors gives the normal in one step — this connects 9.5 directly to 9.6. Always simplify the final plane equation by dividing out common factors.
Vector Field
A vector field is a function that assigns a vector to each point in a region. In the plane, \( \mathbf{F}(x, y) \) assigns a vector to each point \( (x, y) \); in space, \( \mathbf{F}(x, y, z) \) assigns a vector to each point. Graphing many of these vectors gives a picture of a flow — for example, wind currents over the Earth or the gravitational field surrounding it.
Flow Lines (Streamlines)
The path that a small object (like a toy boat in a current) would follow through a vector field is called a flow line or streamline. It is everywhere tangent to the field vectors.
Gravitational Field
By Newton's Law of Gravity, the gravitational force on an object is directed toward the center of the Earth and is inversely proportional to the square of the distance from that center. Placing the origin at the Earth's center, the field is modeled by scaling a unit vector pointing toward the origin by the force magnitude \( GMm/r^{2} \).
Solution
Evaluate at sample points: \( \mathbf{F}(1, 3) = \mathbf{i} + 3\mathbf{j} \), \( \mathbf{F}(3, 3) = 3\mathbf{i} + 3\mathbf{j} \), \( \mathbf{F}(-4, 6) = -4\mathbf{i} + 6\mathbf{j} \), \( \mathbf{F}(-6, -1) = -6\mathbf{i} - \mathbf{j} \), \( \mathbf{F}(6, -6) = 6\mathbf{i} - 6\mathbf{j} \). Plotting these, the vectors point away from the origin, and their magnitude grows with distance from the origin.
Solution
Sample values: \( \mathbf{F}(1, 0) = \langle 0, 1 \rangle \), \( \mathbf{F}(2, 2) = \langle -2, 2 \rangle \), \( \mathbf{F}(3, 0) = \langle 0, 3 \rangle \), \( \mathbf{F}(0, 1) = \langle -1, 0 \rangle \). The field shows the wheel rotating counterclockwise, with points near the edge moving faster than points near the center.
Solution
Every vector is vertical: pointing upward above the \( xy \)-plane and downward below it, with magnitude increasing with distance from the \( xy \)-plane.
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Figure FM.1 — Wind over the surface of the Earth represented by a vector field.
Figure
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Figure FM.6 — The gravitational field, with vectors directed toward the Earth's center and shrinking with distance.
Figure
Instructor Note
Vector fields are your first taste of where this whole chapter is heading — fluid flow, electromagnetic fields, stress fields, and heat flux in later courses are all vector (or tensor) fields. Get comfortable now reading a field's behavior from its formula: \( \langle x, y \rangle \) radiates outward, \( \langle -y, x \rangle \) rotates. That intuition pays off enormously in fluid mechanics and FEA.