Unit 1, lesson 1

Three-Dimensional Coordinates and Vectors

How to locate points in right-handed three-dimensional space, measure distances, write spheres, and describe displacements and forces as vectors with components.

  • Theory part 1
  • Practice part 2
  • Coordinates, Vectors, and Products

Theory, part 1

To tell a drone where to hover you need three numbers, how far east, how far north and how high. The same three numbers can also describe a step from one place to another, or a push with a size and a direction. This lesson sets up the coordinate system that every later lesson works in and the vectors that live in it.

Key terms

Right-handed coordinates

Definition A rectangular coordinate system in space consists of a point OO, the origin, and three mutually perpendicular number lines through OO, the xx, yy and zz axes, all with the same unit of length. A point PP gets the coordinates (a,b,c)(a, b, c) when you reach it from OO by moving aa units parallel to the xx axis, then bb units parallel to the yy axis, then cc units parallel to the zz axis, with negative numbers meaning steps in the negative direction. The system is right-handed when the fingers of your right hand, curled from the positive xx axis toward the positive yy axis through a right angle, leave your thumb pointing along the positive zz axis. This course always uses right-handed coordinates.

In plain words Three perpendicular rulers meet at one point, and every point of space has exactly one triple of readings. The right-hand rule fixes which way the third ruler points, so everyone draws the same picture.

Example To reach P(2,−3,4)P(2, -3, 4), walk 2 units in the positive xx direction, 3 units in the negative yy direction, then climb 4 units. If you swapped the labels of the xx and yy axes and kept the zz axis, curling from the new positive xx toward the new positive yy would leave your right thumb pointing down. That system is left-handed, and in it the component formula for the cross product, which comes later, would give vectors pointing opposite to the right-hand rule.

Coordinate planes and octants

Definition The three coordinate planes are

the xy-plane z=0,the yz-plane x=0,the xz-plane y=0.\begin{gathered} \text{the } xy\text{-plane } z = 0, \\[0.4em] \text{the } yz\text{-plane } x = 0, \\[0.4em] \text{the } xz\text{-plane } y = 0. \end{gathered}

They divide space into eight octants, the regions where all three coordinates are nonzero and each coordinate has a fixed sign. The first octant is the set where x>0x > 0, y>0y > 0 and z>0z > 0. Points that lie on a coordinate plane belong to no octant. When a later problem speaks of a solid or a surface “in the first octant”, it means the closed region x≥0x \ge 0, y≥0y \ge 0, z≥0z \ge 0, which for a solid adds only boundary pieces of zero volume.

In plain words The floor and two walls of a room corner, extended forever, cut space into eight blocks. The signs of the three coordinates tell you which block a point is in.

Example The point (−1,2,−3)(-1, 2, -3) has x<0x < 0, y>0y > 0 and z<0z < 0, so it lies on the negative xx side of the yzyz-plane, on the positive yy side of the xzxz-plane and below the xyxy-plane. The point (4,0,5)(4, 0, 5) lies in the xzxz-plane and is in no octant. Its distance to the xyxy-plane is ∣z∣=5\lvert z \rvert = 5, since dropping straight down changes only zz.

Distance

Definition The distance ∣P1P2∣\lvert P_1 P_2 \rvert between P1(x1,y1,z1)P_1(x_1, y_1, z_1) and P2(x2,y2,z2)P_2(x_2, y_2, z_2) is the nonnegative number whose square is

∣P1P2∣2=(x2−x1)2+(y2−y1)2+(z2−z1)2.\begin{aligned} \lvert P_1 P_2 \rvert^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\[0.4em] &\quad + (z_2 - z_1)^2. \end{aligned}

Writing Δx=x2−x1\Delta x = x_2 - x_1, Δy=y2−y1\Delta y = y_2 - y_1 and Δz=z2−z1\Delta z = z_2 - z_1 for the coordinate differences, the distance d=∣P1P2∣d = \lvert P_1 P_2 \rvert is

d=(Δx)2+(Δy)2+(Δz)2.d = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}.

The formula follows from the Pythagorean theorem used twice, once in a horizontal plane for the xx and yy differences and once in a vertical plane for the result and the zz difference. This needs perpendicular axes with a common unit, which is part of the setup above.

In plain words Square the three coordinate differences, add, and take the square root. It is the flat distance formula with one more term.

Example For P1(1,−2,3)P_1(1, -2, 3) and P2(4,2,15)P_2(4, 2, 15) the differences are 33, 44 and 1212, so

∣P1P2∣=9+16+144=13.\lvert P_1 P_2 \rvert = \sqrt{9 + 16 + 144} = 13.

Spheres

Definition The sphere with center C(h,k,l)C(h, k, l) and radius r>0r > 0 is the set of all points whose distance from CC is rr. Squaring the distance formula gives its equation

(x−h)2+(y−k)2+(z−l)2=r2.\begin{aligned} &(x - h)^2 + (y - k)^2 \\[0.4em] &\quad + (z - l)^2 = r^2. \end{aligned}

An equation x2+y2+z2+Dx+Ey+Fz+G=0x^2 + y^2 + z^2 + Dx + Ey + Fz + G = 0 can be put in this form by completing the square in each variable. The result is a sphere if the right side comes out positive, a single point if it is zero, and no points at all if it is negative.

In plain words A sphere is everything at one fixed distance from a center. Completing squares reveals the center and the radius.

Example Take x2+y2+z2−4x+6y−2z=2x^2 + y^2 + z^2 - 4x + 6y - 2z = 2. Completing the square in each variable,

(x−2)2+(y+3)2+(z−1)2=2+4+9+1=16.\begin{aligned} &(x - 2)^2 + (y + 3)^2 + (z - 1)^2 \\[0.4em] &= 2 + 4 + 9 + 1 = 16. \end{aligned}

This is the sphere with center (2,−3,1)(2, -3, 1) and radius 44. Setting z=0z = 0 shows where it meets the xyxy-plane, namely the circle (x−2)2+(y+3)2=15(x - 2)^2 + (y + 3)^2 = 15 of radius 15\sqrt{15}.

Vectors and components

Definition A vector is a quantity with a magnitude and a direction. It is represented by a directed line segment, an arrow from an initial point to a terminal point, and two arrows with the same length and the same direction represent the same vector. If an arrow representing a\mathbf{a} starts at the origin and ends at (a1,a2,a3)(a_1, a_2, a_3), the numbers a1a_1, a2a_2, a3a_3 are the components of a\mathbf{a} and we write a=⟨a1,a2,a3⟩\mathbf{a} = \langle a_1, a_2, a_3 \rangle. The vector from A(x1,y1,z1)A(x_1, y_1, z_1) to B(x2,y2,z2)B(x_2, y_2, z_2) is

AB→=⟨x2−x1, y2−y1, z2−z1⟩.\overrightarrow{AB} = \langle x_2 - x_1, \ y_2 - y_1, \ z_2 - z_1 \rangle.

The vector OP→\overrightarrow{OP} is the position vector of PP. The zero vector 0=⟨0,0,0⟩\mathbf{0} = \langle 0, 0, 0 \rangle has length zero and no direction.

In plain words A point is a place, a vector is a step. The components say how far the step goes in each axis direction, wherever it starts.

Example From A(1,0,2)A(1, 0, 2) to B(4,−2,3)B(4, -2, 3) the displacement is AB→=⟨3,−2,1⟩\overrightarrow{AB} = \langle 3, -2, 1 \rangle. The same vector carries (0,0,0)(0, 0, 0) to (3,−2,1)(3, -2, 1) and carries (5,5,5)(5, 5, 5) to (8,3,6)(8, 3, 6).

Magnitude

Definition The magnitude, or length, of a=⟨a1,a2,a3⟩\mathbf{a} = \langle a_1, a_2, a_3 \rangle is

∣a∣=a12+a22+a32.\lvert \mathbf{a} \rvert = \sqrt{a_1^2 + a_2^2 + a_3^2}.

It is the distance between the initial and terminal points of any arrow representing a\mathbf{a}. It is never negative, and it is zero only for a=0\mathbf{a} = \mathbf{0}.

In plain words The magnitude is how long the arrow is, with no direction attached.

Example The vector ⟨2,−1,2⟩\langle 2, -1, 2 \rangle has magnitude 4+1+4=3\sqrt{4 + 1 + 4} = 3.

Vector addition and scalar multiplication

Definition For a=⟨a1,a2,a3⟩\mathbf{a} = \langle a_1, a_2, a_3 \rangle, b=⟨b1,b2,b3⟩\mathbf{b} = \langle b_1, b_2, b_3 \rangle and a real number cc,

a+b=⟨a1+b1,a2+b2,a3+b3⟩,c a=⟨ca1,ca2,ca3⟩.\begin{aligned} \mathbf{a} + \mathbf{b} &= \langle a_1 + b_1, a_2 + b_2, a_3 + b_3 \rangle, \\[0.4em] c\,\mathbf{a} &= \langle c a_1, c a_2, c a_3 \rangle. \end{aligned}

Geometrically, a+b\mathbf{a} + \mathbf{b} is the arrow from the tail of a\mathbf{a} to the head of b\mathbf{b} when b\mathbf{b} is placed at the head of a\mathbf{a}. The vector c ac\,\mathbf{a} has length ∣c∣ ∣a∣\lvert c \rvert \, \lvert \mathbf{a} \rvert and points the same way as a\mathbf{a} if c>0c > 0 and the opposite way if c<0c < 0. Two nonzero vectors are parallel when one is a scalar multiple of the other. Subtraction is a−b=a+(−1)b\mathbf{a} - \mathbf{b} = \mathbf{a} + (-1)\mathbf{b}.

In plain words Add vectors by doing one step and then the other. Multiplying by a number stretches or shrinks the step and flips it when the number is negative.

Example With a=⟨1,2,−1⟩\mathbf{a} = \langle 1, 2, -1 \rangle and b=⟨3,0,4⟩\mathbf{b} = \langle 3, 0, 4 \rangle,

a+b=⟨4,2,3⟩,2a−b=⟨−1,4,−6⟩.\begin{aligned} \mathbf{a} + \mathbf{b} &= \langle 4, 2, 3 \rangle, \\[0.4em] 2\mathbf{a} - \mathbf{b} &= \langle -1, 4, -6 \rangle. \end{aligned}

Unit vectors

Definition A unit vector is a vector of length 11. For any nonzero vector a\mathbf{a}, the vector

u=a∣a∣\mathbf{u} = \frac{\mathbf{a}}{\lvert \mathbf{a} \rvert}

is the unit vector in the direction of a\mathbf{a}. The zero vector cannot be normalized, since it has no direction and dividing by 00 is undefined.

In plain words A unit vector keeps only the direction. Divide a vector by its own length and you get it.

Example For a=⟨2,−1,2⟩\mathbf{a} = \langle 2, -1, 2 \rangle with ∣a∣=3\lvert \mathbf{a} \rvert = 3, the unit vector is u=⟨23,−13,23⟩\mathbf{u} = \langle \tfrac{2}{3}, -\tfrac{1}{3}, \tfrac{2}{3} \rangle. The vector of length 66 in the same direction is 6u=⟨4,−2,4⟩6\mathbf{u} = \langle 4, -2, 4 \rangle.

Basis

Definition A linear combination of vectors v1\mathbf{v}_1, v2\mathbf{v}_2, v3\mathbf{v}_3 is a vector c1v1+c2v2+c3v3c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 with real numbers c1c_1, c2c_2, c3c_3. The three vectors form a basis of three-dimensional space when every vector a\mathbf{a} can be written as such a combination in exactly one way, that is, for each a\mathbf{a} there is one and only one choice of the numbers c1c_1, c2c_2, c3c_3. The standard basis vectors are

i=⟨1,0,0⟩,j=⟨0,1,0⟩,k=⟨0,0,1⟩.\begin{gathered} \mathbf{i} = \langle 1, 0, 0 \rangle, \quad \mathbf{j} = \langle 0, 1, 0 \rangle, \\[0.4em] \mathbf{k} = \langle 0, 0, 1 \rangle. \end{gathered}

They are unit vectors along the positive axes, mutually perpendicular, and in this order they point along the positive xx, yy and zz axes of a right-handed system. They form a basis, because

c1i+c2j+c3k=⟨c1,c2,c3⟩,c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k} = \langle c_1, c_2, c_3 \rangle,

so the combination equals ⟨a1,a2,a3⟩\langle a_1, a_2, a_3 \rangle exactly when c1=a1c_1 = a_1, c2=a2c_2 = a_2 and c3=a3c_3 = a_3. The coefficients with respect to i\mathbf{i}, j\mathbf{j}, k\mathbf{k} are just the components.

In plain words Every step in space is some amount east, some amount north and some amount up. The basis vectors are the one-unit steps in those three directions.

Example The vector ⟨3,−2,5⟩\langle 3, -2, 5 \rangle is 3i−2j+5k3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}. Going the other way, 4i−k4\mathbf{i} - \mathbf{k} is ⟨4,0,−1⟩\langle 4, 0, -1 \rangle, with a zero j\mathbf{j} component that is easy to forget. Not every three vectors form a basis. The vectors i\mathbf{i}, j\mathbf{j} and i+j\mathbf{i} + \mathbf{j} all have zz component 00, so no combination of them equals k\mathbf{k}.

Forces

Definition A force is modeled as a vector whose magnitude is its strength, measured in newtons (N), and whose direction is the direction of the push or pull. When several forces act on the same object, their combined effect is the resultant force, their vector sum. The object is in equilibrium when the resultant is the zero vector.

In plain words Forces add like steps. If they cancel out completely, nothing accelerates.

Example Two forces F1=⟨2,−3,5⟩\mathbf{F}_1 = \langle 2, -3, 5 \rangle and F2=⟨1,4,−2⟩\mathbf{F}_2 = \langle 1, 4, -2 \rangle, in newtons, act on a ring. Their resultant is ⟨3,1,3⟩\langle 3, 1, 3 \rangle. A third force keeps the ring in equilibrium exactly when it cancels this sum, so

F3=−(F1+F2)=⟨−3,−1,−3⟩,\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2) = \langle -3, -1, -3 \rangle,

with magnitude 19≈4.36\sqrt{19} \approx 4.36 newtons.

Check yourself

  1. Find the center and radius of the sphere x2+y2+z2+2x−8z=8x^2 + y^2 + z^2 + 2x - 8z = 8, and say whether it meets the xyxy-plane.
Show the answer

Completing squares gives

(x+1)2+y2+(z−4)2=8+1+16=25,\begin{aligned} &(x + 1)^2 + y^2 + (z - 4)^2 \\[0.4em] &= 8 + 1 + 16 = 25, \end{aligned}

so the center is (−1,0,4)(-1, 0, 4) and the radius is 55. The center is 44 units above the xyxy-plane and 4<54 < 5, so the sphere meets that plane, in the circle (x+1)2+y2=9(x + 1)^2 + y^2 = 9.

  1. Find the unit vector in the direction from A(2,1,−1)A(2, 1, -1) to B(8,−1,2)B(8, -1, 2).
Show the answer

The displacement is AB→=⟨6,−2,3⟩\overrightarrow{AB} = \langle 6, -2, 3 \rangle, with length 36+4+9=7\sqrt{36 + 4 + 9} = 7. The unit vector is ⟨67,−27,37⟩\langle \tfrac{6}{7}, -\tfrac{2}{7}, \tfrac{3}{7} \rangle.

  1. A weight of 1010 newtons hangs at rest from two cables. One cable pulls with force ⟨3,2,4⟩\langle 3, 2, 4 \rangle newtons, where the zz axis points up. What force does the other cable exert?
Show the answer

Gravity is ⟨0,0,−10⟩\langle 0, 0, -10 \rangle. Equilibrium needs the three forces to add to 0\mathbf{0}, so the second cable pulls with

−(⟨3,2,4⟩+⟨0,0,−10⟩)=⟨−3,−2,6⟩ newtons.\begin{aligned} &-\big(\langle 3, 2, 4 \rangle + \langle 0, 0, -10 \rangle\big) \\[0.4em] &= \langle -3, -2, 6 \rangle \text{ newtons.} \end{aligned}

How this lesson connects

It also uses Real numbers, square roots, the Pythagorean theorem, and elementary algebra.