Unit 6, lesson 77

Algebraic Vectors and Component Form

How to describe a vector by its horizontal and vertical components, write it with the standard unit vectors i and j, and add, scale and measure vectors with plain arithmetic.

  • Theory part 153
  • Practice part 154
  • Vectors

An arrow drawn on a map is hard to calculate with. Replace it by two numbers, how far it goes east and how far it goes north, and nothing is lost, because its length and its direction can both be recovered from them. Writing a vector as a pair of numbers lets you add, stretch and measure arrows with ordinary arithmetic instead of rulers and protractors.

Key terms

Component form

Definition Fix a coordinate system in the plane. If a vector v\mathbf{v} is represented by the directed line segment from the initial point P(x1,y1)P(x_1, y_1) to the terminal point Q(x2,y2)Q(x_2, y_2), its component form is

v=⟨x2−x1,  y2−y1⟩.\mathbf{v} = \langle x_2 - x_1, \; y_2 - y_1 \rangle.

The numbers a=x2−x1a = x_2 - x_1 and b=y2−y1b = y_2 - y_1 are the horizontal and vertical components of v\mathbf{v}. Two directed line segments represent the same vector exactly when they give the same components, so the pair ⟨a,b⟩\langle a, b \rangle determines the vector completely. The representative that starts at the origin ends at the point (a,b)(a, b) and is called the position vector of that point. Angle brackets keep the vector ⟨a,b⟩\langle a, b \rangle apart from the point (a,b)(a, b), although some books use parentheses for both. The zero vector is 0=⟨0,0⟩\mathbf{0} = \langle 0, 0 \rangle. Vectors in space work the same way with three components ⟨a,b,c⟩\langle a, b, c \rangle.

In plain words Subtract start from finish, one coordinate at a time. The first number says how far the arrow runs to the right (negative means left) and the second how far it runs up (negative means down), wherever the arrow happens to start.

Example The segment from P(−1,2)P(-1, 2) to Q(3,5)Q(3, 5) gives

v=⟨3−(−1),  5−2⟩=⟨4,3⟩.\mathbf{v} = \langle 3 - (-1), \; 5 - 2 \rangle = \langle 4, 3 \rangle.

The segment from (2,−1)(2, -1) to (6,2)(6, 2) also gives ⟨4,3⟩\langle 4, 3 \rangle, so it is the same vector drawn in another place. Both segments have slope 3/43/4, the vertical component over the horizontal one. In the same way a line with slope mm runs in the direction of ⟨1,m⟩\langle 1, m \rangle.

Magnitude from components

Definition The magnitude, or length, of v=⟨a,b⟩\mathbf{v} = \langle a, b \rangle is

∥v∥=a2+b2.\lVert \mathbf{v} \rVert = \sqrt{a^2 + b^2}.

Some books write ∣v∣\lvert \mathbf{v} \rvert instead. The magnitude is never negative, and it is 00 only for the zero vector. In space the length of ⟨a,b,c⟩\langle a, b, c \rangle is a2+b2+c2\sqrt{a^2 + b^2 + c^2}.

In plain words The two components are the legs of a right triangle and the arrow is its hypotenuse, so the Pythagorean theorem gives the length.

Example For ⟨4,3⟩\langle 4, 3 \rangle the length is 16+9=5\sqrt{16 + 9} = 5. For ⟨−2,5⟩\langle -2, 5 \rangle it is 4+25=29≈5.39\sqrt{4 + 25} = \sqrt{29} \approx 5.39. Squaring removes the signs, so ⟨2,−5⟩\langle 2, -5 \rangle is just as long.

Direction angle and resolving a vector

Definition The direction angle of a nonzero vector v=⟨a,b⟩\mathbf{v} = \langle a, b \rangle is the angle θ\theta with 0∘≤θ<360∘0^\circ \le \theta < 360^\circ, measured counterclockwise from the positive xx-axis to the position vector of v\mathbf{v}. Then

v=⟨∥v∥cos⁡θ,  ∥v∥sin⁡θ⟩,\mathbf{v} = \langle \lVert \mathbf{v} \rVert \cos \theta, \; \lVert \mathbf{v} \rVert \sin \theta \rangle,

and tan⁡θ=b/a\tan \theta = b/a whenever a≠0a \ne 0. The tangent repeats every 180∘180^\circ, so the quadrant of θ\theta has to be read from the signs of aa and bb. The zero vector has no direction angle.

In plain words A length and a direction give the components through cosine and sine, and the components give back the length and the direction. Finding the components of an arrow is called resolving the vector.

Example A rope pulls on a sled with a force of 5050 newtons at 30∘30^\circ above the horizontal. Its components are

⟨50cos⁡30∘,  50sin⁡30∘⟩=⟨253,  25⟩≈⟨43.3,  25⟩,\begin{aligned} &\langle 50 \cos 30^\circ, \; 50 \sin 30^\circ \rangle \\[0.4em] &\quad = \langle 25\sqrt{3}, \; 25 \rangle \approx \langle 43.3, \; 25 \rangle, \end{aligned}

so the rope pulls about 43.343.3 newtons forward and 2525 newtons upward. Now go the other way. The vector ⟨−3,3⟩\langle -3, 3 \rangle has length 18=32\sqrt{18} = 3\sqrt{2} and tan⁡θ=−1\tan \theta = -1. Both 135∘135^\circ and 315∘315^\circ have tangent −1-1, and since the vector points left and up, its direction angle is 135∘135^\circ.

Adding vectors by components

Definition For u=⟨u1,u2⟩\mathbf{u} = \langle u_1, u_2 \rangle and v=⟨v1,v2⟩\mathbf{v} = \langle v_1, v_2 \rangle

u+v=⟨u1+v1,  u2+v2⟩andu−v=⟨u1−v1,  u2−v2⟩.\begin{gathered} \mathbf{u} + \mathbf{v} = \langle u_1 + v_1, \; u_2 + v_2 \rangle \\[0.4em] \text{and} \\[0.4em] \mathbf{u} - \mathbf{v} = \langle u_1 - v_1, \; u_2 - v_2 \rangle. \end{gathered}

This agrees with the head to tail rule and the parallelogram law for arrows. Addition is commutative and associative, v+0=v\mathbf{v} + \mathbf{0} = \mathbf{v}, and v+(−v)=0\mathbf{v} + (-\mathbf{v}) = \mathbf{0} where −v=⟨−v1,−v2⟩-\mathbf{v} = \langle -v_1, -v_2 \rangle.

In plain words Add the horizontal parts and add the vertical parts. Two trips combine into one trip whose eastward part is the sum of the two eastward parts, and the same for north.

Example Take east as the first component and north as the second. An airplane flies with velocity ⟨300,400⟩\langle 300, 400 \rangle kilometers per hour relative to the air, a speed of 500500, while the wind blows with velocity ⟨60,−80⟩\langle 60, -80 \rangle. Its velocity over the ground is

⟨300+60,  400+(−80)⟩=⟨360,320⟩,\begin{aligned} &\langle 300 + 60, \; 400 + (-80) \rangle \\[0.4em] &\quad = \langle 360, 320 \rangle, \end{aligned}

and its ground speed is

3602+3202≈481.7 kilometers per hour.\begin{aligned} &\sqrt{360^2 + 320^2} \\[0.4em] &\quad \approx 481.7 \text{ kilometers per hour.} \end{aligned}

Scalar multiplication

Definition For a real number cc, called a scalar, and a vector v=⟨a,b⟩\mathbf{v} = \langle a, b \rangle,

cv=⟨ca,  cb⟩.c\mathbf{v} = \langle ca, \; cb \rangle.

Its length is

∥cv∥=∣c∣ ∥v∥.\lVert c\mathbf{v} \rVert = \lvert c \rvert \, \lVert \mathbf{v} \rVert.

For v≠0\mathbf{v} \ne \mathbf{0} the vector cvc\mathbf{v} points the same way as v\mathbf{v} when c>0c > 0 and the opposite way when c<0c < 0, and 0v=00\mathbf{v} = \mathbf{0}. For all scalars cc and dd and all vectors u\mathbf{u} and v\mathbf{v}

c(u+v)=cu+cv,(c+d)v=cv+dv,c(dv)=(cd)v.\begin{gathered} c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}, \\[0.4em] (c + d)\mathbf{v} = c\mathbf{v} + d\mathbf{v}, \\[0.4em] c(d\mathbf{v}) = (cd)\mathbf{v}. \end{gathered}

In plain words Multiplying by a number stretches or shrinks the arrow, and a negative number also flips it around. Drawn from the same starting point, the new arrow lies on the same line as the old one.

Example Take v=⟨4,−2⟩\mathbf{v} = \langle 4, -2 \rangle, whose length is 20=25\sqrt{20} = 2\sqrt{5}. Then

3v=⟨12,−6⟩,−12v=⟨−2,1⟩.\begin{gathered} 3\mathbf{v} = \langle 12, -6 \rangle, \\[0.4em] -\tfrac{1}{2}\mathbf{v} = \langle -2, 1 \rangle. \end{gathered}

The first is three times as long and points the same way. The second has length 5\sqrt{5}, half of 252\sqrt{5}, and points the opposite way.

Unit vectors

Definition A unit vector is a vector of length 11. For every nonzero vector v\mathbf{v}

u=v∥v∥=1∥v∥ v\mathbf{u} = \frac{\mathbf{v}}{\lVert \mathbf{v} \rVert} = \frac{1}{\lVert \mathbf{v} \rVert}\,\mathbf{v}

is the unit vector in the direction of v\mathbf{v}. The zero vector has no direction, so it has no unit vector. The unit vectors in the plane are exactly the vectors ⟨cos⁡θ,sin⁡θ⟩\langle \cos \theta, \sin \theta \rangle, which end on the unit circle when drawn from the origin. The unit vector in the direction of a nonzero v\mathbf{v} is ⟨cos⁡θ,sin⁡θ⟩\langle \cos \theta, \sin \theta \rangle, where θ\theta is the direction angle of v\mathbf{v}.

In plain words Divide a vector by its own length to keep its direction and make its length exactly 11. Then multiply by whatever length you need.

Example The vector v=⟨−5,12⟩\mathbf{v} = \langle -5, 12 \rangle has length 25+144=13\sqrt{25 + 144} = 13, so the unit vector in its direction is ⟨−5/13, 12/13⟩\langle -5/13, \, 12/13 \rangle. A vector of length 2626 in the same direction is

26⟨−513,1213⟩=⟨−10,24⟩.26 \left\langle -\frac{5}{13}, \frac{12}{13} \right\rangle = \langle -10, 24 \rangle.

Standard unit vectors

Definition The standard unit vectors in the plane are

i=⟨1,0⟩andj=⟨0,1⟩.\mathbf{i} = \langle 1, 0 \rangle \quad \text{and} \quad \mathbf{j} = \langle 0, 1 \rangle.

Every vector in the plane can be written in exactly one way as a combination of them,

⟨a,b⟩=ai+bj,\langle a, b \rangle = a\mathbf{i} + b\mathbf{j},

and the coefficients are the components. In space a third vector k=⟨0,0,1⟩\mathbf{k} = \langle 0, 0, 1 \rangle joins them, i\mathbf{i} and j\mathbf{j} get a third component 00, and ⟨a,b,c⟩=ai+bj+ck\langle a, b, c \rangle = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}.

In plain words The vector i\mathbf{i} is one step right and j\mathbf{j} is one step up. Any vector is some number of steps right plus some number of steps up, and those two counts are its components.

Example The vector 3i−4j3\mathbf{i} - 4\mathbf{j} is ⟨3,−4⟩\langle 3, -4 \rangle. Arithmetic with i\mathbf{i} and j\mathbf{j} follows the usual rules of algebra,

(2i+j)−3(i−2j)=2i+j−3i+6j=−i+7j=⟨−1,7⟩.\begin{aligned} &(2\mathbf{i} + \mathbf{j}) - 3(\mathbf{i} - 2\mathbf{j}) \\[0.4em] &\quad = 2\mathbf{i} + \mathbf{j} - 3\mathbf{i} + 6\mathbf{j} \\[0.4em] &\quad = -\mathbf{i} + 7\mathbf{j} = \langle -1, 7 \rangle. \end{aligned}

Check yourself

  1. Find the component form and the magnitude of the vector from P(2,−3)P(2, -3) to Q(−4,5)Q(-4, 5).
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Subtract start from finish,

⟨−4−2,  5−(−3)⟩=⟨−6,8⟩.\langle -4 - 2, \; 5 - (-3) \rangle = \langle -6, 8 \rangle.

Its magnitude is 36+64=10\sqrt{36 + 64} = 10.

  1. With u=⟨3,−1⟩\mathbf{u} = \langle 3, -1 \rangle and v=⟨−2,4⟩\mathbf{v} = \langle -2, 4 \rangle, compute 2u−3v2\mathbf{u} - 3\mathbf{v} and write it with i\mathbf{i} and j\mathbf{j}.
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Scale first, then subtract,

2u−3v=⟨6,−2⟩−⟨−6,12⟩=⟨12,−14⟩=12i−14j.\begin{aligned} 2\mathbf{u} - 3\mathbf{v} &= \langle 6, -2 \rangle - \langle -6, 12 \rangle \\[0.4em] &= \langle 12, -14 \rangle \\[0.4em] &= 12\mathbf{i} - 14\mathbf{j}. \end{aligned}
  1. Find the component form of the vector with length 66 and direction angle 210∘210^\circ.
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Use the cosine and sine of the direction angle,

⟨6cos⁡210∘,  6sin⁡210∘⟩=⟨6⋅(−32),  6⋅(−12)⟩=⟨−33,  −3⟩.\begin{aligned} &\langle 6 \cos 210^\circ, \; 6 \sin 210^\circ \rangle \\[0.4em] &\quad = \left\langle 6 \cdot \left(-\frac{\sqrt{3}}{2}\right), \; 6 \cdot \left(-\frac{1}{2}\right) \right\rangle \\[0.8em] &\quad = \langle -3\sqrt{3}, \; -3 \rangle. \end{aligned}

Both components are negative, as they must be for an angle in the third quadrant.

How this lesson connects