Unit 6, lesson 76

Geometric Vectors and Vector Addition

What a vector is as an arrow with a magnitude and a direction, why sliding it changes nothing, and how to add vectors head to tail or with the parallelogram law.

  • Theory part 151
  • Practice part 152
  • Vectors

Walk 3 blocks east and then 4 blocks north, and you end up 5 blocks from where you started, not 7. Distances do not simply add, because where you end up also depends on which way each leg points. A vector carries both pieces of information, a size and a direction, which is exactly what displacements, velocities and forces need.

Key terms

Directed line segment

Definition The directed line segment from a point PP to a point QQ is the line segment between them together with a direction, from its initial point PP to its terminal point QQ. It is written PQ→\overrightarrow{PQ} and drawn as an arrow with its head at QQ. Its length is the distance between PP and QQ. The segments PQ→\overrightarrow{PQ} and QP→\overrightarrow{QP} use the same two points but are different, because they point opposite ways.

In plain words It is an arrow. The tail is where you start and the head is where you end up.

Example A drone flies from a warehouse WW to a house HH that is 800 meters away. Its trip is the directed line segment WH→\overrightarrow{WH}, with initial point WW, terminal point HH and length 800 meters. The trip back is HW→\overrightarrow{HW}, with the same length but the opposite direction.

Vector

Definition Two directed line segments are equivalent if they have the same length and point in the same direction, and all segments of length zero are equivalent to each other. A vector is a quantity that has a magnitude and a direction, with one exception, the zero vector 0\mathbf{0}, which has magnitude 00 and no direction. Geometrically a vector is represented by a directed line segment, and equivalent segments represent the same vector, so a vector may be drawn starting at any point. The zero vector is represented by a segment from a point to itself. Vectors are written in bold, such as v\mathbf{v}, or as PQ→\overrightarrow{PQ} for the vector that the segment from PP to QQ represents. A quantity with a size but no direction, such as a mass or a temperature, is called a scalar.

In plain words A vector is an arrow you may slide anywhere, as long as you do not turn it or stretch it. Where it sits does not matter, only how long it is and which way it points.

Example The segment from P(1,2)P(1, 2) to Q(4,6)Q(4, 6) and the segment from R(−2,0)R(-2, 0) to S(1,4)S(1, 4) both move 3 units right and 4 units up. They have the same length and the same direction, so they represent the same vector, and PQ→=RS→\overrightarrow{PQ} = \overrightarrow{RS}. Sliding the first arrow 3 units left and 2 units down puts it exactly on top of the second.

Magnitude

Definition The magnitude of a vector v\mathbf{v}, written ∥v∥\lVert \mathbf{v} \rVert (some books write ∣v∣\lvert \mathbf{v} \rvert), is the length of any directed line segment that represents it. Equivalent segments have equal lengths, so the choice of segment does not matter. The magnitude is never negative, and it is 00 only for the zero vector. If v=PQ→\mathbf{v} = \overrightarrow{PQ} with P=(x1,y1)P = (x_1, y_1) and Q=(x2,y2)Q = (x_2, y_2), the distance formula gives

∥v∥=(x2−x1)2+(y2−y1)2.\lVert \mathbf{v} \rVert = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

In plain words The magnitude is the length of the arrow. For a displacement it is the straight-line distance from start to finish, and for a velocity it is the speed.

Example For P(1,2)P(1, 2) and Q(4,6)Q(4, 6),

∥PQ→∥=(4−1)2+(6−2)2=9+16=5.\begin{aligned} \lVert \overrightarrow{PQ} \rVert &= \sqrt{(4 - 1)^2 + (6 - 2)^2} \\[0.4em] &= \sqrt{9 + 16} = 5. \end{aligned}

If the units are kilometers, this displacement has magnitude 5 kilometers, even if the road actually traveled from PP to QQ was longer.

Direction

Definition Two nonzero vectors have the same direction when their segments are parallel, or lie on one line, and point the same way. They have opposite directions when their segments are parallel but point opposite ways. In the coordinate plane the direction of a nonzero vector v\mathbf{v} is usually given by its direction angle θ\theta. When v\mathbf{v} is drawn with its initial point at the origin, θ\theta is the angle from the positive xx-axis to v\mathbf{v}, measured counterclockwise, with 0∘≤θ<360∘0^\circ \le \theta < 360^\circ. The zero vector has no direction.

In plain words Direction says which way the arrow points. The direction angle measures it just like an angle in standard position.

Example Drawn from the origin, the vector PQ→\overrightarrow{PQ} from (1,2)(1, 2) to (4,6)(4, 6) ends at (3,4)(3, 4) in the first quadrant, so

tan⁡θ=43,θ=arctan⁡43≈53.13∘.\begin{gathered} \tan \theta = \frac{4}{3}, \\[0.8em] \theta = \arctan \frac{4}{3} \approx 53.13^\circ. \end{gathered}

The vector QP→\overrightarrow{QP} ends at (−3,−4)(-3, -4) in the third quadrant. Its reference angle is also about 53.13∘53.13^\circ, so its direction angle is about 180∘+53.13∘=233.13∘180^\circ + 53.13^\circ = 233.13^\circ. Opposite directions always differ by 180∘180^\circ.

Adding vectors head to tail

Definition To add vectors u\mathbf{u} and v\mathbf{v} head to tail, also called tip to tail, draw u\mathbf{u}, then draw v\mathbf{v} with its initial point at the terminal point of u\mathbf{u}. The sum u+v\mathbf{u} + \mathbf{v}, also called the resultant, is the vector from the initial point of u\mathbf{u} to the terminal point of v\mathbf{v}. In terms of points,

PQ→+QR→=PR→.\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}.

Vector addition is associative, so a longer sum is drawn by chaining all the arrows head to tail. Magnitudes satisfy the triangle inequality ∥u+v∥≤∥u∥+∥v∥\lVert \mathbf{u} + \mathbf{v} \rVert \le \lVert \mathbf{u} \rVert + \lVert \mathbf{v} \rVert, with equality exactly when one of the vectors is zero or both have the same direction.

In plain words Do one move, then the other. The sum is the single move that takes you from the very start to the very end.

Example A hiker walks 5 kilometers east and then 12 kilometers north. The two legs meet at a right angle, so the resultant displacement has magnitude

52+122=169=13 kilometers.\begin{aligned} \sqrt{5^2 + 12^2} &= \sqrt{169} \\[0.4em] &= 13 \text{ kilometers.} \end{aligned}

Its direction, measured from east toward north, is arctan⁡125≈67.38∘\arctan \frac{12}{5} \approx 67.38^\circ. The hiker walked 17 kilometers but ended up only 13 kilometers from the start.

Parallelogram law

Definition Let u=PQ→\mathbf{u} = \overrightarrow{PQ} and v=PS→\mathbf{v} = \overrightarrow{PS} be drawn from the same initial point PP, not along one line, and complete the parallelogram PQRSPQRS. Then the diagonal from PP is the sum,

u+v=PR→.\mathbf{u} + \mathbf{v} = \overrightarrow{PR}.

The opposite sides QR→\overrightarrow{QR} and SR→\overrightarrow{SR} represent v\mathbf{v} and u\mathbf{u}, so going around either side gives the same diagonal, and u+v=v+u\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}. If θ\theta is the angle between u\mathbf{u} and v\mathbf{v} at PP, with 0∘<θ<180∘0^\circ < \theta < 180^\circ, then the law of cosines in triangle PQRPQR, whose angle at QQ is 180∘−θ180^\circ - \theta, gives

∥u+v∥2=∥u∥2+∥v∥2+2∥u∥∥v∥cos⁡θ.\begin{aligned} \lVert \mathbf{u} + \mathbf{v} \rVert^2 &= \lVert \mathbf{u} \rVert^2 + \lVert \mathbf{v} \rVert^2 \\[0.4em] &\quad + 2 \lVert \mathbf{u} \rVert \lVert \mathbf{v} \rVert \cos \theta. \end{aligned}

The same formula holds when the vectors lie along one line, with θ=0∘\theta = 0^\circ or θ=180∘\theta = 180^\circ.

In plain words Draw both arrows from one point and build the parallelogram they span. The diagonal from that point is the sum. It agrees with the head to tail rule, because each far side of the parallelogram is a copy of one of the arrows.

Example Two ropes pull on a crate with forces of 30 newtons and 40 newtons, at an angle of 60∘60^\circ to each other. The resultant force has magnitude

302+402+2⋅30⋅40cos⁡60∘=900+1600+1200=3700≈60.8 newtons.\begin{aligned} &\sqrt{30^2 + 40^2 + 2 \cdot 30 \cdot 40 \cos 60^\circ} \\[0.4em] &\quad = \sqrt{900 + 1600 + 1200} \\[0.4em] &\quad = \sqrt{3700} \approx 60.8 \text{ newtons.} \end{aligned}

In the triangle formed by the 30 newton force, the 40 newton force drawn from its head and the resultant, the angle opposite the resultant is 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ. The law of sines then gives sin⁡α=40sin⁡120∘3700≈0.569\sin \alpha = \frac{40 \sin 120^\circ}{\sqrt{3700}} \approx 0.569, so the resultant makes an angle α≈34.7∘\alpha \approx 34.7^\circ with the 30 newton force.

Opposite vectors and subtraction

Definition The opposite of a vector v\mathbf{v}, written −v-\mathbf{v}, has the same magnitude as v\mathbf{v} and the opposite direction, and −0=0-\mathbf{0} = \mathbf{0}. If v=PQ→\mathbf{v} = \overrightarrow{PQ} then −v=QP→-\mathbf{v} = \overrightarrow{QP}, and v+(−v)=PP→=0\mathbf{v} + (-\mathbf{v}) = \overrightarrow{PP} = \mathbf{0}. Subtraction is defined by

u−v=u+(−v).\mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}).

When u\mathbf{u} and v\mathbf{v} are drawn from the same initial point, u−v\mathbf{u} - \mathbf{v} is the vector from the terminal point of v\mathbf{v} to the terminal point of u\mathbf{u}, the other diagonal of the parallelogram.

In plain words Subtracting a vector means adding its reverse. Drawn from one point, the difference is the arrow from the tip of v\mathbf{v} to the tip of u\mathbf{u}, because v+(u−v)=u\mathbf{v} + (\mathbf{u} - \mathbf{v}) = \mathbf{u}.

Example From a campsite CC, hiker AA is 8 kilometers east and hiker BB is 6 kilometers north. With a=CA→\mathbf{a} = \overrightarrow{CA} and b=CB→\mathbf{b} = \overrightarrow{CB}, the vector from BB to AA is

BA→=BC→+CA→=a−b.\overrightarrow{BA} = \overrightarrow{BC} + \overrightarrow{CA} = \mathbf{a} - \mathbf{b}.

It covers 8 kilometers east and 6 kilometers south, so its magnitude is 64+36=10\sqrt{64 + 36} = 10 kilometers and it points arctan⁡68≈36.87∘\arctan \frac{6}{8} \approx 36.87^\circ south of east.

Check yourself

  1. A boat heads due north at 15 kilometers per hour relative to the water, while the current flows due east at 8 kilometers per hour. Find the boat’s actual speed and direction.
Show the answer

The two velocities are perpendicular, so the resultant has magnitude

152+82=289=17 kilometers per hour.\begin{aligned} &\sqrt{15^2 + 8^2} = \sqrt{289} \\[0.4em] &\quad = 17 \text{ kilometers per hour.} \end{aligned}

It points east of north by the angle arctan⁡815≈28.1∘\arctan \frac{8}{15} \approx 28.1^\circ.

  1. Do the segments from A(2,1)A(2, 1) to B(5,3)B(5, 3) and from C(−1,4)C(-1, 4) to D(2,6)D(2, 6) represent the same vector? Find its magnitude, and say how DC→\overrightarrow{DC} is related to it.
Show the answer

Both segments move 3 units right and 2 units up, so they have the same length and direction, and AB→=CD→\overrightarrow{AB} = \overrightarrow{CD}. The magnitude is 9+4=13≈3.61\sqrt{9 + 4} = \sqrt{13} \approx 3.61. The segment from DD to CC has the same length and the opposite direction, so DC→=−AB→\overrightarrow{DC} = -\overrightarrow{AB}.

  1. Two forces of 8 newtons each act on the same point, with an angle of 120∘120^\circ between them. Find the magnitude of the resultant.
Show the answer

The parallelogram law gives

∥u+v∥2=64+64+2⋅8⋅8cos⁡120∘=128−64=64,\begin{aligned} &\lVert \mathbf{u} + \mathbf{v} \rVert^2 \\[0.4em] &\quad = 64 + 64 + 2 \cdot 8 \cdot 8 \cos 120^\circ \\[0.4em] &\quad = 128 - 64 = 64, \end{aligned}

so the resultant is 8 newtons. The diagonal splits the parallelogram, which is a rhombus here, into two equilateral triangles, so the resultant makes a 60∘60^\circ angle with each force.

How this lesson connects