Precalculus · N.VM.B.4 · N.VM.B.4A

Adding and subtracting vectors

One trip is followed by a second trip in a different direction. Which single trip has the same effect as both, and how is it found from the two arrows or from their components?

You need

Component form of a vectorPrecalculus QuadrilateralsGrades 3 to 5

Unlocks

Forces and velocities as vectorsPrecalculus

Where you are 3 in Vector addition and scalar multiplication

Trigonometry7

Polar coordinates and vectors10

Vector addition and scalar multiplication3

Adding and subtracting vectors

Forces and velocities as vectors Scaling a vector

Why this exists

Kele, Tala and Zaid go geocaching: hunting for small containers hidden in a park, using a GPS that gives positions in meters east and north of the trailhead. Kele walks to one cache, then turns and walks to the next. Each walk is a vector, so much east and so much north. Two questions follow at once. Where is Kele now, as a single trip from the trailhead? And which trip takes Tala, who is standing at a different cache, straight to Kele?

The first question asks for the sum of two vectors and the second for their difference. Both have a picture (arrows placed end to end) and a calculation (components added or subtracted), and the two always agree.

Definition

The sum \mathbf{u} + \mathbf{v} is found by the triangle rule, also called the tip-to-tail rule: draw \mathbf{u}, place the tail of \mathbf{v} at the head of \mathbf{u}, and draw the arrow from the tail of \mathbf{u} to the head of \mathbf{v}. That arrow is the sum, also called the resultant of \mathbf{u} and \mathbf{v}. More vectors are added the same way, each tail at the previous head, and the resultant runs from the first tail to the last head.

In components, with \mathbf{u} = \langle a, b \rangle and \mathbf{v} = \langle c, d \rangle,

\mathbf{u} + \mathbf{v} = \langle a + c,\ b + d \rangle

and in \mathbf{i}, \mathbf{j} form the coefficients of \mathbf{i} add, and so do those of \mathbf{j}.

The difference is the sum with the opposite:

\mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}) = \langle a - c,\ b - d \rangle

Intuition

In Figure 1, \mathbf{u} = \langle 6, 1 \rangle and \mathbf{v} = \langle 1, 4 \rangle. Walking \mathbf{u} and then \mathbf{v} ends 6 + 1 = 7 to the right and 1 + 4 = 5 up, so the resultant is \langle 7, 5 \rangle. The horizontal changes add on their own and the vertical changes add on their own, which is why addition by components is the triangle rule written in numbers.

Figure 1The triangle rule along either pair of sides, and the parallelogram rule, all give the same diagonal \mathbf{u} + \mathbf{v} = \langle 7, 5 \rangle.

Walking \mathbf{v} first and then \mathbf{u} traces the dashed sides and ends at the same point. The two routes are the two halves of a parallelogram, and the sum is its diagonal from the shared tail. This is the parallelogram rule: draw \mathbf{u} and \mathbf{v} from one tail, complete the parallelogram, and the diagonal from that tail is \mathbf{u} + \mathbf{v}. The two rules are one rule, since the far sides of the parallelogram are copies of \mathbf{u} and \mathbf{v} (opposite sides of a parallelogram are parallel and equal).

Figure 2Two pictures of \mathbf{u} - \mathbf{v} = \langle 5, -3 \rangle. Left, as \mathbf{u} + (-\mathbf{v}), tip to tail. Right, with \mathbf{u} and \mathbf{v} drawn from one tail, as the arrow from the head of \mathbf{v} to the head of \mathbf{u}.

Subtraction has two pictures (Figure 2). As \mathbf{u} + (-\mathbf{v}), walk \mathbf{u} and then walk \mathbf{v} backward. Or draw both from one tail: \mathbf{u} - \mathbf{v} is the arrow from the head of \mathbf{v} to the head of \mathbf{u}, because it is the trip that, added to \mathbf{v}, gives \mathbf{u}. That second picture is also the other diagonal of the parallelogram in Figure 1. So a parallelogram with sides \mathbf{u} and \mathbf{v} has diagonals \mathbf{u} + \mathbf{v} and \mathbf{u} - \mathbf{v}.

Results

Properties. For all vectors \mathbf{u}, \mathbf{v}, \mathbf{w} and scalars k, m:

Property	Statement
Commutative	\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}
Associative	(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})
Zero vector	\mathbf{u} + \mathbf{0} = \mathbf{u}
Opposite	\mathbf{u} + (-\mathbf{u}) = \mathbf{0}
Distributive over vector sums	k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v}
Distributive over scalar sums	(k + m)\mathbf{v} = k\mathbf{v} + m\mathbf{v}

Head to head. If \mathbf{a} runs from a point O to A and \mathbf{b} runs from O to B, then \overrightarrow{AB} = \mathbf{b} - \mathbf{a}.

Triangle inequality. For all vectors \mathbf{u} and \mathbf{v},

\|\mathbf{u} + \mathbf{v}\| \le \|\mathbf{u}\| + \|\mathbf{v}\|

with equality exactly when one vector is a nonnegative multiple of the other. Also

\big|\, \|\mathbf{u}\| - \|\mathbf{v}\| \,\big| \le \|\mathbf{u} \pm \mathbf{v}\|

Why it is true

Triangle rule and components. Place \mathbf{u} with tail at (x, y); its head is at (x + a, y + b). Place \mathbf{v} there; its head is at (x + a + c, y + b + d). The arrow from the first tail to the last head has components \langle a + c, b + d \rangle.

Head to head. \mathbf{a} + (\mathbf{b} - \mathbf{a}) = \mathbf{b} by the associative, opposite and zero properties, so \mathbf{b} - \mathbf{a} is the trip that continues from A to B.

Properties. Each is the matching fact about numbers, used in both components at once. For the first distributive law,

\begin{aligned} k(\mathbf{u} + \mathbf{v}) &= \langle k(a + c),\ k(b + d) \rangle \\ &= \langle ka + kc,\ kb + kd \rangle = k\mathbf{u} + k\mathbf{v} \end{aligned}

by the distributive property of numbers in the middle step. Commutativity also has a picture: the two routes around the parallelogram in Figure 1.

Triangle inequality. If \mathbf{u} and \mathbf{v} are nonzero and not parallel, the arrows \mathbf{u}, \mathbf{v} and \mathbf{u} + \mathbf{v} drawn tip to tail are the sides of a true triangle, and the Triangle inequality for its side lengths says \|\mathbf{u} + \mathbf{v}\| < \|\mathbf{u}\| + \|\mathbf{v}\|. If they are parallel, all three arrows lie on one line: the lengths add when \mathbf{u} and \mathbf{v} point the same way (the equality case) and partly cancel when they point opposite ways. A zero vector gives equality at once. For the lower bound, \mathbf{u} = (\mathbf{u} + \mathbf{v}) + (-\mathbf{v}), so \|\mathbf{u}\| \le \|\mathbf{u} + \mathbf{v}\| + \|\mathbf{v}\|; swapping the roles of \mathbf{u} and \mathbf{v} gives the other half, and replacing \mathbf{v} by -\mathbf{v} gives the minus sign.

Worked examples

Example 1

Sums, differences and combinations

Let \mathbf{u} = \langle 8, -1 \rangle and \mathbf{v} = \langle 2, 9 \rangle. Find \mathbf{u} + \mathbf{v}, \mathbf{u} - \mathbf{v}, \mathbf{v} - \mathbf{u}, and the vector 3\mathbf{u} - 2\mathbf{v} with its magnitude.

Solution

Add or subtract matching components:

\begin{aligned} \mathbf{u} + \mathbf{v} &= \langle 8 + 2,\ -1 + 9 \rangle = \langle 10, 8 \rangle \\ \mathbf{u} - \mathbf{v} &= \langle 8 - 2,\ -1 - 9 \rangle = \langle 6, -10 \rangle \\ \mathbf{v} - \mathbf{u} &= \langle 2 - 8,\ 9 - (-1) \rangle = \langle -6, 10 \rangle \end{aligned}

\mathbf{v} - \mathbf{u} = -(\mathbf{u} - \mathbf{v}): subtraction is not commutative, and reversing the order reverses the arrow. For the combination, scale first, then subtract: 3\mathbf{u} = \langle 24, -3 \rangle and 2\mathbf{v} = \langle 4, 18 \rangle, so

\begin{aligned} 3\mathbf{u} - 2\mathbf{v} &= \langle 20, -21 \rangle \\ \|3\mathbf{u} - 2\mathbf{v}\| &= \sqrt{400 + 441} = \sqrt{841} = 29 \end{aligned}

Example 2

The resultant of two walks

Kele walks 600 meters with direction angle 30^\circ, then 400 meters with direction angle 120^\circ. Find the single trip from start to finish: its distance and direction angle.

Solution

Write each walk in components using \langle r\cos\theta, r\sin\theta \rangle:

\begin{aligned} \mathbf{a} &= \langle 600\cos 30^\circ,\ 600\sin 30^\circ \rangle \\ &= \langle 300\sqrt{3},\ 300 \rangle \\ \mathbf{b} &= \langle 400\cos 120^\circ,\ 400\sin 120^\circ \rangle \\ &= \langle -200,\ 200\sqrt{3} \rangle \end{aligned}

The resultant \mathbf{r} = \mathbf{a} + \mathbf{b} is

\begin{aligned} \mathbf{r} &= \langle 300\sqrt{3} - 200,\ 300 + 200\sqrt{3} \rangle \\ &\approx \langle 319.6, 646.4 \rangle \end{aligned}

Squaring and adding, the \displaystyle \sqrt{3} terms cancel:

\begin{aligned} \|\mathbf{r}\|^2 &= 310{,}000 - 120{,}000\sqrt{3} \\ &\quad + 210{,}000 + 120{,}000\sqrt{3} \\ &= 520{,}000 \end{aligned}

so \displaystyle \|\mathbf{r}\| = \sqrt{520{,}000} = 200\sqrt{13} \approx 721.1 meters. Both components are positive, so \mathbf{r} points into quadrant I and its direction angle is \displaystyle \theta = \tan^{-1}\frac{646.4}{319.6} \approx 63.7^\circ (see Inverse trigonometric ratios).

Check. The walks differ by 90^\circ, so they meet at a right angle and \displaystyle \sqrt{600^2 + 400^2} = \sqrt{520{,}000}. The angle at the start is \displaystyle \tan^{-1}\frac{400}{600} \approx 33.7^\circ past 30^\circ, again 63.7^\circ. Kele walked 1000 meters but ended only about 721 meters away.

Example 3

From one cache to another

Tala’s cache is at \mathbf{a} = \langle 700, 200 \rangle and Zaid’s at \mathbf{b} = \langle 300, 600 \rangle, both measured in meters from the trailhead. Find the trip from Tala to Zaid, its length and its direction angle.

Solution

Both vectors share the trailhead as tail, so the trip from Tala’s head to Zaid’s head is \mathbf{b} - \mathbf{a} = \langle 300 - 700,\ 600 - 200 \rangle = \langle -400, 400 \rangle. Its length is \displaystyle \sqrt{160{,}000 + 160{,}000} = 400\sqrt{2} \approx 565.7 meters. Equal sizes with x negative and y positive put it at 135^\circ, straight northwest.

Check. \mathbf{a} + (\mathbf{b} - \mathbf{a}) is \langle 700 - 400,\ 200 + 400 \rangle, which is \langle 300, 600 \rangle = \mathbf{b}.

Example 4

Solving for an unknown vector

Find \mathbf{x} with 3\mathbf{x} + \langle 2, -7 \rangle = \langle -10, 8 \rangle.

Solution

Add the opposite \langle -2, 7 \rangle to both sides. On the left the associative property regroups the sum as 3\mathbf{x} + (\langle 2, -7 \rangle + \langle -2, 7 \rangle), the opposite property turns the bracket into \mathbf{0}, and the zero property leaves 3\mathbf{x}:

3\mathbf{x} = \langle -10 - 2,\ 8 + 7 \rangle = \langle -12, 15 \rangle

Multiply both sides by \displaystyle \frac{1}{3}; since \displaystyle \frac{1}{3}(3\mathbf{x}) = (\frac{1}{3} \cdot 3)\mathbf{x} = \mathbf{x}, this gives \mathbf{x} = \langle -4, 5 \rangle.

Check. 3\langle -4, 5 \rangle + \langle 2, -7 \rangle is \langle -12 + 2,\ 15 - 7 \rangle, which is \langle -10, 8 \rangle.

Example 5

The fourth corner and the diagonals

Parallelogram ABCD has A(-2, 1), B(5, 3) and D(0, 6). Find C, and write both diagonals as vectors.

Solution

Let \mathbf{u} = \overrightarrow{AB} = \langle 7, 2 \rangle and \mathbf{v} = \overrightarrow{AD} = \langle 2, 5 \rangle. Opposite sides are equal vectors, so \overrightarrow{BC} = \mathbf{v} and C is B moved by \mathbf{v}: C = (5 + 2, 3 + 5) = (7, 8). By the parallelogram rule,

\begin{aligned} \overrightarrow{AC} &= \mathbf{u} + \mathbf{v} = \langle 9, 7 \rangle \\ \overrightarrow{DB} &= \mathbf{u} - \mathbf{v} = \langle 5, -3 \rangle \end{aligned}

the second by head to head: from the head of \mathbf{v} (at D) to the head of \mathbf{u} (at B).

Check. C - A = \langle 9, 7 \rangle and B - D = \langle 5, -3 \rangle, read straight from the coordinates.

Example 6

Testing the triangle inequality

Let \mathbf{u} = \langle 8, 6 \rangle and \mathbf{v} = \langle 12, -9 \rangle. Compare \|\mathbf{u} + \mathbf{v}\| and \|\mathbf{u} - \mathbf{v}\| with the bounds. Then do the same for \mathbf{u} and \mathbf{w} = 2\mathbf{u}.

Solution

\displaystyle \|\mathbf{u}\| = \sqrt{64 + 36} = 10 and \displaystyle \|\mathbf{v}\| = \sqrt{144 + 81} = 15, so both lengths must lie between 15 - 10 = 5 and 10 + 15 = 25. Indeed \mathbf{u} + \mathbf{v} = \langle 20, -3 \rangle has length \displaystyle \sqrt{409} \approx 20.2, and \mathbf{u} - \mathbf{v} = \langle -4, 15 \rangle has length \displaystyle \sqrt{241} \approx 15.5.

With \mathbf{w} = \langle 16, 12 \rangle, the vectors point the same way, and \mathbf{u} + \mathbf{w} = \langle 24, 18 \rangle has length 30 = 10 + 20: the upper bound is reached. And \mathbf{u} - \mathbf{w} = -\mathbf{u} has length 10 = 20 - 10, the lower bound.

Example 7

Why the diagonals bisect each other

Prove with vectors that the diagonals of every parallelogram bisect each other.

Solution

Let ABCD be a parallelogram with \mathbf{u} = \overrightarrow{AB} and \mathbf{v} = \overrightarrow{AD}, so \overrightarrow{AC} = \mathbf{u} + \mathbf{v} and \overrightarrow{BD} = \mathbf{v} - \mathbf{u}. Let M be the midpoint of \overline{AC} and N that of \overline{BD}. Half of a vector reaches the midpoint, so

\begin{aligned} \overrightarrow{AM} &= \tfrac{1}{2}(\mathbf{u} + \mathbf{v}) \\ &= \tfrac{1}{2}\mathbf{u} + \tfrac{1}{2}\mathbf{v} \\ \overrightarrow{AN} &= \mathbf{u} + \tfrac{1}{2}(\mathbf{v} - \mathbf{u}) \\ &= \left(1 - \tfrac{1}{2}\right)\mathbf{u} + \tfrac{1}{2}\mathbf{v} \end{aligned}

The first line uses k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v}; the second uses it again and then (k + m)\mathbf{u} = k\mathbf{u} + m\mathbf{u} to combine \displaystyle \mathbf{u} - \frac{1}{2}\mathbf{u}. Both equal \displaystyle \frac{1}{2}\mathbf{u} + \frac{1}{2}\mathbf{v}, so the same trip from A reaches M and N: they are one point, on both diagonals, halfway along each.

Check. In Example 5, the midpoint of \overline{AC} is (2.5, 4.5) and the midpoint of \overline{BD} is \displaystyle \left(\frac{5 + 0}{2}, \frac{3 + 6}{2}\right) = (2.5, 4.5).

Boundaries

Same kind, same size. Only vectors of the same kind add: two displacements in meters, not a displacement and a velocity, and two vectors with two components each, not one with two and one with three.

Lengths do not add. \|\mathbf{u} + \mathbf{v}\| = \|\mathbf{u}\| + \|\mathbf{v}\| only when one is a nonnegative multiple of the other; otherwise the resultant is shorter, as in Example 2.

Subtraction is neither commutative nor associative. \mathbf{v} - \mathbf{u} = -(\mathbf{u} - \mathbf{v}), and (\mathbf{u} - \mathbf{v}) - \mathbf{w} \ne \mathbf{u} - (\mathbf{v} - \mathbf{w}) unless \mathbf{w} = \mathbf{0}.

The parallelogram needs two directions. When \mathbf{u} and \mathbf{v} are parallel the parallelogram collapses onto a line; the triangle rule and components still work.

Where this goes next

With sums and scalar multiples together, every k\mathbf{u} + m\mathbf{v} is a linear combination, and a\mathbf{i} + b\mathbf{j} is now a true sum. The next lesson, forces and velocities, uses the resultant for physical quantities: a plane’s own velocity plus the wind’s, or several pushes on one object. The dot product later measures the angle between two vectors.

Exercises

Adding and subtracting by components

Let \mathbf{p} = \langle -3, 7 \rangle and \mathbf{q} = \langle 6, -1 \rangle.

Find \mathbf{p} + \mathbf{q}.
Find \mathbf{p} - \mathbf{q}.
Find \mathbf{q} - \mathbf{p}.
Find 2\mathbf{p} + 5\mathbf{q}.
Find (3\mathbf{i} - 8\mathbf{j}) - (-2\mathbf{i} + 4\mathbf{j}) and its magnitude.
Find \|\mathbf{p} + \mathbf{q}\| and \|\mathbf{p}\| + \|\mathbf{q}\|, each to two decimal places.

Arrows between points

\overrightarrow{PQ} = \langle 3, -9 \rangle and \overrightarrow{QR} = \langle -7, 6 \rangle. Find \overrightarrow{PR} and its length.
\overrightarrow{RS} = \langle 5, 4 \rangle and \overrightarrow{RT} = \langle -2, 6 \rangle. Find \overrightarrow{ST}.
Parallelogram ABCD has A(-4, 1), B(1, -1) and C(4, 3). Find D.
For the parallelogram in Exercise 9, write \overrightarrow{AC} and \overrightarrow{BD} in terms of \mathbf{u} = \overrightarrow{AB} and \mathbf{v} = \overrightarrow{BC}, and find their lengths.

Resultants

Kele walks 500 meters east, then 1200 meters north. Find the distance and direction angle of the resultant, the angle to one decimal place.
\mathbf{u} has magnitude 10 and direction angle 60^\circ; \mathbf{v} has magnitude 10 and direction angle 180^\circ. Find \mathbf{u} + \mathbf{v}, its magnitude and its direction angle.
Zaid walks 300 meters with direction angle 45^\circ, then 400 meters with direction angle 135^\circ. Find the distance and direction angle of the resultant, the angle to one decimal place.
Tala walks \langle 1.2, 3.5 \rangle kilometers, then \langle 2.8, -0.9 \rangle kilometers. Which trip brings her back to the start, and how long is it, to two decimal places?

Equations and properties

Solve \mathbf{x} + \langle 3, -8 \rangle = \langle -1, 2 \rangle.
Solve 2\mathbf{x} - \langle 6, 1 \rangle = 4\mathbf{x} + \langle -4, 9 \rangle.
Simplify 3(\mathbf{u} - 2\mathbf{v}) - 2(\mathbf{u} - 4\mathbf{v}).
Find scalars a and b with a\langle 1, 3 \rangle + b\langle 2, -1 \rangle = \langle 8, 3 \rangle.

The triangle inequality

\|\mathbf{u}\| = 7 and \|\mathbf{v}\| = 3. Find the largest and smallest possible values of \|\mathbf{u} + \mathbf{v}\|, and say when each happens.
Can a geocaching loop of three straight legs, 2, 3 and 6 kilometers long, end where it started? Use vectors to explain.

Reasoning

Explain why, with \mathbf{u} and \mathbf{v} drawn from the same tail, \mathbf{u} - \mathbf{v} runs from the head of \mathbf{v} to the head of \mathbf{u}, not the other way.
Prove k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v} for \mathbf{u} = \langle a, b \rangle, \mathbf{v} = \langle c, d \rangle, naming the property of numbers you use.
Prove (k + m)\mathbf{v} = k\mathbf{v} + m\mathbf{v} by components.
Prove the parallelogram law \|\mathbf{u} + \mathbf{v}\|^2 + \|\mathbf{u} - \mathbf{v}\|^2 = 2\|\mathbf{u}\|^2 + 2\|\mathbf{v}\|^2, and say what it means for the diagonals and sides of a parallelogram.
Zaid claims \|\mathbf{u} - \mathbf{v}\| = \|\mathbf{u}\| - \|\mathbf{v}\|. Give a counterexample and state the correct relation.

Answers

\langle 3, 6 \rangle. \langle -3 + 6,\ 7 - 1 \rangle.
\langle -9, 8 \rangle. \langle -3 - 6,\ 7 - (-1) \rangle.
\langle 9, -8 \rangle, the opposite of Exercise 2.
\langle 24, 9 \rangle. 2\mathbf{p} = \langle -6, 14 \rangle and 5\mathbf{q} = \langle 30, -5 \rangle.
5\mathbf{i} - 12\mathbf{j}, magnitude 13. 3 - (-2) = 5, -8 - 4 = -12, and \displaystyle \sqrt{25 + 144} = 13.
\displaystyle \|\mathbf{p} + \mathbf{q}\| = \sqrt{45} = 3\sqrt{5} \approx 6.71, and \displaystyle \|\mathbf{p}\| + \|\mathbf{q}\| = \sqrt{58} + \sqrt{37} \approx 13.70. The resultant is shorter, as the triangle inequality requires.
\overrightarrow{PR} = \langle -4, -3 \rangle, length 5. \overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{QR}, and \displaystyle \sqrt{16 + 9} = 5.
\langle -7, 2 \rangle. Same tail R, so \overrightarrow{ST} = \overrightarrow{RT} - \overrightarrow{RS}.
D(-1, 5). \overrightarrow{AD} = \overrightarrow{BC} = \langle 3, 4 \rangle, added to A.
\overrightarrow{AC} = \mathbf{u} + \mathbf{v} = \langle 8, 2 \rangle, length \displaystyle 2\sqrt{17}; \overrightarrow{BD} = \mathbf{v} - \mathbf{u} = \langle -2, 6 \rangle, length \displaystyle 2\sqrt{10}. Here \mathbf{u} = \langle 5, -2 \rangle, \mathbf{v} = \langle 3, 4 \rangle, and \overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = -\mathbf{u} + \mathbf{v}.
1300 meters at about 67.4^\circ. \langle 500, 1200 \rangle has length \displaystyle \sqrt{250{,}000 + 1{,}440{,}000} = 1300, and \displaystyle \tan^{-1}\frac{1200}{500} \approx 67.4^\circ.
\displaystyle \langle -5, 5\sqrt{3} \rangle, magnitude 10, direction angle 120^\circ. \displaystyle \mathbf{u} = \langle 5, 5\sqrt{3} \rangle, \mathbf{v} = \langle -10, 0 \rangle; \displaystyle \sqrt{25 + 75} = 10, and \displaystyle \cos 120^\circ = -\frac{1}{2}, \displaystyle \sin 120^\circ = \frac{\sqrt{3}}{2} match \displaystyle \frac{-5}{10} and \displaystyle \frac{5\sqrt{3}}{10}.
500 meters at about 98.1^\circ. The legs are \displaystyle \langle 150\sqrt{2}, 150\sqrt{2} \rangle and \displaystyle \langle -200\sqrt{2}, 200\sqrt{2} \rangle, sum \displaystyle \langle -50\sqrt{2}, 350\sqrt{2} \rangle, length \displaystyle \sqrt{5000 + 245{,}000} = 500. It points into quadrant II, so the angle is 180^\circ - \tan^{-1} 7 \approx 98.1^\circ.
\langle -4, -2.6 \rangle kilometers, about 4.77 kilometers. The first two legs sum to \langle 4, 2.6 \rangle, and the closing leg is its opposite; \displaystyle \sqrt{16 + 6.76} = \sqrt{22.76}.
\mathbf{x} = \langle -4, 10 \rangle. Add \langle -3, 8 \rangle to both sides.
\mathbf{x} = \langle -1, -5 \rangle. Subtracting 4\mathbf{x} and adding \langle 6, 1 \rangle gives -2\mathbf{x} = \langle 2, 10 \rangle.
\mathbf{u} + 2\mathbf{v}. Distribute: 3\mathbf{u} - 6\mathbf{v} - 2\mathbf{u} + 8\mathbf{v}, then combine with (k + m)\mathbf{v} = k\mathbf{v} + m\mathbf{v}.
a = 2, b = 3. Components give a + 2b = 8 and 3a - b = 3; then b = 3a - 3, so 7a - 6 = 8.
Largest 10, when \mathbf{v} points the same way as \mathbf{u}; smallest 4, when it points the opposite way. The triangle inequality gives |7 - 3| \le \|\mathbf{u} + \mathbf{v}\| \le 7 + 3.
No. A closed loop has \mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}, so \mathbf{c} = -(\mathbf{a} + \mathbf{b}) and \|\mathbf{c}\| \le \|\mathbf{a}\| + \|\mathbf{b}\|. The longest leg, 6, exceeds 2 + 3 = 5.
The arrow from the head of \mathbf{v} to the head of \mathbf{u} is the trip \mathbf{w} with \mathbf{v} + \mathbf{w} = \mathbf{u} (tip to tail). Adding -\mathbf{v} to both sides gives \mathbf{w} = \mathbf{u} - \mathbf{v}. The reverse arrow is \mathbf{v} - \mathbf{u} = -(\mathbf{u} - \mathbf{v}).
k(\mathbf{u} + \mathbf{v}) = \langle k(a + c), k(b + d) \rangle, which by the distributive property of numbers is \langle ka + kc, kb + kd \rangle. That splits as \langle ka, kb \rangle + \langle kc, kd \rangle, which is k\mathbf{u} + k\mathbf{v}.
With \mathbf{v} = \langle a, b \rangle, (k + m)\mathbf{v} = \langle (k + m)a, (k + m)b \rangle. By the distributive property of numbers this is \langle ka + ma, kb + mb \rangle, which splits as \langle ka, kb \rangle + \langle ma, mb \rangle = k\mathbf{v} + m\mathbf{v}.
With \mathbf{u} = \langle a, b \rangle, \mathbf{v} = \langle c, d \rangle, the left side is (a + c)^2 + (b + d)^2 + (a - c)^2 + (b - d)^2. The cross terms 2ac, 2bd cancel against -2ac, -2bd, leaving 2a^2 + 2b^2 + 2c^2 + 2d^2. Since the diagonals are \mathbf{u} + \mathbf{v} and \mathbf{u} - \mathbf{v}: the squares of the two diagonals add up to the squares of all four sides.
With \mathbf{u} = \langle 1, 0 \rangle and \mathbf{v} = \langle 0, 1 \rangle, \displaystyle \|\mathbf{u} - \mathbf{v}\| = \sqrt{2} but \|\mathbf{u}\| - \|\mathbf{v}\| = 0. The correct relation is the inequality \|\mathbf{u} - \mathbf{v}\| \ge \big|\, \|\mathbf{u}\| - \|\mathbf{v}\| \,\big|, with equality only when the vectors point the same way (or one is zero).