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Vectors: adding, scaling, length, components

Geometry · level 8, lesson 4 of 6

A question to start

A swimmer heads straight across a river, swimming at 1 metre per second. The river flows downstream at 1 metre per second. Where does she end up, and how fast is she actually moving?

Not 2 metres per second, and not straight across. She moves at about 1.41 metres per second, diagonally, landing as far downstream as the river is wide. Adding a speed "across" to a speed "along" does not work like adding ordinary numbers, because each one has a direction. Quantities like these are called vectors, and they have their own arithmetic. It is the arithmetic of navigation, forces, winds and flight, and it turns many geometry problems into short calculations.

The idea

Scalars and vectors

A scalar is a quantity that is just a size: a temperature of 20°C, a mass of 3 kg, a distance of 5 km. A vector has a size and a direction: "5 km towards the north-east", a push of 10 newtons downwards, a wind of 30 km/h from the west. Speed is a scalar; velocity (speed in a direction) is a vector.

A vector is drawn as an arrow. Its length shows the size, which is called the magnitude, and its arrowhead shows the direction. Names are written with an arrow on top, \(\vec v\), or in bold, \(\mathbf{v}\). The arrow from point \(A\) to point \(B\) is written \(\overrightarrow{AB}\).

Where the arrow starts does not matter. Two arrows with the same length and the same direction are the same vector, wherever they are drawn. "Walk 3 km north" is the same instruction whether you start in Toronto or on Mars.

Components

On a grid, a vector is described by how far it goes across and how far up: its components. We write \(\vec v = (3, 4)\) for "3 across and 4 up" (some books write it as a column, or as \(3\hat\imath + 4\hat\jmath\), where \(\hat\imath = (1, 0)\) and \(\hat\jmath = (0, 1)\) are the unit steps along the axes).

From the components, Pythagoras gives the magnitude:

\[|\vec v| = \sqrt{v_1^2 + v_2^2} \qquad \text{so} \qquad |(3, 4)| = \sqrt{9 + 16} = 5\]

From a magnitude \(r\) and a direction angle \(\theta\) (measured anticlockwise from the positive \(x\)-axis), the unit circle of lesson 2 gives the components:

\[\vec v = (r\cos\theta,\ r\sin\theta)\]

because the unit circle point \((\cos\theta, \sin\theta)\) has length 1 and the right direction, and multiplying by \(r\) stretches it to length \(r\). To go back from components to an angle, use \(\tan\theta = \dfrac{v_2}{v_1}\), but check the quadrant from the signs, since \(\tan^{-1}\) only returns angles between \(-90°\) and \(90°\).

Adding vectors

To add \(\vec u + \vec v\), place them tip to tail: draw \(\vec u\), then start \(\vec v\) where \(\vec u\) ends. The sum, also called the resultant, is the arrow from the very start to the very end. It is the single trip that does the same as the two trips one after the other.

u = (3, 1)v = (1, 2)u + v = (4, 3)

In components, you simply add across to across and up to up:

\[(u_1, u_2) + (v_1, v_2) = (u_1 + v_1,\ u_2 + v_2)\]

Why? The total distance across is the across part of the first trip plus the across part of the second, and the same for up. The drawing also shows that \(\vec u + \vec v = \vec v + \vec u\): going \(\vec u\) then \(\vec v\) or \(\vec v\) then \(\vec u\) goes round the two sides of a parallelogram to the same corner. For this reason the sum is also called the parallelogram law.

The magnitudes do not simply add. \(|(3, 1)| + |(1, 2)| \approx 3.16 + 2.24 = 5.40\), but \(|(4, 3)| = 5\). The straight trip is never longer than the two-leg trip: \(|\vec u + \vec v| \le |\vec u| + |\vec v|\), the triangle inequality, with equality only when the two point the same way.

Scaling and subtracting

Multiplying a vector by a number \(k\) (a scalar) multiplies its length by \(|k|\), keeping the direction if \(k > 0\) and reversing it if \(k < 0\). In components, \(k(v_1, v_2) = (kv_1, kv_2)\). So \(2(3, 4) = (6, 8)\), twice as long; and \(-\vec v\) is the same length pointing the opposite way.

Two non-zero vectors are parallel exactly when one is a scalar multiple of the other.

Subtraction is adding the opposite: \(\vec u - \vec v = \vec u + (-\vec v)\). It has a picture worth remembering: if two arrows start at the same point, \(\vec u - \vec v\) is the arrow from the tip of \(\vec v\) to the tip of \(\vec u\) (because \(\vec v + (\vec u - \vec v) = \vec u\)).

A unit vector has length 1. Dividing any non-zero vector by its length gives the unit vector in its direction: \(\hat v = \dfrac{\vec v}{|\vec v|}\). For \((3, 4)\) that is \(\left(\tfrac35, \tfrac45\right)\).

Position vectors

Fix an origin \(O\). Every point \(A\) then has a position vector \(\vec a = \overrightarrow{OA}\), whose components are just \(A\)'s coordinates. The arrow from \(A\) to \(B\) is "back from \(A\) to \(O\), then out to \(B\)":

\[\overrightarrow{AB} = \vec b - \vec a\]

The midpoint of \(AB\) is halfway along: \(\vec a + \tfrac12(\vec b - \vec a) = \tfrac12(\vec a + \vec b)\). This is the midpoint formula from level 6, now with a reason behind it.

Three dimensions

In space, a vector has three components, \((v_1, v_2, v_3)\): across, along and up. Everything above works the same, and the length comes from Pythagoras used twice (the diagonal of a box):

\[|\vec v| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]

Worked examples

1. Where is the rover? A rover drives 300 m due east, then turns and drives 200 m in the direction 60° anticlockwise from east. How far is it from its start, and in which direction?

  1. First leg: \((300, 0)\). Second leg: \((200\cos 60°,\ 200\sin 60°) = (100,\ 173.2)\).
  2. Add: \((400,\ 173.2)\).
  3. Distance: \(\sqrt{400^2 + 173.2^2} \approx 435.9\) m.
  4. Direction: \(\tan\theta = \dfrac{173.2}{400}\), so \(\theta \approx 23.4°\) north of east.
  5. Check with the cosine law of level 7: the angle between the legs at the turn is \(180° - 60° = 120°\), and \(\sqrt{300^2 + 200^2 - 2 \times 300 \times 200\cos 120°} = \sqrt{190\,000} \approx 435.9\). They agree.

2. Crossing a river. A boat points straight across a river 120 m wide, moving at 4 m/s through the water. The current flows at 1.5 m/s downstream. Find the boat's real velocity, the crossing time and how far downstream it lands.

  1. Take "across" as the \(y\)-direction and "downstream" as \(x\): boat \((0, 4)\), current \((1.5, 0)\), sum \((1.5, 4)\).
  2. Speed \(\sqrt{1.5^2 + 4^2} = \sqrt{18.25} \approx 4.27\) m/s, at \(\tan^{-1}\tfrac{1.5}{4} \approx 20.6°\) off the straight-across line.
  3. Only the across part gets it over: time \(= \tfrac{120}{4} = 30\) s.
  4. In that time the current carries it \(1.5 \times 30 = 45\) m downstream.

3. Calculating with components. Let \(\vec u = (2, -1)\) and \(\vec v = (3, 4)\).

  1. \(\vec u + \vec v = (5, 3)\).
  2. \(3\vec u - 2\vec v = (6, -3) - (6, 8) = (0, -11)\).
  3. \(|\vec u - \vec v| = |(-1, -5)| = \sqrt{1 + 25} = \sqrt{26} \approx 5.10\).
  4. Unit vector along \(\vec v\): \(\tfrac15(3, 4) = (0.6,\ 0.8)\).

4. A proof with vectors. Prove that the diagonals of a parallelogram cut each other in half.

  1. Let the parallelogram be \(OABC\) with \(O\) at the origin, \(\overrightarrow{OA} = \vec a\) and \(\overrightarrow{OC} = \vec c\). Opposite sides of a parallelogram are equal and parallel, so \(\overrightarrow{AB} = \vec c\) and \(B\) is at \(\vec a + \vec c\).
  2. The midpoint of diagonal \(OB\) is \(\tfrac12(\vec a + \vec c)\).
  3. The midpoint of diagonal \(AC\) is \(\tfrac12(\vec a + \vec c)\), by the midpoint formula.
  4. They are the same point, so each diagonal passes through the other's midpoint. ∎

Compare this with a proof by congruent triangles: the vector proof needs no diagram-reading at all.

Try it

  1. Let \(\vec u = (-3, 5)\) and \(\vec v = (4, 1)\). Find \(\vec u + \vec v\), \(2\vec u - 3\vec v\) and \(|\vec v|\).
  2. \(A = (1, 2)\) and \(B = (7, -6)\). Find \(\overrightarrow{AB}\), its length, and the midpoint of \(AB\).
  3. A force of 50 newtons acts in the direction 210° (anticlockwise from the positive \(x\)-axis). Find its components.
  4. For which \(k\) is \((k, 6)\) parallel to \((2, 3)\)?
  5. A plane heads due north at 250 km/h through the air, and a wind blows towards the east at 60 km/h. Find its speed over the ground and its direction.
  6. Find the unit vector in the direction of \((5, -12)\).
  7. Prove with vectors that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long.
  8. Find the distance between the points \((1, 2, 3)\) and \((4, 6, 15)\) in space.
Answers
  1. \(\vec u + \vec v = (1, 6)\). \(2\vec u - 3\vec v = (-6, 10) - (12, 3) = (-18, 7)\). \(|\vec v| = \sqrt{16 + 1} = \sqrt{17} \approx 4.12\).
  2. \(\overrightarrow{AB} = (7 - 1, -6 - 2) = (6, -8)\), with length \(\sqrt{36 + 64} = 10\). Midpoint \(\tfrac12\big((1, 2) + (7, -6)\big) = (4, -2)\).
  3. \((50\cos 210°,\ 50\sin 210°) = \left(-25\sqrt3,\ -25\right) \approx (-43.3,\ -25)\). Both negative: it points down and to the left, in quadrant III, as 210° should.
  4. \((k, 6) = t(2, 3)\) needs \(3t = 6\), so \(t = 2\) and \(k = 2t = 4\).
  5. \((60, 250)\) in (east, north): speed \(\sqrt{60^2 + 250^2} = \sqrt{66\,100} \approx 257\) km/h, at \(\tan^{-1}\tfrac{60}{250} \approx 13.5°\) east of north.
  6. \(|(5, -12)| = 13\), so the unit vector is \(\left(\tfrac{5}{13}, -\tfrac{12}{13}\right)\).
  7. Put one corner at \(O\) and the others at \(\vec a\) and \(\vec b\). The midpoints of the two sides from \(O\) are \(\tfrac12\vec a\) and \(\tfrac12\vec b\). The segment between them is \(\tfrac12\vec b - \tfrac12\vec a = \tfrac12(\vec b - \vec a)\), which is half of the third side \(\vec b - \vec a\), in the same direction: parallel and half as long. ∎
  8. The vector between them is \((3, 4, 12)\), with length \(\sqrt{9 + 16 + 144} = \sqrt{169} = 13\).

Common mistake

Adding magnitudes. A 3 km walk and a 4 km walk at right angles take you 5 km from the start, not 7. Add the components, then find the length.

Losing the quadrant. For \((-3, -3)\), \(\tan^{-1}\tfrac{-3}{-3} = \tan^{-1} 1 = 45°\), but the vector points down and left, at 225°. Always check the signs of the components against the angle you report.

Where it leads