Vector arithmetics (HL)

This discusses addition and subtraction of vectors, which are reminiscent of addition, subtraction of complex numbers.

Contents

Additive inverse

This is a fancy name for 0v\bm 0 - \bm v. In general

AB=BA\overrightarrow{AB} = -\overrightarrow{BA}

which makes AB\overrightarrow{AB} and BA\overrightarrow{BA} additive inverses of each other.

Furthermore, in components

(x1,y1,z1)=(x1,y1,z1)-(x_1, y_1, z_1) = (-x_1, -y_1, -z_1)

Properties

First, recall the definition of displacement vectors from before.

AB=OBOA\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}
Displacement vector AB is OB - OA.
final minus initial: AB is OB minus OA

which is the vector equivalent to the change = final  initial\text{change = final } - \text{ initial} from chemistry and physics.

This says adding up the displacements result in the total (net) displacement.

AB+BC=AC\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}

Tip: In triangle ABC{ABC}, AB+BC+CA=0\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0, meaning if you traverse a triangle, you end up where you started.

The associative, commutative, distributive and identity properties hold:

(u+v)+w=u+(v+w)(\bm u + \bm v) + \bm w = \bm u + (\bm v + \bm w)
u+v=v+u\bm u + \bm v = \bm v + \bm u
k(u+v)=ku+kvk(\bm u + \bm v) = k\bm u + k\bm v
v+0=v\bm v + \bm 0 = \bm v

The sum or difference of vectors is a displacement vector (as opposed to a position vector).

Add vectors algebraically

(x1,y1,z1)+(x2,y2,z2)=(x1+x2,y1+y2,z1+z2)(x_1, y_1, z_1)+(x_2, y_2, z_2) = (x_1 + x_2,\, y_1 + y_2,\, z_1 + z_2)

Add vectors geometrically

Treat all vectors to be added as displacement vectors. Translate vectors add needed, without rotating or stretching them.

For AB\overrightarrow{AB}, ie a vector from AA to BB, AA is the tail while BB is the head.

The following are different methods that produce the same answers for addition, or subtraction.

parallelogram addition

using parallelogram, sum is from where tails meet to where heads meet
Parallelogram addition
  1. Translate the vectors so that their tails coincide
  2. The vectors form a vertex of a parallelogram.
  3. Sum is from this tail to opposite vertex of the parallelogram.

triangle subtraction

using triangle, difference b minus a is from head of a to head of b
Triangle subtraction

For subtracting ba\bm b - \bm a

  1. Translate the vectors so that their tails coincide
  2. Difference is from head of a\bm a to head of b\bm b

head-to-tail addition/subtraction

connecting each vector's head to next vector's tail, the sum is tail of first vector to head of last vector
Head-to-tail addition
  1. Translate the vectors so one head aligns with the other’s tail
  2. Sum is from first tail to second head.

Note for subtraction, add the additive inverse of the second vector.

for subtraction, reverse each vector being subtracted, then add using head-to-tail addition above
b - a = b + (-a)

Scalar multiplication

k(l,m,n)=(kl,km,kn)k(l, m, n) = (kl, km, kn)

(3,4,5)(3, -4, 5) and (9,12,15)(-9, 12, -15) are scalar multiples of each other, because

93=124=155=3\frac{-9}{3} = \frac{12}{-4} = \frac{-15}{5} = -3

If the ratios had yielded different numbers, then the two vectors would not have been scalar multiples.

Magnitude

The magnitude of a vector v=(x1,y1,z1)\bm v = (x_1, y_1, z_1) is

v=x12+y12+z12\lvert\bm v\rvert = \sqrt{x_1^2 + y_1^2 + z_1^2}

You may be tempted to say u+v\lvert \bm u + \bm v \rvert is u+v\lvert\bm u\rvert + \lvert\bm v\rvert but this is very very wrong. This is supposed to be finding the third side in a triangle. Easiest way is to add the vectors before finding the magnitude.

Unit vector

The unit vector is

v^=vv=1x12+y12+z12(x1,y1,z1)\bm{\hat v} = \frac{\bm v}{\lvert\bm v\rvert} = \frac{1}{\sqrt{x_1^2 + y_1^2 + z_1^2}}(x_1, y_1, z_1)

i\bm i, j\bm j, k\bm k are the unit vectors for the xx, yy, and zz-axes. The axes are perpendicular to each other and intersect at the origin (0,0,0)(0, 0, 0).

Example: Rescale the displacement vector a=(3,2,6)\bm a = (-3, 2, 6) so that it is in the same direction but with a magnitude of 2121.


The magnitude of a\bm a is

a=(3)2+22+62=9+4+36=49=7\begin{align*} \left|\bm a\right| &= \sqrt{(-3)^2 + 2^2 + 6^2} \\ &= \sqrt{9 + 4 + 36} \\ &= \sqrt{49} \\ &= 7 \end{align*}

What we want is 2121 times the unit vector, such that

21aa=21aa=217a=3a=(9,6,18)\begin{align*} 21\frac{\bm a}{\left|\bm a\right|} &= \frac{21}{\left|\bm a\right|}\cdot\bm a \\ &= \frac{21}{7} \cdot \bm a\\ &= 3 \bm a \\ &= (-9, 6, 18) \qed \end{align*}