1.1 - Scalars and Vectors

Definitions

Scalars have magnitude only, so just a size.

Vectors have both magnitude and direction.

Examples: distance and speed are scalars. Position, displacement, velocity, and acceleration are vectors.

Here’s why the difference matters. If you walk 3 m east and then 3 m west, you’ve covered a distance of 6 m, but your displacement is zero, because you ended up where you started. A speedometer reading of 25 m/s is a speed. Add a direction (25 m/s north) and it becomes a velocity.

In one dimension, the sign is the direction. Pick which way is positive, and anything pointing the other way gets a negative sign. If east is

+x

, then

v_x = -4.0 \text{ m/s}

means 4.0 m/s to the west. A negative velocity doesn’t mean “slow.” It only tells you the direction.

Drawing Vectors

Vectors are drawn as arrows, where: - The head is the tip of the arrow. - The tail is the starting point.

When you draw vectors, keep their relative sizes accurate. If you’re working on a grid, use the grid spacing to hold the right proportions. This matters a lot for force vectors, where the relative sizes have to be correct.

All of these vector ideas carry over into three dimensions:

x

y

, and

z

Unit Vector Notation

Vectors can be written using unit vector notation. A unit vector has a length of 1, and the number in front of it scales the vector to the magnitude you want.

The unit vectors are: -

\hat{i}

for the x-direction -

\hat{j}

for the y-direction -

\hat{k}

for the z-direction

Example of Vector Notation

\vec{r} = 5\hat{i} + 6\hat{j} - 2\hat{k}

, that means: - x-component: 5 - y-component: 6 - z-component: -2

Alternative Notation

As a shorthand, you can also list a vector’s components inside angle brackets:

\vec{r} = \langle 5, 6, -2 \rangle

That said, the

\hat{i}

\hat{j}

\hat{k}

notation is the standard for AP Physics C.

Magnitude and Direction

Unit vector notation isn’t the only way to describe a vector. You can also give its magnitude (its length, which is never negative) and its direction (usually an angle measured counterclockwise from the

+x

axis). You need to be able to switch between the two forms.

Components to Magnitude and Direction

The components

A_x

and

A_y

are the legs of a right triangle, and the vector is the hypotenuse. Using the right-triangle relationships on the equation sheet (

a^2 + b^2 = c^2

and

\tan\theta = \frac{a}{b}

|\vec{A}| = \sqrt{A_x^2 + A_y^2} \qquad \tan\theta = \frac{A_y}{A_x}

In three dimensions, the magnitude just picks up one more term:

|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}

Watch the quadrant. Your calculator’s

\tan^{-1}

only gives angles between

-90^\circ

and

+90^\circ

, so it can’t tell up-and-left from down-and-right. Always sketch the vector first. For

\vec{D} = -3.0\hat{i} + 4.0\hat{j}

|\vec{D}| = \sqrt{(-3.0)^2 + (4.0)^2} = 5.0 \text{ m}

The calculator gives

\tan^{-1}\left(\frac{4.0}{-3.0}\right) = -53^\circ

, but

\vec{D}

points up and to the left (the second quadrant). Add

180^\circ

: the real direction is

127^\circ

from the

+x

axis.

Magnitude and Direction to Components

If a vector has magnitude

A

and makes an angle

\theta

with the

+x

axis, its components are:

A_x = A\cos\theta \qquad A_y = A\sin\theta

Example: a ball leaves a launcher at 12.0 m/s,

30.0^\circ

above the

+x

axis.

v_x = (12.0)\cos 30.0^\circ = 10.4 \text{ m/s} \qquad v_y = (12.0)\sin 30.0^\circ = 6.00 \text{ m/s}

\vec{v} = 10.4\hat{i} + 6.00\hat{j}

m/s. Check:

\sqrt{10.4^2 + 6.00^2} = 12.0

m/s.

Position Vectors

A position vector starts at a point and points in a specific direction. We write it as

\vec{r}

More precisely, the position vector always starts at the origin and ends at the object’s location. That’s why its components are just the object’s coordinates: an object at

(6.0, -8.0)

m has

\vec{r} = 6.0\hat{i} - 8.0\hat{j}

The unit vector

\hat{r}

(read “r-hat”) points in the same direction as

\vec{r}

but has a length of exactly 1 and no units. You get it by dividing the vector by its own magnitude:

\hat{r} = \frac{\vec{r}}{|\vec{r}|}

For the object above,

|\vec{r}| = \sqrt{6.0^2 + (-8.0)^2} = 10

m, so

\hat{r} = \frac{6.0\hat{i} - 8.0\hat{j}}{10} = 0.60\hat{i} - 0.80\hat{j}

Check:

\sqrt{0.60^2 + 0.80^2} = 1

. You’ll see

\hat{r}

again in force laws that point along the line between two objects, like gravity (Topic 2.6).

Vector Addition

To add vectors, add their matching components:

\vec{c} = \vec{a} + \vec{b} = (a_x + b_x)\hat{i} + (a_y + b_y)\hat{j} + (a_z + b_z)\hat{k}

Graphically, adding vectors means placing them tip to tail: draw

\vec{A}

, start

\vec{B}

at the tip of

\vec{A}

, and the resultant

\vec{C} = \vec{A} + \vec{B}

runs from the tail of

\vec{A}

to the tip of

\vec{B}

. Adding components does the same thing with numbers.

Example: a robot moves

\vec{A} = 4.0\hat{i} + 1.0\hat{j}

m, then

\vec{B} = -2.0\hat{i} + 3.0\hat{j}

m. Find its total displacement in unit vector notation and as a magnitude and direction.

\vec{C} = (4.0 + (-2.0))\hat{i} + (1.0 + 3.0)\hat{j} = 2.0\hat{i} + 4.0\hat{j} \text{ m}

|\vec{C}| = \sqrt{2.0^2 + 4.0^2} = 4.5 \text{ m} \qquad \theta = \tan^{-1}\left(\frac{4.0}{2.0}\right) = 63^\circ

The robot ends up 4.5 m from where it started,

63^\circ

above the

+x

axis. That’s less than

|\vec{A}| + |\vec{B}|

(about 7.7 m), because the two moves partly cancel in the

x

direction.

Vector Subtraction

Subtracting is the same as adding the inverse. To subtract a vector, flip the signs of all its components:

\vec{c} = \vec{a} - \vec{b} = \vec{a} + (-\vec{b}) = (a_x - b_x)\hat{i} + (a_y - b_y)\hat{j} + (a_z - b_z)\hat{k}

If a component is positive, it becomes negative, and if it’s negative, it becomes positive.

Example Problems

Given Vectors

\vec{a} = 4\hat{i} + 3\hat{j} - 2\hat{k}

\vec{b} = -3\hat{i} + 2\hat{j} + 1\hat{k}

Part A: Find $\vec{a} + \vec{b}$

\vec{c} = \vec{a} + \vec{b}

Separate by components:

\vec{c} = (a_x + b_x)\hat{i} + (a_y + b_y)\hat{j} + (a_z + b_z)\hat{k}

\vec{c} = (4 + (-3))\hat{i} + (3 + 2)\hat{j} + (-2 + 1)\hat{k}

\vec{c} = 1\hat{i} + 5\hat{j} - 1\hat{k}

Or just:

\vec{c} = \hat{i} + 5\hat{j} - \hat{k}

(when the coefficient is 1, people often leave it off).

Part B: Find $\vec{a} - \vec{b}$

\vec{c} = \vec{a} - \vec{b}

This is the same as

\vec{a} + (-\vec{b})

, so flip all the signs of

\vec{b}

\vec{c} = (4 - (-3))\hat{i} + (3 - 2)\hat{j} + (-2 - 1)\hat{k}

\vec{c} = (4 + 3)\hat{i} + (3 - 2)\hat{j} + (-2 - 1)\hat{k}

\vec{c} = 7\hat{i} + 1\hat{j} - 3\hat{k}

Or:

\vec{c} = 7\hat{i} + \hat{j} - 3\hat{k}

Part C: Find $\vec{b} - \vec{a}$

\vec{c} = \vec{b} - \vec{a}