Vectors

### Vectors

The concept of a vector is one of the most important in physics. Vectors represent all kinds of important quantities like velocity, acceleration, momentum and force.

A vector is a directed line segment. It is drawn as an arrow (right), and has only two important aspects, its length and its direction.

The length or magnitude of a vector represents the size of a physical quantity, like a force or speed (speed is the magnitude of velocity).

The direction of a vector (where the arrow points) is the direction of action of the physical quantity. For example, it might be the direction of an applied pushing force.

The only two features of a vector that are important are length (which captures the magnitude or size of the quantity) and direction. As long as length and direction are preserved, a vector can be moved anywhere in a coordinate system, purely for convenience.

### Vector translation

Above we noted that the only important things about a vectors are its length and direction. It doesn't matter where it is located on the plane (or in space). In fact, we are free to move vectors where ever we'd like, just for the sake of convenience, without changing their meaning.

Vectors A, B and C on the left could be force vectors, velocity vectors, acceleration vectors ... you name it. Often it's convenient to translate vectors to the origin.

Play the animation to translate all three vectors to the origin. None of the meaning of the vectors is altered in any way.

Vectors are of little use unless we can add them. Below are the two principal methods of vector addition.

#### Method 1: Tip to tail

The easiest way to add vectors is the tip-to-tail (or head-to-tail) method. Remember that the only two important things about vectors are length and direction. Therefore we can move any vector to any location in the plane as we like and, as long as we don't change the length or direction, it remains the same vector

Adding by the tip-to-tail method means to move one vector so that its tail lies on the tip of the first vector. The resultant vector, A+B - the sum of the two - is simply the new vector drawn from the origin of the first vector to the arrow of the second.

Run this animation to add vectors A and B to find the resultant vector A+B. The resultant vector (just a vector sum) is often labeled "R".

Any number of vectors can be added in this way by just chaining them together, arrow of the current vector to origin of the next, and drawing in the vector R.

#### Method 2: Parallelogram

The parallelogram method of vector addition is shown on below. Notice that it's the same thing as tip-to-tail, but in this case vector B is moved down so that the vectors are tail-to-tail or origin-to-origin.

The resultant vector is the diagonal of the parallelogram formed by two copies of each vector.

Notice that the head-to-tail method of vector addition is embedded within the parallelogram method (twice). Look for it.

#### Which is better?

Once in a while the parallelogram method is more convenient, but the head-to-tail method is usually the place to start. It's much more useful for adding more than two vectors, and it's the method we'll almost always use to program computers to do vector addition.

### Adding more than two vectors

Using the head-to-tail method to add more than two vectors is very easy, graphically or numerically (we'll look at numerical addition below). One thing that often confuses students is whether, during the addition process, vectors can cross. They can.

Below is an example of a four-vector addition done in head-to-tail fashion, in which vectors cross. It's fine, no problem.

Adding more than two vectors with the parallelogram method is more cumbersome. You'd have to do the first addition to come up with a resultant vector, then add that to the next vector to find the new resultant, and so on.

Another thing that the head to tail method shows is that vector addition is commutative. Look at the six ways to add three vectors in the panel below. All six yield, graphically, the same resultant vector R.

X

### Commutative property

The commutative property applies to both addition and multiplication. It says that order of pure addition or pure multiplication doesn't matter:

#### a · b   =   b · a

Notice that regardless of the order of addition of vectors, the resultant or sum vector (in magenta) is the same – vector addition is commutative.

### A thought experiment ...

Here's a thought experiment. Imagine a bowling ball suspended in the middle of a room by 1000 bungee cords, each attached somewhere on a wall, the ceiling or the floor. Each cord, of course, is exerting a force on the ball, and each can be represented by a vector - only in three dimensions instead of two - that's legal. The length of the vector is proportional to the strength of the force and the direction is the direction of the pull.

Now the ball isn't moving, so there can't be any net force on it, otherwise it would be accelerating.

The net force is the sum of all force vectors. That means that the vector sum of all 1000 vectors has to be zero. So if we join all of those vectors together, tip-to-tail on a 3-D grid, the tip of the last vector will touch the tail of the first - no matter what order of addition, and the resultant vector will have a length of zero. That's cool.

The four scenarios below might help you to visualize vector addition. In all cases, the airplane has a forward velocity vector of 150 Km/h (the plane is traveling – or trying to travel – at 150 Km/h in the forward direction). As we would expect, a tailwind adds to the overall velocity of an airplane and a headwind subtracts. You can experience this by flying to the west coast and back on a commercial plane.

The westbound trip can take 6 hours or more while the eastbound flight often takes substantially less time.

A cross wind (coming in at 90˚ to the direction of travel) shifts the course of the plane away from the wind. In the case of wind blowing at an odd angle to the forward velocity vector, use the law of cosines to solve for the resultant velocity vector.

### Resolving vectors into components

We know that we can add two vectors to get another. Very often (quite often, really) we need to find two convenient vectors that add to a vector of interest. This is called resolving a vector into components.

The example at right shows a vector drawn on a Cartesian plane. By drawing the dashed lines from the tip of the vector to the axes, at right angles, we come up with the vectors Vx and Vy, which sum to vector V.

As we'll see, resolving V into two vectors that lie along our coordinate axes will be a big help in solving some problems.

### Using trigonometry to resolve vectors

When the length of a vector and the angle it makes with either axis is known, we can use trigonometry to find the lengths fo the components along each axis.

Using either of the right triangles drawn by the dashed line (left), it's easy to see that

$$sin(\theta) = \frac{V_y}{V} \phantom{000} \text{and} \phantom{000} cos(\theta) = \frac{V_x}{V},$$

so we can derive the relationships shown.

It might be helpful at this point to review your trig. and the special triangles, the 45-45-90 triangle and the 30-60-90 triangle, for the angles of which we can find convenient exact solutions for the sine and cosine functions.

### Making the coordinate system work for you

There are many problems in physics and other fields where changing from one coordinate system to a more convenient one makes solving a problem simpler and more intuitive. One example is a mass on an inclined plane. The problem is shown in panel 1. The only force that makes the ball roll down the ramp is the force of gravity, Fg. The trouble is that Fg is at an odd angle to the ramp. But we can impose a different coordinate system upon the problem, a more convenient one in which the x-axis is classed along the ramp and the y-axis perpendicular to it (panel 2). In this case, we can resolve the Fg vector into its Fgx and Fgy components (panel 3).

Panels 4, 5 & 6 show ramps of increasing steepness. Notice that the component of Fg pointing down the ramp in each increases with the steepness. The larger force accounts for the fact that the ball will roll faster as the ramp steepens.

By the way, no matter what kind of crazy coordinate system you impose on this problem, the ball will still roll down the ramp like it always did. Nature couldn't give one whit about your coordinate system. Your choice of coordinate system is made to make your mathematical modeling of the situation easier, or even just possible.

### Practice problems

1. Sketch head-to-tail additions of the following vector pairs. Roll over or tap the images to see the solution.

1. A boat travels at 5 knots across a river with a current of 1 knot (a knot is one nautical mile per hour). If the intended direction of the boat is due north (the river runs east-west), find the actual course (in degrees from north) and the speed of the boat as it moves.

2. Use trigonometry to find the x- and y-components of these vectors (roll over for solutions):

1. The figure below shows an airplane flying due west (270˚ compass bearing — see the compass "rose" on the right). Find the actual speed and direction of the plane if it encounters a wind of (a) 30 mi./h from the north (0˚), (b) 30 mi./h from the southwest (225˚).

Solution

Part (a)

\begin{align} R &= \sqrt{165^2 + 30^2} \\ &= 167.7 \frac{mi}{h} \; \text{ (a little faster)} \\ \\ \theta &= tan^{-1} \left( \frac{130}{165} \right) = 10.3˚ \\ \\ \text{course } &= 270˚ - 10.3˚ = 259.7˚ \end{align}

Part (b) Begin with the law of cosines

\begin{align} R^2 &= 30^2 + 165^2 - 2(30)(165) cos(45˚) \\ &= 30^2 + 165^2 - 2(30)(165) \frac{\sqrt{2}}{2} \\ \\ &= 23,175 \\ \\ R &= \sqrt{23175} = 152.2 \; \frac{mi}{h} \end{align}

The plane is a little slower into a headwind. For the angle, we use the law of sines:

$$\frac{sin(\theta)}{30} = \frac{sin(45)}{152.2}$$

\begin{align} \theta &= sin^{-1}\left( \frac{30}{152.2} sin(45˚) \right) \\ \\ &= 8.01˚ \\ \\ \text{course } &= 270˚ + 8˚ = 278˚ \end{align}

### Special triangles can help

Often we work with angles that are special fractions of a circle. It's a good thing to memorize the dimensions of two special triangles with hypotenuses of lenght 1, the 30-60-90 triangle and the 45-45-90 triangle. I can't overemphasize how much knowing these comes in handy later in math and physics.

Doing math with vectors graphically really helps with learning, but in order to really do any serious computations with vectors, we need to learn how to manipulate them numerically.

We'll begin by noticing that there are two ways we can describe a vector on the plane:

• specify the beginning and end points of the vector
• translate the vector to the origin [where one endpoint is (0, 0)], then the vector is specified only be giving the endpoint.

Take the vectors A and B (black) in the figure below. It's easy to translate them to the origin by simply subtracting the coordinates of the beginning of the vector (the dot) from both coordinates. For example, to translate vector A to the origin, we subtract the coordinate (-4, -2) from each coordinate. Then we get

• start: (-4,-2) - (-4,-2) = ((-4+4), (-2+2)) = (0, 0)
• end: (-2,4) - (-4,-2) = ((-2+4),(4+2)) = (2, 6)

This result is shown in the magenta vector A. You should work your way through the same translation for the B vector.

We can think of vector translation another way, too. Suppose we specify the two ends of some vector A (below) with two vectors from the origin, v1 & v2. Now if we subtract v1 from v2 (or add the negative of v1 to v2) we will get vector A moved to the origin. Check out the graph below to see how it works. You should be familiar with both ways of translating vectors to the origin mathematically.

We can also flip vectors by 180˚ numerically. This just involves flipping the signs of all coordinates. For example, in the graph below, vector A is flipped around to -A by transforming its beginning coordinate (-1, 3) to (1, -3) and its end coordinate (6, 6) to (-6, -6).

These vectors can be translated to the orign graphically or numerically to show that they lie on the same line but point in opposite directions. They also have the same magnitude because the Pythagorean theorem, which we use to calculate vector lengths, depends only on the squares of the coordinates, therefore sign is unimportant.

Finally, we can add vectors numerically. To do so we simply add coordinates. For a 2-dimensional vector, that means

(x1, y1) + (x2, y2) = (x1+x2, y1+y2)

Vectors can be added and then translated to the origin, or translated to the origin first, then added.

Vector translation and addition of vectors are commutative (can be done in either order).

#### Vector translation

A vector with beginning at $(x_1, \, y_1)$ and end at $(x_2, \, y_2)$ can be translated to the origin by subtracting $(x_1, \, y_1)$ from each coordinate.

Vectors $\vec{A} = (a, \, b)$ and $\vec{B} = (c, \, d)$ are added by adding respective coordinates: $\vec{A} + \vec{B} = (a+c, \, b+d)$

Vector subtraction is just the same as adding the negative of a vector.

The negative of vector $(a, \, b)$ is $(-a, \, -b).$

### Example 1

Add vectors   $\vec{A} = (-3, \, 3)$   and   $\vec{B} = (1, \, 6).$ These vectors both originate from   $(0, 0).$

Solution: Adding the vectors is straightforward:

$$\vec{A} + \vec{B} = (-3 + 1, \, 3 + 6) = (-2, \, 9)$$

All of these vectors originate from $(0, 0).$ The graphical result is shown. A parallelogram is drawn in to help you see the addition.

### Example 2

Add vector A with endpoints (1, 1) & (-2, 3) to vector B with endpoints (-2, -2) & (3, 4). Translate the result to the origin.

Solution: The coordinates of the endpoints of our sum vector are

start = (1, 1) + (-2, -2) = (-1, -1)

end = (-2, 3) + (3, 4) = (1, 7)

So the coordinates of the two ends of our vector are (-1, -1) and (1, 7). We can translate this vector to the origin by subracting (-1, -1) from each to get:

start = (-1, -1) - (-1, -1) = (0, 0)

end = (1, 7) - (-1, -1) = (2, 8)

Now we can show that we get the same result by first translating each vector to the origin and then adding the resulting vectors:

A = (-2, 3) - (1, 1) = (-3, 2), where the start coordinate just turns into the origin: (1, 1) - (1, 1) = (0, 0).

and

B = (3, 4) - (-2, -2) = (5, 6), where the start coordinate is again (-2, -2) - (-2, -2) = (0, 0)

The sum of our two translated vectors is

(-3, 2) + (5, 6) = (2, 8), just what we got on the first try. Easy peasy.

Below: vectors A & B add to the magenta vector.

In the graph below, A & B are translated to the origin. The sum vector is easier to visualize in that view. The original vectors are in green for comparison.

### Practice problems

Calculate the sum of the vectors and translate to the origin if necessary. Problems 1-3 give vectors from the origin (0, 0). Problems 4-6 include the start and endpoints (respectively) of the vectors.

 1 (-2, -5) & (5, 4) Solution \begin{align} \bar{v}_1 + \bar{v}_2 &= (-2 + 5, -5 + 4) \\ &= \bf (3, -1) \end{align} 2 (2, -3) & (-2, -6) Solution \begin{align} \bar{v}_1 + \bar{v}_2 &= (2 + (-2), -3 + (-6)) \\ &= (2 - 2, -3 - 6) \\ &= \bf (0, -9) \end{align} 3 (7, 1) & (1, 5) Solution \begin{align} \bar{v}_1 + \bar{v}_2 &= (7 + 1, 1 + 5) \\ &= \bf (8, 6) \end{align}

 4 (-2, -2) to (3, 4) and (-1, 7) to (2, 2) Solution First translate the vectors to the origin. We need to "zero-out" the origin of each vector, by adding (2, 2) to the start and end of the first, and (1, -7) to the start and end of the second: \begin{align} \bar{v}_1 &= (2 + 3, 2 + 4) = (5, 6) \\ \bar{v}_2 &= (2 + 1, 2 - 7) = (3, -5) \end{align} Now add the translated vectors: \begin{align} \bar{v}_1 + \bar{v}_2 &= (5 + 3, 6 - 5) \\ &= \bf (8, 1) \end{align} 5 (4, 5) to (-4, -4) and (-2, -4) to (2, 6) Solution First translate the vectors to the origin. We need to "zero-out" the origin of each vector, by adding (-4, -5) to the start and end of the first, and (2, 4) to the start and end of the second: \begin{align} \bar{v}_1 &= (-4 - 4, -4 - 5) = (-8, -9) \\ \bar{v}_2 &= (2 + 2, 6 + 4) = (4, 10) \end{align} Now add the translated vectors: \begin{align} \bar{v}_1 + \bar{v}_2 &= (-8 + 4, -9 + 10) \\ &= \bf (-4, 1) \end{align} 6 (7, -1) to (2, 5) and (4, 4) to (6, -1) Solution First translate the vectors to the origin. We need to "zero-out" the origin of each vector, by adding (-7, 1) to the start and end of the first, and (-2, -5) to the start and end of the second: \begin{align} \bar{v}_1 &= (2 - 7, 5 + 1) = (-5, 6) \\ \bar{v}_2 &= (6 - 4, -1 - 4) = (2, -5) \end{align} Now add the translated vectors: \begin{align} \bar{v}_1 + \bar{v}_2 &= (-5 + 2, 6 - 5) \\ &= \bf (-3, 1) \end{align}

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