1. Vectors

1.1. Vectors and scalars

Physical quantities whose magnitude is given by a real value are called scalars. Examples of scalar quantities are time, temperature and pressure. Each scalar has units and if the units of two different scalars are compatible, they can be combined with the same rules as for real numbers: for example, we can add a mass, in kg, with other masses also in kg; we can divide a distance in km by a time in hours, to give a speed in km/h, and so on.

In contrast to scalars, a vector is a physical quantity that in addition to its magnitude it also has a direction. The magnitude of the vector is sometimes also called norm or module. A typical example of a vector is a force; the magnitude of a force is measured in units of force, i. e. newton (N) in the international system of units. The direction of the force is the direction in which it is exerted. We will denote vectors by letters with an arrow over them, for instance a force $\overrightarrow{F}$, and their magnitude by enclosing the symbol inside vertical bars: $|\overrightarrow{F}|$. In cases when $F$ cannot be confused with other scalar quantities, we will simply write the magnitude of $\overrightarrow{F}$ as $F$. The magnitude of a vector is a scalar.

Figure 1.1 shows three forces; the weight $\overrightarrow{W}$ of a 10 kg sphere, which has a magnitude of 98 N and points downward in the vertical, and the forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ exerted by one the feet of a woman and a man on the ground.

Two vectors are considered equal if their magnitude and direction are the same. In the example in Figure 1.1, the two forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ might be equal if they have the same magnitude and direction.

1.2 Addition of vectors

The sum of two vectors is defined as another vector which can be obtained by placing one of the vectors after the other and then joining the initial point of the first vector with the final point of the second one. The right-hand side of Figure 1.2 shows the sum $\overrightarrow{A}+\overrightarrow{B}$ of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, which was obtained by displacing $\overrightarrow{B}$ until its initial point is at the final point of $\overrightarrow{A}$.

In the case of forces, the vector sum of two forces gives a force that produces the same effect as the two forces combined.

Example 1.1

Two forces are applied on the two sides of a string that passes through a pulley, as shown in the figure.

If the friction on the pulley's axle is negligible, the magnitude of the forces on both sides of the string are equal (and is called the tension on the string) and in this case it was measured to be 400 N. Find the total force exerted by the string on the pulley.

Solution: The following figure shows the sum $\overrightarrow{F}$ of the two forces, obtained by placing ${\overrightarrow{T}}_1$ after ${\overrightarrow{T}}_2$. We know that the distances PR and RS are both 400 (in units of N), PR makes an angle of 34° with the vertical and RS is horizontal.

The right-hand side of the figure above shows a copy of the triangle PQR reduced by a factor of 400. Since the hypotenuse of that smaller right-angled triangle is 1, the other two sides are sin 34° and cos 34°. Enlarging the small triangle by 400, we thus obtain the lengths PQ and QR:

and the distance QS is then $\mathrm{Q}\mathrm{R}+\mathrm{R}\mathrm{S}=623.68$. The magnitude of the total force is the hypotenuse of the triangle PQS which is obtained from the Pythagorean theorem:

It is easy to see that the direction of the total force $\overrightarrow{F}$ will have to be 28° above the horizontal, since it has to make the same angle with the directions of ${\overrightarrow{T}}_1$ and ${\overrightarrow{T}}_2$, because they both have the same magnitude. If you are not convinced you can compute the angle RSP.

The addition of vectors has the following four important properties:

The commutative law can be proven by completing a parallelogram with the two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ as shown in Figure 1.3. The sum of the vectors is the diagonal of the parallelogram, which is obtained by either placing $\overrightarrow{B}$ after $\overrightarrow{A}$ or by placing $\overrightarrow{A}$ after $\overrightarrow{B}$.

The proof of the associative law is given in Figure 1.4, which shows three vectors $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$, together with the sums $\overrightarrow{A}+\overrightarrow{B}$ and $\overrightarrow{B}+\overrightarrow{C}$. The sum of the three vectors can then be simply written as $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$ without any need for parentheses. Regarding Figure 1.4, notice that even though the vectors seem to be on the same plane, they don't have to; in fact, $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$ could even be along three of the edges of a tetrahedron.

The vector zero, $\overrightarrow{0}$, is the only vector that when added to any vector $\overrightarrow{A}$ will not change that vector. It is unique and it has magnitude equal to zero; it is also the only vector that has no direction. For any vector $\overrightarrow{A}$ there is always a unique vector, called its negative, such that when added to $\overrightarrow{A}$ the result will be the vector zero. The negative of $\overrightarrow{A}$ is denoted $-\overrightarrow{A}$ and it is easy to see that it must be a vector with the same magnitude $|\overrightarrow{A}|$, but in the opposite direction to $\overrightarrow{A}$ (see Figure 1.5).

In the case of forces, the uniqueness of the negative means that the only way to cancel the effect of a force is by adding another force with the same magnitude but opposite direction.

The subtraction of two vectors can be defined as the sum of the first vector and the negative of the second one:

Thus, $\overrightarrow{A}-\overrightarrow{B}$ can be obtained by placing $-\overrightarrow{B}$ at the final point of $\overrightarrow{A}$, as in the left-hand side of Figure 1.6. The right-hand side of Figure 1.6 shows that $\overrightarrow{A}-\overrightarrow{B}$ can also be obtained as one of the diagonals of the parallelogram obtained with $\overrightarrow{A}$ and $\overrightarrow{B}$. As we saw before, the diagonal that begins from the common point of the two vectors is $\overrightarrow{A}+\overrightarrow{B}$; the other diagonal, from the endpoint of $\overrightarrow{B}$ to the end point of $\overrightarrow{A}$, is $\overrightarrow{A}-\overrightarrow{B}$.

Example 1.2

A box weighing 50 N is hung from the ceiling by two strings, as shown in the figure. Find the forces in the two strings.

Solution: Since the block remains at rest, its weight must be canceled by the sum of the forces in the two strings. Thus, if ${\overrightarrow{T}}_1$ and ${\overrightarrow{T}}_2$ are the forces exerted by the two strings and $\overrightarrow{W}$ is the weight of the box, we have:

The left-hand side of the following image shows those three forces and the right-hand side shows a smaller triangle, pqr, similar to the forces triangle PQR.

The sides pq and qr are reduced copies of the two strings on the left and the right. The side qr was given the length of the string on the right divided by 20 cm, to make the triangle simpler; thus qr moves 2 units to the right and 1 unit up. Since qp must also move 2 units to the right, qp must be the string on the left divided by 10 cm and p is 2 units below q. Hence, pr is 3 units and the lengths of the other two sides can be computed with the Pythagorean theorem:

Since $|-\overrightarrow{W}|=50$ N, we just have to multiply pq and qr by 50/3 N to obtain $T_1$ and $T_2$:

and the directions of $\overrightarrow{T_1}$ and $\overrightarrow{T_2}$ are pointing upwards along the two strings.

1.3 Scalar multiplication

When we add a vector $\overrightarrow{A}$ with itself $n$ times, the result is a vector in the same direction of $\overrightarrow{A}$ but with magnitude $n$ times bigger and we write the result as $n\overrightarrow{A}$. If $n$ is a negative integer, we interpret $n\overrightarrow{A}$ as summing $|n|$ times the negative of the vector, $-\overrightarrow{A}$, thus leading to a vector in the opposite direction of $\overrightarrow{A}$ and with magnitude $|n|$ times bigger than the magnitude of $\overrightarrow{A}$.

The generalization to any scalar $k$ is that the product $k\overrightarrow{A}$ gives a vector with magnitude $|k|A$, in the same direction of $\overrightarrow{A}$, when $k$ is positive, or in the opposite direction when $k$ is negative. When $k=0$, the product $k\overrightarrow{A}$ gives the vector zero, $\overrightarrow{0}$.

The following table shows the important properties of the product of scalars with vectors.

We can also divide a vector by a scalar, as in $\overrightarrow{A}/k$, which means that we are multiplying vector $\overrightarrow{A}$ by $1/k$. In the case of the weight $\overrightarrow{W}$ of an object of mass $m$, the weight divided by the mass,

is always the same vector $\overrightarrow{g}$, independently of the mass of the object, which is the acceleration of gravity. It has magnitude of approximately 9.8 m/s², in the vertical direction and pointing down. This is simply a consequence of Newton's second law, which states that a force $\overrightarrow{F}$ acting on an object of mass $m$ produces an acceleration $\overrightarrow{a}$ given by the relation:

The weight $\overrightarrow{W}$ is the force produced by the gravitational attraction of Earth on an object and $\overrightarrow{g}$ is the acceleration it produces near the surface of the Earth. A word of caution: even though we can write equation (1.2) as $\overrightarrow{a}=\overrightarrow{F}/m$, don't even think of writing it as $m=\overrightarrow{F}/\overrightarrow{a}$; a vector cannot be divided by another vector to give a scalar! If we were given two vectors $\overrightarrow{F}$ and $\overrightarrow{a}$ and were asked to find the value of $m$ in equation (1.2), we should first make sure that the problem has solution, by checking that the two vectors have the same direction; having done that, we would then proceed to say that the magnitude of the vectors in both sides of equation (1.2) must be equal and therefore,

where we are dividing two scalars, and not two vectors. Another important point about scalar multiplication is that it is not just the product of a number by a vector, as in $\overrightarrow{A}+\overrightarrow{A}=2\overrightarrow{A}$. For instance, the product $m\overrightarrow{a}$ in equation (1.2) changes a vector $\overrightarrow{a}$, in the space of accelerations, into a vector in a different vector space, the space of forces.

1.4 Dot product

The dot product between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, denoted by $\overrightarrow{A}·\overrightarrow{B}$, gives a scalar, equal to the product of the magnitudes of the vectors times the cosine of the angle between their directions:

Figure 1.7 shows two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ and the angle $\theta$ between their directions. The projection of $\overrightarrow{A}$ in the direction of $\overrightarrow{B}$ is equal to $A\cos \theta$; notice that it can be negative when $\theta >\pi /2$, namely when that projection points in the opposite direction of $\overrightarrow{B}$. The projection of $\overrightarrow{B}$ in the direction of $\overrightarrow{A}$ is $B\cos \theta$. Therefore, the dot product $\overrightarrow{A}·\overrightarrow{B}$ is also equal to the projection of one of the vectors along the direction of the other, times the magnitude of that other vector. The dot product is also referred to as scalar product.

Since the angle $\theta$ between two vectors is within 0 and $\pi$ radians, the cosine of that angle can be within ─1 and 1, so the dot product $\overrightarrow{A}·\overrightarrow{B}$ will be within $-AB$ and $AB$. It will be positive if $\theta$ is an acute angle (less than $\pi /2$), negative if that angle is obtuse (greater than $\pi /2$) and zero when the two vectors are perpendicular (see Figure 1.8).

If two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ have the same direction, then $\overrightarrow{A}·\overrightarrow{B}=AB$, while if they have opposite directions then $\overrightarrow{A}·\overrightarrow{B}=-AB$ and if they are perpendicular, $\overrightarrow{A}·\overrightarrow{B}=0$.

The dot product has the following properties:

Solution: Let us consider a triangle with sides $A$, $B$ and $C$ and an angle $\theta$ between the sides of lengths $A$ and $B$. We can define three vectors $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$ along the triangle sides, as in the following figure:

We then have $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$ and applying the properties of the dot product, the square of the magnitude of $\overrightarrow{C}$ is,

Which leads to the law of cosines:

With the law of cosines we don't have to try to arrange right-angled triangles to apply the Pythagorean theorem, as we did in the solution of example 1.1. We could have applied the law of cosines directly to triangle PRS, where the angle PRS is 124°, to obtain $F$ immediately:

1.5 Cross product

The cross produt between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is defined as another vector $\overrightarrow{C}$, denoted as,

with magnitude equal to the product of the magnitudes of the two vectors, times the sine of the angle between their directions:

The direction of $\overrightarrow{A}\times \overrightarrow{B}$ is, by definition, perpendicular to both vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ and following the right-hand rule: if you point the four fingers of your right hand (excluding the thumb) in the direction of vector $\overrightarrow{A}$, in such way that they can turn towards the direction of vector $\overrightarrow{B}$, your thumb will then point in the direction of $\overrightarrow{A}\times \overrightarrow{B}$, as shown in Figure 1.9. Notice that any two vectors for which $\sin \theta$ is different from zero will always define a plane, and $\overrightarrow{A}\times \overrightarrow{B}$ will be perpendicular to that plane. Furthermore, equation (1.6) implies that the magnitude $|\overrightarrow{A}\times \overrightarrow{B}|$ is the area of the parallelogram generated by vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ (shaded area in the figure).

Some important properties of the cross product are the following:

Solution: For a triangle with sides $A$, $B$ and $C$ and angles $\alpha$, $\beta$ and $\theta$, let us define vectors $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$ as in the following figure:

We then have,

multiplying, with cross product, each term of that equation by $\overrightarrow{A}$ and using properties of the cross product we get,

If instead of multiplying by $\overrightarrow{A}$ we now multiply by $\overrightarrow{B}$ we get,

Dividing the last two equations above by $ABC$, we arrive at the law of sines:

Problems

1. Proof that the diagonals of a rhombus are perpendicular.
   Help If the sides of the rhombus are vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, the two diagonals will be $\overrightarrow{A}+\overrightarrow{B}$ and $\overrightarrow{A}-\overrightarrow{B}$ and they will be perpendicular if their dot product is equal to zero.
2. Knowing that the three boxes remain at rest and the weight $W_2$ in the figure is 64 N, compute the weights $W_1$ and $W_3$. Answer $W_1=31$ N and $W_3=49$ N. Help The tension on each string is equal to the weight of the box attached to it. Then notice that the forces on the strings form a triangle with angles of 28°, 48° and 104°; knowing that the tension in the string attached to $W_2$ has module 64 N, use the law of sines to find the tensions on the other two strings.
3. Proof the following identities:
   1. $\overrightarrow{A}·\overrightarrow{B}={\displaystyle \frac{1}{4}}[|\overrightarrow{A}+\overrightarrow{B}{|}^2-|\overrightarrow{A}-\overrightarrow{B}{|}^2]$ Help On the right-hand side, simplify $|\overrightarrow{A}+\overrightarrow{B}{|}^2=(\overrightarrow{A}+\overrightarrow{B})·(\overrightarrow{A}+\overrightarrow{B})$ and $|\overrightarrow{A}-\overrightarrow{B}{|}^2=(\overrightarrow{A}-\overrightarrow{B})·(\overrightarrow{A}-\overrightarrow{B})$ and you should get just $\overrightarrow{A}·\overrightarrow{B}$.
   2. $|\overrightarrow{A}\times \overrightarrow{B}{|}^2+\overrightarrow{A}·{\overrightarrow{B}}^2=A^2{B}^2$     Help On the left-hand side, notice that $|\overrightarrow{A}\times \overrightarrow{B}{|}^2=(AB\sin \theta {)}^2$ and $\overrightarrow{A}·{\overrightarrow{B}}^2=(AB\cos \theta {)}^2$ and use the identity ${\sin }^2\theta +{\cos }^2\theta =1$ to obtain the result on the right.