Vector addition and subtraction, the scalar product (dot product) and the vector product (cross product). This page supports various multimedia tutorials in Physclips, including initially Constant Acceleration and Circular Motion.

cartoon of man looking for his bag A vector has magnitude and direction

    Let's start with displacement, which is a vector. That means that it is specified by both a magnitude and a direction. If I move my bag 10 metres North, it is not in the same position as it would be if I were to displace it 10 metres East - I'd certainly notice the difference when I went to look for my bag.

    The magnitude of the displacement is just how far the object is from the origin of reference – 10 m in my examples above. The direction may be specified in any convenient way. Here are some examples of vectors:

    • 10 metres North (a displacement)
    • 15 kilometres per hour, directly towards the opponent's goal (a velocity), and
    • 9.8 m.s−2 down (an acceleration).

    Directions are sometimes expressed in terms of North and South, East and West, up and down. We might also use "radially outwards from the centre of the circle" or "parallel to the initial direction" etc. In very many physics problems, we define a system of axes and use them.

    To distinguish a vector, some books used bold face. In handwriting, we usually use underlining. Here we use both. The symbol r is commonly used for a displacement, just as v is used for velocity and a is used for acceleration.

An example

    graph of vector (x,y) = (2,1) Let's consider a displacement in the x,y plane. Suppose I displace an object from the origin (0,0), to the point (2.0,1.0), where the units are metres. The purple arrow shows this displacement. We'll call this displacement r.

    Magnitude and direction: How far have we moved it? From Pythagoras' theorem, the distance moved is ((1.0 m)2 + (2.0 m)2)½ = 2.2 m. In what direction have we moved it? We could describe this in several ways, including this: at angle θ from the x axis (in the positive mathematical sense). From trigonometry, tan θ = 1.0/2.0 so θ = 27. In other words, we could describe it as

      r  =  2.2 metres at +27 from the x axis.
    If our y and x axes were the North and East on a map, we should similarly say r  =  2.2 metres at +27 North of East.

    Note that, in both cases, we have given a magnitude and direction. This is necessary: because r is a vector, it has magnitude and direction. There are two pieces of information on the left hand side of the equation and so must there be on the right.

    To write the magnitude of a vector, we simply use normal type face. In this example, note carefully the difference

      r  =  2.2 metres,
      r  =  2.2 metres at +27 from the x axis.
    Sometimes we may write the magitude of a vector like this: r  =  |r|   (=  magnitude of r).


Unit vectors

Vectors in three dimensions

Vector addition and subtraction

Relative motion and moving coordinates

sketch of cyclist in a headwind

    What does a head wind feel like?

      Suppose that the wind is coming from the East (ie towards the West), at 5 m.s−1. So the velocity of the wind, relative to the ground, is vw = 5 m.s−1 West. If you are bicycling East at 10 m.s−1 (ie vb = 10 m.s−1 East), what will be the velocity of the wind, relative to you? How strong is the wind that you feel on your face?

      The velocity of the wind with respect to the ground, vw is usually called the true wind, and the velocity relative to an observer moving over the ground is called the apparent wind, vaw. So this question would often be expressed thus: what is the apparent wind? Or, alternatively, what is the velocity of the wind, vaw, in a coordinate system moving with you?

      Here, it's easy: the wind is travelling at vaw with respect to you, and you are travelling at vb with respect to the ground, so the true wind is:

        vw  =  vaw + vb .     So:
        vaw  =  vwvb.
      In the case of the cyclist going East and the wind going West, we could write the wind's velocity as
        vw  =  5 m.s−1 West  =  − 5 m.s−1 East

      so the cyclist has a head wind of 15 m.s−1 – an answer that you probably did in your head already.

sketch of cyclist in a crosswind

    What does a cross wind feel like?

      What if you are travelling North, with the wind from the East, with the same speeds as before? What will be the apparent wind then?

      The same colour coding is used as in the previous example. Using the right angle triangle, we calculate the angle at the top: it is tan−1vw/vb. Note that, if vw < vb, which is often the case, the apparent wind due to a side wind can be reasonably close to an apparent head wind, as is shown here.

      Cyclists often say "Tailwinds are rare, and downhill tailwinds don't happen." Using vector subraction, can you explain this belief?

      Addition and subtraction of velocities is also very important to sailors, first because the velocity of the boat over the water adds to that of the water over the ground to give the boat's velocity with respect to the ground. More importantly, the relative velocity of the wind, as measured on the boat, is the 'true wind' (ie the velocity of the wind with respect to the land) minus the velocity of the boat with respect to the land. See The physics of sailing for examples. Also see Plane on a conveyor belt.

vectors in moving and stationary frames

Position, velocity and acceleration in different frames

    Let's derive that more generally: suppose that we measure position r in a frame of reference fixed on the ground. Let the origin of a frame moving at constant velocity with respect to the ground be rf with respect to the frame fixed on the ground. So the velocity of the moving frame is vf = drf/dt. Let position in the moving frame be r', so

      r  =  rf + r' .
    Provided that time can be measured by synchronised clocks in both frames (but see Relativity for details), we can take derivatives of both sides
      v  =  dr/dt  =  drf/dt + dr'/dt  = vf + v' ,    or      v'  =  v − vf ,

    where v' is the velocity measured in the moving frame (called the dashed frame). Taking time derivatives again
      a  =  dv/dt  =  dvf/dt + dv'/dt  = af + a' ,
    where a' is the acceleration measured in the moving frame. However, we assumed that the moving frame had a constant velocity, so af is zero, so
      a = a
    If we observe the same forces in both frames, we see that, subject to the assumptions we've noted, the accelerations are the same and so the laws of mechanics are the same. This brings up the topic of relativity, in both Galilean and Einsteinien form, which we discuss in much greater detail in this link.

The scalar product (the dot product)

The vector product (the cross product)

sketch of spanner with force F applied at displacement r

    Now consider the torque τ produced by a force F about an axis from which it is displaced by r. (See the Rotation chapter on Physclips.) The magnitude of that torque is proportional to r and proportional to F, so again we need a product of two vectors. The magnitude τ of the torque is also proportional to sin θ, where θ in this case is between r and F. (Notice that it, in both these cases, we need sin θ, whereas the scalar product had cos θ). The torque τ is a vector: torques in different directions in general cause rotations about different axes. That direction of the torque (which is often but not always parallel to the axis about which it causes rotation) is at right angles to both r and F.

sketch of a cross b

    So we define the vector product thus:
      |a X b|  =  ab sin θ, and
      a X b is at right angles to a and b in a right handed sense.

      a X b is pronounced "a cross b".


sketch of a cross b
    The "right handed sense" is needed because there are two directions at right angles to a and b. So, if the thumb of your right hand is in the direction of a and your right forefinger in the direction of b, then your right middle finger is in the direction of a X b. Unless your hands are extraordinarily flexible, the thumb and two fingers can only be mutually perpendicular in one way.
    Another way to define the direction, which doesn't require hands, is that, if the plane of a the blade of a screwdriver is rotated from a to b, then a normal (right-handed!) screw advances in the direction of a X b. Another couple of of mnemonics are suggested by the names "TIM" and "NED", which could be remembered as representing the equations Thumb X Index  =  Middle as in the photograph or, using cardinal directions, North X East  =  Down. (So you can trust these mnemonics, you should also satisfy yourself that, for the right hand, Thumb  =  Index X Middle and that North  =  East X Down.)

picture of unit vectors

    We'll get some practice with that result, because, looking again at the unit vectors, we can write some very useful equations.

      |i X i|  =  1*1 sin 0  =  0.    So:

      i X i  =  j X j  =  k X k =  0.

    Looking at the directions, we can also see that:
      i X j  =  k,    j X k  =  i      and that     k X i  =  j,    
    but notice that
      j X i  =  − k,    k X j  =  − i      and that     i X k  =  − j.    
    Now we come to apply the cross product to vectors in general, calling them again a and b. In many cases, we can simply evaluate the magnitude |a X b|  =  ab sin θ, and take the direction as perpendicular to a and to b in the right handed sense. If, however, we have the components of a and b, then we can write the cross product explicitly. (Be warned, however, that 3 components times 3 components will give us a long equation.)

      a X b  =  (ax i + ay j + az kX (bx i + by j + bz k)

      = (axbx) i X i + (ayby) j X j + (azbz) k X k
           + (axby) i X j + (aybz) j X k + (azbx) k X i
           + (aybx) j X i + (azby) k X j+ (axbz) i X k,      so, collecting terms and using the simplifications given above:
      a X b  = (axby − aybx)k + (aybz − azby)i + (azbx − axbz)j

    The symmetry of this expression gives a mnemonic for writing it down, which is easily seen if you write down the symbols in this array:
      ax    ay    az    ax

      bx    by    bz    bx

      i      j      k      i      j

    and take the diagonals between the elements in the top two rows and multiply them by a term in the third row. But enough algebra! Now it's time to have a look at the Rotation chapter on Physclips, or the section on electric motors and generators, in order to see the examples that required cross products, and to see how they work in the real world.

    The physics of sailing has some interesting examples of vector addition and subtraction.

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