A girl at Chennai airport pulls her suitcase by its long handle. She pulls with \(50\,\text{N}\) along the handle, which makes \(37°\) with the floor, and the suitcase rolls \(10\,\text{m}\). A pull of \(50\,\text{N}\) over \(10\,\text{m}\) suggests \(500\,\text{J}\) of work. Yet her pull transfers less energy than that. Part of the pull only lifts on the handle, and that part moves the suitcase nowhere.
A product that is a number
Two vectors can be multiplied in two ways. The first way gives a scalar. It is called the dot product, or the scalar product, and it is written with a dot. It multiplies the two magnitudes and the cosine of the angle between the vectors, with the vectors drawn tail to tail. Since the cosine is at most 1, the dot product is largest when the vectors are parallel.
The dot product of two vectors \(\vec A\) and \(\vec B\) is the scalar \(\vec A\cdot\vec B = AB\cos\theta\). Here \(A\) and \(B\) are their magnitudes, and \(\theta\) is the angle between them when they are drawn from one point.
A length times a shadow
The cosine has a picture. Drop a perpendicular from the tip of \(\vec A\) to the line of \(\vec B\). The piece of that line between the tail and the foot of the perpendicular is the projection of \(\vec A\) on \(\vec B\), its shadow. Its signed length is \(A\cos\theta\). So \(\vec A\cdot\vec B\) is \(B\) times the shadow of \(\vec A\), or equally \(A\) times the shadow of \(\vec B\). The order does not matter: \(\vec A\cdot\vec B = \vec B\cdot\vec A\).
The sign tells how the two vectors lean. Below \(90°\) the shadow falls along \(\vec B\), and the dot product is positive. At exactly \(90°\) there is no shadow, and the dot product is zero. Beyond \(90°\) the shadow falls behind the tail, and the dot product is negative. A vector dotted with itself gives \(\vec A\cdot\vec A = A^2\), because \(\cos 0° = 1\).
Work is a dot product
Work is the plainest dot product in physics. Only the part of a force along the displacement transfers energy, and that part is \(F\cos\theta\). So the work of a constant force is \(W = Fd\cos\theta = \vec F\cdot\vec d\). The part of the force at right angles to the motion does no work at all.
The girl pulls her suitcase with \(50\,\text{N}\) along a handle at \(37°\) to the floor, and it moves \(10\,\text{m}\) along the floor. Take \(\cos 37° = 0.8\). Find the work done by her pull, in J.
For two vectors \(\vec A\) and \(\vec B\), the dot product \(\vec A\cdot\vec B\) is negative. What does this tell you?
The dot product from components
Often the components are known and the angle is not. The unit vectors make the dot product easy. Each unit vector has magnitude 1, and the three are at right angles to one another. So \(\hat i\cdot\hat i = 1\) while \(\hat i\cdot\hat j = 0\), and the same holds for every pair. The derivation below uses these facts to reach the component form.
\[ \vec A\cdot\vec B = (A_x\hat i + A_y\hat j + A_z\hat k)\cdot(B_x\hat i + B_y\hat j + B_z\hat k) \]
Why. write both vectors in components.
\[ \vec A\cdot\vec B = A_xB_x\,\hat i\cdot\hat i + A_xB_y\,\hat i\cdot\hat j + \dots \]
Why. the dot product may be multiplied out term by term, like ordinary brackets, giving nine products.
\[ \hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1 \]
Why. each unit vector has length 1 and makes \(0°\) with itself.
\[ \hat i\cdot\hat j = \hat j\cdot\hat k = \hat k\cdot\hat i = 0 \]
Why. the unit vectors are at right angles, and \(\cos 90° = 0\). Six of the nine products vanish.
\[ \vec A\cdot\vec B = A_xB_x + A_yB_y + A_zB_z \]
Read it. only the three products of matching components remain.
Putting the two forms together gives the angle between two vectors: \(\cos\theta = \vec A\cdot\vec B/(AB)\). For \(\vec A = 3\hat i + 4\hat j\) and \(\vec B = 4\hat i + 3\hat j\), the dot product is \(12 + 12 = 24\). Both magnitudes are 5, so \(\cos\theta = 24/25 = 0.96\) and \(\theta = 16.3°\).
Find the angle between \(\vec A = 3\hat i + 4\hat j\) and \(\vec B = 4\hat i + 3\hat j\), in degrees.
The component along any direction
A component along \(x\) is a dot product with \(\hat i\): \(A_x = \vec A\cdot\hat i\). The same works along any direction. The component of \(\vec A\) along a unit vector \(\hat n\) is \(\vec A\cdot\hat n\). The part of the force \(\vec F = 3\hat i + 4\hat j\) N along the displacement \(\vec d = 4\hat i - \hat j\) m is \(\vec F\cdot\vec d/d = 8/\sqrt{17} = 1.94\,\text{N}\). Electric flux uses the same idea. The flux through a flat area is \(\Phi = \vec E\cdot\vec A\), where the area vector points along the normal to the surface.
The test for perpendicular vectors
Two vectors that are not zero are perpendicular exactly when their dot product is zero. This test needs no drawing. For \(\vec A = 3\hat i + 4\hat j\), the vector \(4\hat i - 3\hat j\) is perpendicular, because \(12 - 12 = 0\). Physics uses the test to check a result. A magnetic force is always at right angles to the velocity, so \(\vec F\cdot\vec v = 0\), and the magnetic force does no work.
Which of these vectors is perpendicular to \(\vec A = 3\hat i + 4\hat j\)?
The electric field in a region is \(\vec E = (3\hat i + 4\hat j)\,\text{N/C}\). Find the flux through a flat area of \(2\,\text{m}^2\) whose normal points along \(\hat j\), in N m²/C.
This is the worked example "Work from components" of Work, Kinetic Energy and Power, lesson 1, done with the tool. A force \(\vec F = (3\hat i + 4\hat j)\,\text{N}\) acts on a particle while it moves from the point \((1, 2)\,\text{m}\) to the point \((5, 1)\,\text{m}\). Find the work done by the force.
The displacement is final minus initial: \(\vec d = (5 - 1)\hat i + (1 - 2)\hat j = (4\hat i - \hat j)\,\text{m}\).
The work is the dot product in components: \(W = F_xd_x + F_yd_y = 3 \times 4 + 4 \times (-1) = 12 - 4 = 8\,\text{J}\), the same answer as the physics chapter.
The \(y\) part of the force did negative work, because the particle moved down while that part of the force points up.
No angle was needed. The components did all the work of the cosine.
A charge moves with \(\vec v = (3\hat i + 4\hat j)\times10^4\,\text{m/s}\), and the magnetic force on it is \(\vec F = (8\hat i - 6\hat j)\times10^{-3}\,\text{N}\). How much power does the force deliver, \(P = \vec F\cdot\vec v\)?
The angle in \(AB\cos\theta\) is the angle between the vectors drawn from one point. Two arrows in a figure that meet tip to tail show the other angle, \(180° - \theta\). A dot product is a number, so it cannot be undone. From \(\vec A\cdot\vec B = 12\) and \(\vec A\) you cannot find \(\vec B\), since every vector with the same shadow on \(\vec A\) gives 12. And \(\vec A\cdot\vec B = 0\) does not mean that one vector is zero. The two vectors may simply be perpendicular.
Josiah Willard Gibbs taught physics at Yale in the United States. In 1881 he printed a short pamphlet on vector analysis for his own students, and he added to it in 1884. In it he wrote the two products of vectors with a dot and a cross, the notation of this chapter. In England, Oliver Heaviside reached almost the same system on his own, while he rewrote the equations of electromagnetism in a shorter form. Their notation spread through physics after a student of Gibbs published it as a textbook in 1901.
The suitcase of the opening has its answer. The work is \(\vec F\cdot\vec d = 50 \times 10 \times \cos 37° = 400\,\text{J}\), not \(500\,\text{J}\). The \(40\,\text{N}\) part of the pull along the floor did all of it. The \(30\,\text{N}\) part lifted on the handle and was perpendicular to the motion, so it did none.
Work, Kinetic Energy and Power defines work as \(\vec F\cdot\vec d\) and power as \(\vec F\cdot\vec v\). Work along a curve is a sum of small dot products. Electric Flux and Gauss's Law defines flux as \(\vec E\cdot\vec A\), and magnetic flux through a coil is \(\vec B\cdot\vec A\), the start of electromagnetic induction. The energy of an electric or magnetic dipole in a field is a dot product, \(-\vec p\cdot\vec E\) or \(-\vec m\cdot\vec B\). Ampere's law adds \(\vec B\cdot d\vec l\) round a loop. Magnetic Force on Moving Charges uses \(\vec F\cdot\vec v = 0\) to show that the magnetic force does no work.
- Multiplying the magnitudes and forgetting the cosine.\(\vec A\cdot\vec B = AB\) only when the vectors are parallel. In general multiply by \(\cos\theta\), or use the components.
- Reading a zero dot product as a zero vector.A zero dot product between non-zero vectors means they are perpendicular.
- Writing a vector answer for a dot product.A dot product is a scalar. Work, power and flux carry no direction.