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Dot and cross products computationally

Multiplying vectors is not one single operation. The dot product combines two vectors to produce a scalar. The cross product combines two three-dimensional polar vectors to produce an axial vector perpendicular to both. Their different output types express different physical questions: alignment for the dot product and oriented perpendicular area for the cross product.

The dot product measures alignment

For vectors \(\mathbf a\) and \(\mathbf b\), the geometric definition is

\[\mathbf a\cdot\mathbf b=|\mathbf a|\,|\mathbf b|\cos\theta\]

Here \(\theta\) is the smaller angle from zero to \(\pi\) between nonzero vectors. The result is positive for an acute angle, zero for a right angle, and negative for an obtuse angle. The result has the product of the input units. Dotting a force in newtons with a displacement in metres produces newton-metres, the units of work.

In an orthonormal Cartesian basis, the component rule is

\[\mathbf a\cdot\mathbf b=a_xb_x+a_yb_y+a_zb_z\]

This is the sum of pairwise component products, not the array of pairwise products. Both inputs must have the same dimension and be expressed in the same basis before the component formula is used.

import numpy as np

force_n = np.array([3.0, 4.0, 0.0])
displacement_m = np.array([2.0, 0.0, 0.0])
work_j = np.dot(force_n, displacement_m)
print(f"Work: {work_j:.1f} J")

The dot product is commutative and distributive:

\[\mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a,\qquad \mathbf a\cdot(\mathbf b+\mathbf c)=\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c\]

Dotting a vector with itself gives its squared magnitude:

\[\mathbf a\cdot\mathbf a=|\mathbf a|^2\]

The left side can therefore never be negative for an ordinary Euclidean vector. A negative self dot product signals an implementation error, incompatible data, or use of a different metric that must be stated explicitly.

Projection and work

The scalar component of \(\mathbf a\) along a nonzero vector \(\mathbf b\) is

\[a_{\parallel b}=\mathbf a\cdot\hat{\mathbf b}=\frac{\mathbf a\cdot\mathbf b}{|\mathbf b|}\]

The vector projection is

\[\operatorname{proj}_{\mathbf b}\mathbf a=\frac{\mathbf a\cdot\mathbf b}{\mathbf b\cdot\mathbf b}\mathbf b\]

Both formulas require a nonzero direction vector. The vector projection has the same units as \(\mathbf a\) because the scale factor carries the necessary unit ratio.

Mechanical work by a constant force over a displacement is

\[W=\mathbf F\cdot\Delta\mathbf r=F\,\Delta r\cos\theta\]

Only the force component parallel to the displacement contributes. A perpendicular force does zero work during that displacement even though it may change the direction of motion.

For example, let

\(\mathbf F=(6,2)\,\mathrm N\)

and \(\Delta\mathbf r=(3,-1)\,\mathrm m\). Then

\[W=(6)(3)+(2)(-1)=16\,\mathrm J\]

The unit is joules because a newton-metre is a joule in this scalar work context.

Recovering an angle safely

For two nonzero vectors,

\[\theta=\cos^{-1}\!\left(\frac{\mathbf a\cdot\mathbf b}{|\mathbf a|\,|\mathbf b|}\right)\]

Floating-point roundoff may produce a cosine slightly above one or below negative one, even for valid inputs. Clamp only after verifying that both magnitudes are meaningfully nonzero. Clamping a wildly invalid value would conceal bad data.

def angle_between(first, second):
    denominator = np.linalg.norm(first) * np.linalg.norm(second)
    if denominator < 1e-12:
        raise ValueError("The angle needs two nonzero vectors.")
    cosine = np.clip(np.dot(first, second) / denominator, -1.0, 1.0)
    return np.arccos(cosine)


angle = angle_between(force_n, displacement_m)
print(f"Angle: {np.degrees(angle):.2f} degrees")

The cross product measures oriented area

For ordinary three-dimensional vectors, the cross product satisfies

\[|\mathbf a\times\mathbf b|=|\mathbf a|\,|\mathbf b|\sin\theta\]

Its magnitude equals the area of the parallelogram spanned by the two inputs. Its direction is perpendicular to their plane, chosen by the right-hand rule. In Cartesian components,

\[\mathbf a\times\mathbf b=(a_yb_z-a_zb_y)\hat{\mathbf i}+(a_zb_x-a_xb_z)\hat{\mathbf j}+(a_xb_y-a_yb_x)\hat{\mathbf k}\]

The middle component is a frequent source of sign mistakes. Writing each output component separately makes review easier.

lever_arm_m = np.array([2.0, 0.0, 0.0])
force_n = np.array([0.0, 3.0, 0.0])
torque_nm = np.cross(lever_arm_m, force_n)
print("Torque (N m):", torque_nm)
assert np.allclose(torque_nm, [0.0, 0.0, 6.0])

The cross product is anti-commutative:

\[\mathbf a\times\mathbf b=-(\mathbf b\times\mathbf a)\]

Swapping the inputs reverses the direction. A vector crossed with itself is zero, and parallel or antiparallel vectors also have zero cross product. Zero does not distinguish parallel from antiparallel, so use the dot product too when that distinction matters.

Torque and angular momentum

Torque about a chosen origin is

\[\boldsymbol\tau=\mathbf r\times\mathbf F\]

The position vector \(\mathbf r\) runs from the chosen origin to the force application point. The torque magnitude is force times the perpendicular lever arm. A force directed exactly toward or away from the origin produces zero torque about that origin.

Angular momentum of a particle about an origin is

\[\mathbf L=\mathbf r\times\mathbf p\]

Torque and angular momentum are axial vectors. Under a reflection of the coordinate system they transform differently from polar vectors such as displacement and force. This handedness dependence is why a cross-product implementation must know whether its basis is right-handed. A left-handed graphics coordinate convention can reverse the expected sign if treated as right-handed physics coordinates.

Two-dimensional cross calculations

Two vectors confined to the horizontal-vertical plane have a cross product along the perpendicular axis:

\[(\mathbf a\times\mathbf b)_z=a_xb_y-a_yb_x\]

Some software calls this scalar a cross2D result. It is better understood as the signed perpendicular component of the three-dimensional cross product after setting the missing input components to zero. A positive value means the turn from \(\mathbf a\) to \(\mathbf b\) follows the positive orientation of the stated basis. Its absolute value is the parallelogram area.

This signed area is useful for orientation tests, polygon area, and detecting on which side of a directed line a point lies. Near collinearity, the result is sensitive to roundoff, so geometric algorithms need a scale-aware tolerance or robust predicate.

A worked cross-product example

Let

\(\mathbf r=(2,0,0)\,\mathrm m\)

and \(\mathbf F=(0,3,0)\,\mathrm N\). Then

\[\boldsymbol\tau=\mathbf r\times\mathbf F=(0,0,6)\,\mathrm{N\,m}\]

The magnitude also follows from \(rF\sin90^\circ=6\,\mathrm{N\,m}\). Reversing the operand order would produce the negative perpendicular direction and would not represent the stated torque definition.

Numerical and modeling checks

Useful tests for both operations include:

For almost parallel vectors, calculating an angle with inverse cosine can lose precision because the cosine changes very little near zero angle. In three dimensions, an angle can often be obtained more robustly with \(\operatorname{atan2}(|\mathbf a\times\mathbf b|,\mathbf a\cdot\mathbf b)\). The inputs still need compatible scaling and nonzero magnitudes.

Dot and cross products are compact formulas, but their meaning depends on output type, units, basis, handedness, and operand order. Preserving those facts in code makes the result usable as physics rather than merely as array arithmetic.

A worked Python notebook

A force of 3 newtons acts through a displacement of 2 metres at a variable angle. The question for this calculation is precise: how does work change when we change angle between force and displacement, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

The force is constant along the displacement and the angle refers to the same coordinate frame. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0 to 3.14159 rad. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for angle between force and displacement, expressed in rad. The output is work, expressed in J. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = 6 \cos{\left(q \right)}\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 1.0472 rad, the expression gives 3 J. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("6*cos(q)", locals={"q": q})
reference_input = 1.0471975511965979
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "rad")
print("Predicted work:", reference_output, "J")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 6 J. The midpoint, at 1.5708 rad, gives 3.67394e-16 J. The upper endpoint gives -6 J. Read these together with the assumptions: they describe different choices of angle between force and displacement within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.0, 1.5707963267948966, 3.141592653589793])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} rad -> {output_value:.6g} J")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows angle between force and displacement; the vertical axis shows work. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.0, 3.141592653589793, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Angle between force and displacement (rad)")
ax.set_ylabel("Work (J)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does work change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 1.0472 rad is -5.19615 in those ratio units. Locally, the output decreases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = - 6 \sin{\left(q \right)}\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 1.0629055144645467
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "J")
print("Full expression:", full_prediction, "J")
print("Approximation difference:", full_prediction - linear_prediction, "J")

Here the changed input is 1.06291 rad. The tangent estimate is 2.91838 J, while the full expression gives 2.91801 J. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 3.141592653589793
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of angle between force and displacement inside the stated interval, predict whether work will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: The force is constant along the displacement and the angle refers to the same coordinate frame. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of J: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(3.141592653589793) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 3.141592653589793, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

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