Why Cross Products Need Both Components and Direction
A dot product combines two vectors to give a scalar. A cross product combines two vectors in three-dimensional space to give another vector. Its direction is essential: changing the order of the input vectors can change which way the result points.
This tutorial focuses on computing that vector from components, interpreting its orientation, and checking the result. The checks are especially useful because a single sign error in a component can give a plausible-looking vector that is not perpendicular to the inputs.
Definition in Components
Let \(\mathbf{a}=(a_1,a_2,a_3)\) and \(\mathbf{b}=(b_1,b_2,b_3)\) be vectors in a right-handed three-dimensional coordinate system. Their cross product, written \(\mathbf{a}\times\mathbf{b}\), is defined by the following component formula.
The formula has a specific order. The first component uses coordinates 2 and 3, the second uses coordinates 3 and 1, and the third uses coordinates 1 and 2. Each component is a difference of two products. Keeping the order of those products is important for getting the signs right.
For a compact mnemonic, the formula is sometimes written as a determinant expansion:
Here \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are the unit vectors along the positive \(x\)-, \(y\)-, and \(z\)-axes, respectively. The minus sign on the middle term is part of the expansion. The component formula above is often the clearest way to avoid losing it.
Orientation and the Right-Hand Rule
In a right-handed coordinate system, the positive coordinate directions are oriented so that curling the fingers of your right hand from \(\mathbf{i}\) toward \(\mathbf{j}\) points your thumb along \(\mathbf{k}\). This convention determines the direction associated with the cross product.
The right-hand rule is not a separate calculation that replaces the component formula. The formula gives the components; the rule explains how to interpret the direction those components specify. For example, a positive third component points partly in the positive \(z\)-direction, while a negative third component points partly in the negative \(z\)-direction. A vector can have contributions from more than one coordinate direction.
The order of the inputs matters for orientation. The right-hand rule starts with the first vector and curls toward the second. Reversing that order reverses the direction of the rotation and therefore the direction of the cross product. The following examples show how the component signs express that orientation.
Worked Examples
Worked Example: The Coordinate Directions
Find \(\mathbf{i}\times\mathbf{j}\), where \(\mathbf{i}=(1,0,0)\) and \(\mathbf{j}=(0,1,0)\), and interpret its direction.
Set \(\mathbf{a}=\mathbf{i}\) and \(\mathbf{b}=\mathbf{j}\). Thus \(a_1=1\), \(a_2=0\), \(a_3=0\), \(b_1=0\), \(b_2=1\), and \(b_3=0\). Applying the three component formulas gives
Therefore, \(\mathbf{i}\times\mathbf{j}=(0,0,1)=\mathbf{k}\). The right-hand rule agrees: curl from the positive \(x\)-direction toward the positive \(y\)-direction, and the thumb points along the positive \(z\)-axis.
Check perpendicularity using dot products. Since \(\mathbf{k}=(0,0,1)\),
Both dot products are zero, so the result is perpendicular to both inputs.
Worked Example: Computing a Cross Product with Three Nonzero Components
Let \(\mathbf{a}=(2,-1,3)\) and \(\mathbf{b}=(1,4,-2)\). Compute \(\mathbf{a}\times\mathbf{b}\), interpret its direction, and check perpendicularity.
Substitute the components in order. The first component is
The second component is
The third component is
Consequently,
This vector points in the direction specified by curling the right-hand fingers from \(\mathbf{a}\) toward \(\mathbf{b}\). Its negative first component, positive second component, and positive third component show that it points partly toward negative \(x\), positive \(y\), and positive \(z\).
Now check the result against each input. For \(\mathbf{a}\),
For \(\mathbf{b}\),
Both checks give zero. Thus the computed vector is perpendicular to both inputs, as required.
Worked Example: Reversing the Order Changes the Orientation
Take \(\mathbf{a}=(0,1,0)=\mathbf{j}\) and \(\mathbf{b}=(1,0,0)=\mathbf{i}\). Compute \(\mathbf{a}\times\mathbf{b}\) and compare its direction with the result when the inputs are in the opposite order.
For the stated order, the components are
Therefore, \(\mathbf{j}\times\mathbf{i}=(0,0,-1)=-\mathbf{k}\). The right-hand rule points downward along the negative \(z\)-axis when the fingers curl from \(\mathbf{j}\) toward \(\mathbf{i}\).
For comparison, compute in the reversed order, \(\mathbf{i}\times\mathbf{j}\). The component calculations are
Thus \(\mathbf{i}\times\mathbf{j}=(0,0,1)=\mathbf{k}\), pointing upward. The two answers point in opposite directions, matching the change in order.
Check the first result: \((0,0,-1)\cdot(0,1,0)=0\), and \((0,0,-1)\cdot(1,0,0)=0\). Check the reversed result: \((0,0,1)\cdot(1,0,0)=0\), and \((0,0,1)\cdot(0,1,0)=0\). Each result is perpendicular to both inputs.
Why the Result Is Perpendicular
The perpendicularity check follows directly from the component formula. Recall that the dot product of \((x_1,x_2,x_3)\) and \((y_1,y_2,y_3)\) is \(x_1y_1+x_2y_2+x_3y_3\). We can now calculate the dot product of \(\mathbf{a}\) with the defined cross product.
To prove the first perpendicularity, substitute the three components of \(\mathbf{a}\times\mathbf{b}\) into the dot product with \(\mathbf{a}\):
Every term cancels with an equal term of opposite sign. To prove perpendicularity to \(\mathbf{b}\), calculate separately:
Again, all terms cancel in pairs. Since a zero dot product means the vectors are perpendicular, the result follows for both inputs. This also provides a practical check: after computing a cross product, take its dot product with each original vector. Both values should be zero.
A Common Sign Error
The second component is a frequent source of mistakes. Its correct form is \(a_3b_1-a_1b_3\), not \(a_1b_3-a_3b_1\). Reversing those two terms changes the sign of that component.
For the vectors in the second example, the incorrect calculation would be
This would give the incorrect vector \((-10,-7,9)\). Check it against \(\mathbf{a}=(2,-1,3)\):
Since this dot product is not zero, the incorrect vector is not perpendicular to \(\mathbf{a}\), so it cannot be the cross product. The correction is to use the prescribed second component order:
Then the corrected vector \((-10,7,9)\) has zero dot product with both inputs. A good habit is to write all three component formulas before substituting numbers, then use both dot products to check the finished result.
Check Your Understanding
For each question, pay attention to both the component order and the orientation of the result.
- Use the component formula to find \((1,0,0)\times(0,0,1)\).
- Use the right-hand rule to determine whether \(\mathbf{j}\times\mathbf{k}\) points along positive or negative \(\mathbf{i}\).
- Find \((1,2,0)\times(0,1,1)\), then check its dot product with each input vector.
- Why is checking only one of the two dot products insufficient to verify that a proposed result is perpendicular to both inputs?
- What change in direction should you expect when the order of the two input vectors is reversed?