Cross Product

Cross product or vector product is a binary operation on two vectors in a three-dimensional oriented Euclidean vector space. Cross product, also called the vector cross product, is a mathematical operation performed on two vectors in three-dimensional space.

In this article, we will understand the meaning of cross product, its definition, the formula of the cross product, the cross product of perpendicular vectors, the cross product of parallel vectors, the right-hand rule cross product and the properties of the cross product.

What is Cross Product?

Cross-product is a mathematical operation that takes two vectors in three-dimensional space and produces a third vector that is perpendicular to both input vectors. It is denoted by the symbol “×” and is used to calculate the area of a parallelogram formed by the two vectors and the direction of the resulting vector is determined by the right-hand rule.

Cross Product Definition

Cross product of two independent vectors, A and B, denoted as A × B, yields a new vector that is perpendicular to both A and B and is normal to the plane containing these vectors.

Magnitude of the cross product is given by the product of the magnitudes of A and B and the sine of the angle θ between them.

Symbolically, if A = |A| and B = |B|, the cross product is expressed as:

[Tex]\vec{A} \times \vec{B} = |\vec{A}| \cdot |\vec{B}| \cdot \sin \theta[/Tex]

If two vectors lie in the X-Y plane, their cross-product results in a vector along the Z-axis, perpendicular to the XY plane. The symbol “×” is used between the original vectors, and the resultant vector is represented as ([Tex]\vec{c}[/Tex]) in the equation ([Tex]\vec{a} \times \vec{b} = \vec{c}[/Tex]).

Cross-Product

Formula of Cross Product

Formula for cross product of two vectors ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]) in three-dimensional space is:

[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} [/Tex]

This formula involves finding the determinant of a 3×3 matrix, where the first row represents the unit vectors ([Tex]\hat{i}[/Tex]), ([Tex]\hat{j}[/Tex]), and ([Tex]\hat{k}[/Tex]), the second row represents the components of vector ([Tex]\vec{A}[/Tex]), and the third row represents the components of vector ([Tex]\vec{B}[/Tex]).

Cross Product of Perpendicular Vectors

Cross product of two perpendicular vectors, the resulting vector is perpendicular to both input vectors and its magnitude is equal to the product of the magnitudes of the input vectors.

Suppose, two perpendicular vectors ([Tex]\vec{a}[/Tex]) and ([Tex]\vec{b}[/Tex]). Their cross product, denoted as ([Tex]\vec{a} \times \vec{b}[/Tex]), yields a vector that is perpendicular to both ([Tex]\vec{a}[/Tex]) and ([Tex]\vec{b}[/Tex]). This means that the resulting vector is normal to the plane formed by ([Tex]\vec{a}[/Tex]) and ([Tex]\vec{b}[/Tex]).

Mathematically, if the vectors ([Tex]\vec{a}[/Tex]) and ([Tex]\vec{b}[/Tex]) are perpendicular, their cross product is:

[Tex]\vec{a} \times \vec{b}[/Tex] = |[Tex]\vec{a}[/Tex]|·|[Tex]\vec{b}[/Tex]|·sin θ·[Tex]\vec{n}[/Tex]

where,

• [Tex]|\vec{a}|[/Tex] and [Tex]|\vec{b}|[/Tex] are the magnitudes of ([Tex]\vec{a}[/Tex]) and ([Tex]\vec{b}[/Tex]) respectively
• (θ) is angle between the two vectors
• ([Tex]\vec{n}[/Tex]) is a unit vector perpendicular to the plane formed by ([Tex]\vec{a}[/Tex]) and ([Tex]\vec{b}[/Tex])

For example, two perpendicular vectors ([Tex]\vec{a}[/Tex] = ❬ 1, 0, 0 ❭) and ([Tex]\vec{b}[/Tex] = ❬ 0, 1, 0 ❭). Their cross product is:

[Tex]\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix}[/Tex]

= ([Tex]\hat{i}[/Tex] · (0 · 0) – [Tex]\hat{j}[/Tex] · (1 · 0) + [Tex]\hat{k}[/Tex] · (1 · 0))

= [Tex]\hat{i}[/Tex] · 0 – [Tex]\hat{j}[/Tex] · 0 + [Tex]\hat{k}[/Tex] · 0

= ❬0, 0, 0❭

The resulting vector (❬ 0, 0, 0 ❭) is a zero vector, indicating that it is perpendicular to both ([Tex]\vec{a}[/Tex]) and ([Tex]\vec{b}[/Tex]), as expected when dealing with perpendicular vectors.

Cross Product of Parallel Vectors

Cross product of parallel vectors is always zero. This is because parallel vectors lie on the same line or are collinear, and the angle between them is either 0 or 180 degrees. In both cases, the sine of the angle is zero, resulting in a cross product of zero. Mathematically, if ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]) are parallel vectors, then:

[Tex]\vec{A} \times \vec{B} = |\vec{A}| \cdot |\vec{B}| \cdot \sin \theta = 0[/Tex]

where (θ) is the angle between the vectors.

Therefore, the cross product of parallel vectors is the zero vector. When computing the cross product of parallel vectors, the resulting vector is a zero vector. This implies that its magnitude is zero.

For example, consider two parallel vectors ([Tex]\vec{a}[/Tex] = ❬ 1, 2, 3 ❭) and ([Tex]\vec{b}[/Tex] = ❬ 2, 4, 6 ❭). Their cross product is:

[Tex]\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 4 & 6 \end{vmatrix} = \langle 0, 0, 0 \rangle[/Tex]

This result shows that the cross product of parallel vectors is a zero vector, indicating that the vectors are parallel and do not create a plane.

Right-Hand Rule Cross Product

Right-hand rule is a convention used to determine the direction of the resulting vector when performing a cross product.

1. Align your right hand’s index finger in the direction of the first vector ([Tex]\vec{A}[/Tex]) and your middle finger in the direction of the second vector ([Tex]\vec{B}[/Tex]).
2. Your thumb will then point in the direction of the resulting vector ([Tex]\vec{A} \times \vec{B}[/Tex]).

This convention ensures a consistent way to determine the direction of the resulting vector, especially when dealing with physical quantities like torque or magnetic fields.

For example, if vector ([Tex]\vec{A}[/Tex]) points towards the north and vector ([Tex]\vec{B}[/Tex]) points towards the east, then the resulting vector ([Tex]\vec{A} \times \vec{B}[/Tex]) will point upwards according to the right-hand rule.

Matrix Representation of Cross Product

Cross product of two vectors can also be represented using a matrix.

Let ([Tex]\vec{A}[/Tex] = ❬ a₁, a₂, a₃ ❭) and ([Tex]\vec{B}[/Tex] = ❬ b₁, b₂, b₃ ❭) be two vectors. Then, the cross product ([Tex]\vec{A} \times \vec{B}[/Tex]) can be represented as:

[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}[/Tex]

Expanding the determinant:

[Tex]\vec{A} \times \vec{B}[/Tex] = ([Tex]\hat{i}[/Tex] · (a₂ · b₃ – a₃ · b₂) – ([Tex]\hat{j}[/Tex] · (a₁ · b₃ – a₃ · b₁) + ([Tex]\hat{k}[/Tex] · (a₁ · b₂ – a₂ · b₁)

= ([Tex]\hat{i}[/Tex] · (a₂b₃ – a₃b₂) – ([Tex]\hat{j}[/Tex] · (a₁b₃ – a₃b₁) + ([Tex]\hat{k}[/Tex] · (a₁b₂ – a₂b₁)

= ❬ (a₂b₃ – a₃b₂), -(a₁b₃ – a₃b₁), (a₁b₂ – a₂b₁) ❭

Triple Cross Product

Triple cross product is a mathematical operation involving three vectors in three-dimensional space. Given three vectors ([Tex]\vec{A}[/Tex]), ([Tex]\vec{B}[/Tex]), and ([Tex]\vec{C}[/Tex]), the triple cross product is denoted as ([Tex]\vec{A} \times \vec{B}) \times \vec{C}[/Tex]).

To compute the triple cross product, we first calculate the cross product of vectors ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]), resulting in a new vector. Then, we take the cross product of this new vector and vector ([Tex]\vec{C}[/Tex]).

Mathematically, the triple cross product can be represented as:

[Tex](\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} – (\vec{A} \cdot \vec{B})\vec{C}[/Tex]

This operation is used in various mathematical and physical contexts, such as in rotational dynamics, electromagnetism, and fluid mechanics, where it helps determine angular momentum, magnetic fields, and vorticity, respectively.

Cross Product Properties

The various properties of cross product are listed below:

Orthogonality: The cross product of two non-zero vectors, ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]), results in a vector ([Tex]\vec{C}[/Tex]) that is perpendicular to both ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]).

Magnitude: The magnitude of the cross product, denoted as ([Tex]|\vec{A} \times \vec{B}|[/Tex]), is given by the product of the magnitudes of ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]) multiplied by the sine of the angle (θ) between them:

[Tex]|\vec{A} \times \vec{B}| = |\vec{A}| \cdot |\vec{B}| \cdot \sin \theta[/Tex]

Direction: The direction of the cross product is determined by the right-hand rule, ensuring that the resultant vector is orthogonal to the plane formed by ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]).

Anticommutativity: The cross product is anticommutative, meaning ([Tex]\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})[/Tex]).

Linearity: The cross product follows the distributive law and is linear, satisfying ([Tex]\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}[/Tex]).

Zero Cross Product for Parallel Vectors: If ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]) are parallel (or collinear), then ([Tex]\vec{A} \times \vec{B} = \vec{0}[/Tex])

Cross product of the unit vectors:

• [Tex]\vec{i}×\vec{i}~=~\vec{j}×\vec{j}~=~\vec{k}×\vec{k}~=~0[/Tex]

Cross product of the unit vectors:

• [Tex]\vec{i}×\vec{j}~=~\vec{k}\\\vec{j}×\vec{k}~=~\vec{i}\\\vec{k}×\vec{i}~=~\vec{j}[/Tex]

Application of Cross Product

Application of cross product are as follow:

• Torque calculation in physics and engineering, where it’s used to determine rotational force.
• Magnetic field calculation in electromagnetism, where it helps find the direction of magnetic fields around current-carrying wires.
• Angular momentum calculation in rotational motion problems, indicating the rotational motion’s intensity.
• Essential in vector algebra for determining perpendicular vectors and calculating areas of parallelograms.
• Crucial in computer graphics for determining surface normals, aiding in realistic lighting effects in 3D rendering.

Some Important Results

• Area of a Parallelogram with adjacent sides [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex] is [Tex]|\vec{a}×\vec{b}|[/Tex]
• Area of a Triangle with adjacent sides [Tex]\vec{a}[/Tex] and [Tex]\vec{b}[/Tex] is [Tex]1/2|\vec{a}×\vec{b}|[/Tex]

Examples on Cross Product

Example 1: Calculate the cross product of the vectors ([Tex]\vec{A}[/Tex] = ❬ 2, -1, 3 ❭) and ([Tex]\vec{B}[/Tex] = ❬ -3, 4, 1 ❭).

Solution:

To calculate cross product of two vectors ( [Tex]\vec{A}[/Tex]) and ( [Tex]\vec{B}[/Tex]), denoted as ([Tex]\vec{A} \times \vec{B}[/Tex]) ,we can use the following formula:

[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}[/Tex]

Given ( [Tex]\vec{A}[/Tex] = ❬ 2, -1, 3 ❭) and ( [Tex]\vec{B}[/Tex] = ❬ -3, 4, 1 ❭), we can substitute these values into the determinant:

[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ -3 & 4 & 1 \end{vmatrix}[/Tex]

Expanding determinant:

[Tex]\vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix} – \hat{j} \begin{vmatrix} 2 & 3 \\ -3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix} [/Tex]

Now, compute determinants:

[Tex]\begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix}[/Tex] = (-1)(1) – (3)(4) = -1 – 12 = -13

[Tex]\begin{vmatrix} 2 & 3 \\ -3 & 1 \end{vmatrix}[/Tex] = (2)(1) – (3)(-3) = 2 + 9 = 11

[Tex]\begin{vmatrix} 2 & -1 \\ -3 & 4 \end{vmatrix}[/Tex] = (2)(4) – (-1)(-3) = 8 – 3 = 5

Finally, substitute the determinants into the cross product:

[Tex]\vec{A} \times \vec{B} = \hat{i}(-13) – \hat{j}(11) + \hat{k}(5)[/Tex]

[Tex]\vec{A} \times \vec{B} = -13\hat{i} – 11\hat{j} + 5\hat{k}[/Tex]

So, cross product of ( [Tex]\vec{A}[/Tex]) and ( [Tex]\vec{B}[/Tex]) is ( [Tex]\vec{A} \times \vec{B}[/Tex] = ❬ -13, -11, 5 ❭).

Example 2: Determine the area of the parallelogram formed by the vectors ([Tex]\vec{A}[/Tex] = ❬ 3, 1, -2 ❭) and ([Tex]\vec{B}[/Tex] = ❬ 2, -1, 3 ❭).

Solution:

To determine the area of the parallelogram formed by two vectors ( [Tex]\vec{A}[/Tex] ) and ( [Tex]\vec{B}[/Tex] ), we can use the magnitude of their cross product ( [Tex]|\vec{A} \times \vec{B}|[/Tex] ). The formula for the magnitude of the cross product is:

[Tex]|\vec{A} \times \vec{B}| = |\vec{A}| \cdot |\vec{B}| \cdot \sin(\theta)[/Tex]

Where ( |[Tex]\vec{A}[/Tex]| ) and ( |[Tex]\vec{B}[/Tex]| ) are the magnitudes of vectors ( [Tex]\vec{A}[/Tex]) and ( [Tex]\vec{B}[/Tex]) respectively, and (θ) is the angle between the two vectors.

Given ( [Tex]\vec{A}[/Tex] = ❬ 3, 1, -2 ❭) and ( [Tex]\vec{B}[/Tex] = ❬ 2, -1, 3 ❭), we can compute their magnitudes:

[Tex]|\vec{A}| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{9 + 1 + 4} [/Tex]= √14

[Tex]|\vec{B}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9}[/Tex] = √14

Next, we calculate the cross product \( \vec{A} \times \vec{B} \) and its magnitude:

[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -1 & 3 \end{vmatrix}[/Tex]

= [Tex]\hat{i} \begin{vmatrix} 1 & -2 \\ -1 & 3 \end{vmatrix} – \hat{j} \begin{vmatrix} 3 & -2 \\ 2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 2 & -1 \end{vmatrix}[/Tex]

[Tex]= \hat{i}(1 \cdot 3 – (-2) \cdot (-1)) – \hat{j}(3 \cdot 3 – (-2) \cdot 2) + \hat{k}(3 \cdot (-1) – 1 \cdot 2) [/Tex]

[Tex]= \hat{i}(3 – 2) – \hat{j}(9 + 4) + \hat{k}(-3 – 2) [/Tex]

[Tex]= \hat{i} + \hat{j} – 5\hat{k} [/Tex]

The magnitude of ( [Tex]\vec{A} \times \vec{B}[/Tex] ) is:

[Tex]|\vec{A} \times \vec{B}| = \sqrt{1^2 + 1^2 + (-5)^2} = \sqrt{1 + 1 + 25}[/Tex] = √27

Now, we can calculate the area of the parallelogram:

Area = |[Tex]\vec{A} \times \vec{B}[/Tex]| = √27

Therefore, the area of the parallelogram formed by the vectors ( [Tex]\vec{A}[/Tex]) and ( [Tex]\vec{B}[/Tex] ) is √27 square units.

Example 3: Given that the cross product of two vectors ([Tex]\vec{A} \times \vec{B}[/Tex]) yields a vector ([Tex]\vec{C}[/Tex] = ❬ 4, 5, -3 ❭), if the magnitude of ([Tex]\vec{A}[/Tex]) is (|[Tex]\vec{A}[/Tex]| = 2) and the magnitude of ([Tex]\vec{B}[/Tex]) is (|[Tex]\vec{B}[/Tex]| = 3), find the angle between ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]).

Solution:

To find the angle between two vectors ( [Tex]\vec{A}[/Tex] ) and ( [Tex]\vec{B}[/Tex]) given their magnitudes and the cross product ( [Tex]\vec{C} = \vec{A} \times \vec{B} [/Tex]), we can use the formula for the magnitude of the cross product:

[Tex]|\vec{A} \times \vec{B}| = |\vec{A}| \cdot |\vec{B}| \cdot \sin(\theta)[/Tex]

Given that ( |[Tex]\vec{A}[/Tex]| = 2 ), ( [Tex]|\vec{B}|[/Tex] = 3 ), and ( [Tex]\vec{C}[/Tex] = ❬ 4, 5, -3 ❭), we can calculate the magnitude of ( [Tex]\vec{C}[/Tex]):

[Tex]|\vec{C}| = \sqrt{4^2 + 5^2 + (-3)^2} = \sqrt{16 + 25 + 9}[/Tex] = √50

Now, we can substitute the given values into the formula for the magnitude of the cross product:

[Tex]|\vec{A} \times \vec{B}| = |\vec{A}| \cdot |\vec{B}| \cdot \sin(\theta) [/Tex]

√50 = 2 · 3 · sin(θ)

√50 = 6 · sin(θ)

To find (θ), we divide both sides by 6:

[Tex]\sin(\theta) = \frac{\sqrt{50}}{6}[/Tex]

Now, we can find ( θ ) by taking the inverse sine (arcsine) of both sides:

[Tex]\theta = \arcsin\left(\frac{\sqrt{50}}{6}\right) [/Tex]

Using a calculator, we find:

[Tex]\theta \approx \arcsin\left(\frac{\sqrt{50}}{6}\right) \approx 57.19^\circ[/Tex]

Therefore, the angle between vectors ( [Tex]\vec{A}[/Tex] ) and ( [Tex]\vec{B}[/Tex] ) is approximately ( 57.19°).

Practice Questions on Cross Product

Q1: Given two vectors ( [Tex]\vec{A} [/Tex]= ❬ 1, 2, -3 ❭) and ( [Tex]\vec{B}[/Tex] = ❬ 2, -1, 4 ❭ ), find ( [Tex]\vec{A} \times \vec{B} [/Tex]).

Q2: Find the area of the parallelogram formed by the vectors ( [Tex]\vec{A}[/Tex] = ❬ 3, -2, 1 ❭) and ( [Tex]\vec{B}[/Tex] = ❬ 1, 4, -2 ❭).

Q3: If ( [Tex]\vec{A} \times \vec{B}[/Tex] = ❬ 2, -3, 5 ❭), and ( |[Tex]\vec{A}[/Tex]| = 4 ) and ( |[Tex]\vec{B}[/Tex]| = 3 ), find the angle between ( [Tex]\vec{A}[/Tex] ) and ( [Tex]\vec{B}[/Tex] ).

Q4: Determine the cross product of the vectors ( [Tex]\vec{A} [/Tex]= ❬ -2, 5, 1 ❭) and ( [Tex]\vec{B}[/Tex] = ❬ 3, -1, 2 ❭).

Q5: Given that the cross product of two vectors ( [Tex]\vec{A} \times \vec{B}[/Tex] ) yields ( [Tex]\vec{C}[/Tex] = ❬ 4, -3, 2 ❭), and ( |[Tex]\vec{A}[/Tex]| = 2 ) and ( |[Tex]\vec{B}[/Tex]| = 3 ), find the magnitude of ( [Tex]\vec{C}[/Tex] ).

FAQs on Cross Product

What is Cross Product?

A cross product is a mathematical operation between two vectors in three-dimensional space, resulting in a third vector that is perpendicular to both input vectors.

What is Cross Product Formula?

Formula for the cross product of two vectors ([Tex]\vec{A}[/Tex]) and ([Tex]\vec{B}[/Tex]) is given by:

[Tex]\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}[/Tex]

What is the difference between crosproduct and dot product?

The cross product of two vectors yields a vector that is perpendicular to both input vectors, while the dot product results in a scalar quantity representing the projection of one vector onto the other.

What is cross product of two vectors?

Cross product of two vectors, denoted as ([Tex]\vec{A} \times \vec{B}[/Tex]), is a vector that is perpendicular to both input vectors and is normal to the plane containing these vectors. Its magnitude is equal to the product of the magnitudes of the input vectors and the sine of the angle between them.

What is A×B Cross Product?

Notation (A × B) represents the cross product of two vectors (A) and (B). This operation yields a new vector that is perpendicular to both (A) and (B), and its magnitude is equal to the product of the magnitudes of (A) and (B) multiplied by the sine of the angle between them.