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Cross product is a binary operation on two vectors in a three-dimensional Euclidean vector space. The cross-product of two vectors is calculated using the right-hand rule. The right-hand rule is primarily the result of any two perpendicular vectors to the other two vectors. A cross-product can also be used to calculate the magnitude of the resulting vector.Different types of vectors are defined in vector algebra and is an important concept that falls under Vector Algebra. Addition, Subtraction, Multiplication and much more can be performed with the help of vectors. Two vectors having the same direction or the exact opposite direction, or if either one is having zero length then their cross product is zero.
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Key Takeaways: Vector, Scalar, Cross product, Right-hand thumb rule, Magnitude, Algebra, Binary operations, Addition, Subtraction, Multiplication
What is Cross Product?
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If A and B are two independent vectors, the result of their cross product (A × B) is perpendicular to both vectors and normal to the plane in which they are both located. It is represented by the following:
A × B = |A| |B| sin θ

Cross Product
For example, if we have two vectors in the X-Y plane, their cross product will result in a resultant vector in the direction of the Z-axis, which is perpendicular to the XY plane. Between the original vectors, the symbol is used. The k product, often known as the cross product of two vectors, looks like this:

Formula
If θ is the angle between the given two vectors A and B, then the formula for the cross product of vectors is given by:
A × B = |A| |B| sin θ
Or,
\(\vec{A} \times \vec{B} = \|\vec{A}\| \|\vec{B}\| \sin \theta \ \hat{n}\)
Here,
\(\vec{A}, \vec{B}\) are the two vectors.
\(\|\vec{A}\|, \|\vec{B}\|\) are the magnitudes of given vectors.
θ is the angle between two vectors and \(\hat{n}\) is the unit vector perpendicular to the plane containing the given two vectors, in the direction given by the right-hand rule.
\(\vec{A} \times \vec{B} = |\vec{a}| \ |\vec{b}| \sin \theta \ \hat{n}\)
\(\vec{A} \times \vec{B}=i\left(a_{2} b_{3}-a_{3} b_{2}\right)+j\left(a_{1} b_{3}-a_{3} b_{1}\right)+k\left(a_{1} b_{2}-a_{2} b_{1}\right)\)
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What is the Cross Product of Two Vectors?
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Cross product is a sort of vector multiplication, conducted between two vectors of different nature or kinds. Both the direction and magnitude of a vector are present. We can multiply two or more vectors via cross product and dot product. The resultant vector is called the cross product of two vectors or the vector product when two vectors are multiplied with each other and the product of the vectors is also a vector quantity. The resultant vector is perpendicular to the plane containing the two provided vectors.
Cross Product of Two Vectors Proof
Let \(A = a\hat{i} + b\hat{j} +c\hat{k}\) and \(B= d\hat{i} + e\hat{j} +f\hat{k}\) are the two given vectors, then
\(\|A \times B\|=\sqrt{(b f-e c)^{2}+(d c-f a)^{2}+(a e-b d)^{2}}\)
\(= \sqrt{b^{2} f^{2}+e^{2} c^{2}-2 b c e f+d^{2} c^{2}+f^{2} a^{2}-2 a c d f+a^{2} e^{2}+b^{2} d^{2}-2 a b d e}\)
\(= \sqrt{\left(a^{2}+b^{2}+c^{2}\right)\left(d^{2}+e^{2}+f^{2}\right)\left\{1-\frac{((a i+b j+c k) \cdot(d i+c j+f k))^{2}}{\left(a^{2}+b^{2}+c^{2}\right)\left(d^{2}+c^{2}+f^{2}\right)}\right\}}\)
\(=\|A\| \cdot\|B\| \sqrt{1-c o s^{2} \theta}\)
\(=\|A\| \cdot\|B\| \sin \theta\)
This demonstrates that the area of the parallelogram generated by the use of two specified vectors is the magnitude of the cross product.
What is a Triple Cross Product?
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The triple product is the product of three vectors. In other words, it is the cross product of one vector with the cross products of two other vectors.
A × (B × C) = (A . C) B − (A . B) C
(A × B) × C = −C × (A × B) = −(C . B) A + (C . A) B
Cross Product in Spherical Coordinates
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The resultant vector of two vectors' cross product is perpendicular to both vectors and normal to the plane in which they are located. In a three-dimensional system, spherical coordinates can be used to represent the same trick. Any vector in a three-dimensional system can be defined as the radical distance r, which is the distance from a fixed point to the origin, the polar angle, and the azimuth angle.
The direction of the cross product
The right-hand thumb rule determines the direction of the cross product of two non-zero parallel vectors a and b. Point your index finger along with vector a and your middle finger along vector b with your right hand, then point your thumb in the direction of the cross product.
Read More: Angle Between Two Vectors
Right-Hand Rule - Cross Product of Two Vectors
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The right-hand rule can be used to determine the direction of the unit vector. We can stretch our right hand so that the index finger points in the direction of the first vector and the middle finger points in the direction of the second vector using this rule. The direction or unit vector n is then indicated by the right hand's thumb. We may simply demonstrate that the cross-product of vectors is not commutative using the right-hand rule. If we have two vectors A and B, the right-hand rule diagram looks like this:

Right-hand Rule
Cross Product Properties
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We may utilize properties to find the cross-product of two vectors. To obtain the cross-product of two vectors, features such as anti-commutative property and zero vector property are important. Among other properties are the Jacobi property and the distributive property. The following are the properties of cross-product:
- \(\vec{i} \times \vec{j} = \vec{k}\)
- \(\vec{j} \times \vec{k} = \vec{i}\)
- \(\vec{k} \times \vec{i} = \vec{j}\)
- \(\vec{j} \times \vec{i} = -\vec{k}\)
- \(\vec{k} \times \vec{j} = -\vec{i}\)
- \(\vec{i} \times \vec{k} = -\vec{j}\)
Cross Product of Perpendicular Vectors
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The area of a rectangle with sides X and Y is represented by the cross product of two vectors, which is equal to the product of their magnitudes. The cross product formula becomes: if two vectors are perpendicular to one other.
θ = 90 degrees
We know that sin 90° = 1
So,
\(\vec{X} \times \vec{Y} = |\vec{X}| \cdot |\vec{Y}| \sin \theta\)
\(\vec{X} \times \vec{Y} = |\vec{X}| \cdot |\vec{Y}| \sin 90^{\circ}\)
\(\vec{X} \times \vec{Y} = |\vec{X}| \cdot |\vec{Y}|\),
which is equal to the area of a triangle.
Hence, the cross product of the perpendicular vectors become
\(\vec{X} \times \vec{Y} = |\vec{X}| \cdot |\vec{Y}|\)
Cross Product of Parallel Vectors
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If both vectors are parallel or opposite to each other, the cross-product of two vectors is zero. When two vectors are parallel or opposed to one another, their product is a zero vector. Two vectors have the same sense of direction.
θ = 90 degrees
As we know, sin 0° = 0 and sin 90° = 1
So,
\(\vec{X} \times \vec{Y} = |\vec{X}| \cdot |\vec{Y}| \sin \theta\)
\(\vec{X} \times \vec{Y} = |\vec{X}| \cdot |\vec{Y}| \sin 0^{\circ}\)
\(\vec{X} \times \vec{Y} = |\vec{X}| \cdot |\vec{Y}| \times 0\)
Hence, the cross product of the parallel vectors become
\(\vec{X} \times \vec{Y} = 0\), which is a unit vector.
Things To Remember
- The Cross product of two vectors is always a vector quantity.
- In a vector product, the resulting vector contains a negative sign if the order of vectors are changed.
- The direction is always perpendicular to the plane containing |A| and |B|
- The Cross product of any two linear vectors is always a null vector.
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Sample Questions
Ques. Find the cross product of the given two vectors: \(\vec{X} = 5\hat{i} + 6\hat{j} + 2\hat{k}\) and \(\vec{Y} = \hat{i} + \hat{j} + \hat{k}\) (3 marks)
Ans. Given,
\(\vec{X} = 5\hat{i} + 6\hat{j} + 2\hat{k}\)
\(\vec{Y} = \hat{i} + \hat{j} + \hat{k}\)
We have to write the given vectors in determinant form to find the cross product of two vectors. We can find the cross product of two vectors using the determinant form.
\(\vec{X} \times \vec{Y} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 5 & 6 & 2 \\ 1 & 1 & 1 \end{vmatrix}\)
By expanding,
\(\vec{X} \times \vec{Y} = (6-2)\vec{i} - (5-2)\vec{j} + (5-6)\vec{k}\)
Hence,
\(\vec{X} \times \vec{Y} = 4\vec{i} - 3\vec{j} - \vec{k}\)
Ques. Find \(\vec{a} \times \vec{b}\) if \(\vec{a} = 2\hat{i} + \hat{k}\) and \(\vec{b} = \hat{i} + \hat{j} + \hat{k}\) (3 marks)
Ans. Given,
\(\vec{a} = 2\hat{i} + \hat{k}\)
\(\vec{b} = \hat{i} + \hat{j} + \hat{k}\)
So,
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix}\)
\(=\hat{i}(0-1) - \hat{j}(2-1) + \hat{k}(2-0)\)
\(= -\hat{i}-\hat{j}+2\hat{k}\)
Ques. Two vectors have their scalar magnitude as |a| = 2√3 and |b| = 4, while the angle between the two vectors is 60°. Calculate the cross product of two vectors. (2 marks)
Ans. Given sin 60° = √3/2
The cross product of the two vectors is given by,
\(\vec{A} \times \vec{B} = |\vec{a}| \ |\vec{b}| \sin \theta \ \hat{n}\)
= 2√3 × 4 × √3/2 \(\hat{n}\)
= 12 \(\hat{n}\)
Ques. If \(\vec{a}\) = (2, -4, 4) and \(\vec{b}\) = (4, 0, 3), find the angle between them. (4 marks)
Ans.
\(\vec{a}\) = 2i – 4j + 4k
\(\vec{b}\) = 4i + 0j + 3k
The magnitude of \(\vec{a}\) is
|a| = √(22 + 42 + 42) = √36 = 6
The magnitude of \(\vec{b}\) is
|b| = √(42 + 02 + 32) = √25 = 5
As per the cross product formula, we have
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 4 \\ 4 & 0 & 3 \end{vmatrix}\)
\(=\hat{i}[(-4 \times 3) - (4 \times 0)] - \hat{j}[(3 \times 2) - (4 \times 4)] + \hat{k}[(2 \times 0) - (-4 \times 4)]\)
\(= -12\hat{i}+10\hat{j}+16\hat{k}\)
\(\vec{a} \times \vec{b}\) = (-12, 10, 16)
The length of the \(\vec{c}\) is
|c| = √(-122 + 102 + 162)
= √(144 + 100 + 256)
= √500
= 10√5
\(\vec{a} \times \vec{b} = |a| \ |b| \sin \theta\)
\(\sin \theta = \frac{\vec{a} \times \vec{b}}{|a| \ |b|}\)
\(\sin \theta = \frac{10\sqrt{5}}{6 \times 5}\)
\(\sin \theta = \frac{\sqrt{5}}{3}\)
θ = sin-1(√5/3)
θ = sin-1(0.74)
θ = 48°
Ques. Find the cross product of two vectors \(\vec{a}\) = (3, 4, 5) and \(\vec{b}\) = (7, 8, 9) (3 marks)
Ans. The cross product is given as,
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 7 & 8 & 9 \end{vmatrix}\)
\(=[(4 \times 9) - (5 \times 8)]\hat{i} - [(3 \times 9) - (5 \times 7)]\hat{j} + [(3 \times 8) - (4 \times 7)]\hat{k}\)
\(= (36-40)\hat{i}- (27-35)\hat{j}+(24-28)\hat{k}\)
\(= -4\hat{i}+8\hat{j}-4\hat{k}\)
Ques. Given \(\vec{a} = (\hat{i} + 3\hat{j} - 2\hat{k}) \times (-\hat{i} + 3\hat{k})\). Find the magnitude of \(\vec{a}\). (4 marks)
Ans.
\(\vec{a} = (\hat{i} + 3\hat{j} - 2\hat{k}) \times (-\hat{i} + 3\hat{k})\)
\(\vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix}\)
\(=\hat{i}(9-0) - \hat{j}(3-2) + \hat{k}(0+3)\)
\(= 9\hat{i}-\hat{j}+3\hat{k}\)
\(|\vec{a}| = \sqrt{9^2 + 1^2 + 3^2}\)
\(=\sqrt{91}\)
Ques. Find a vector of magnitude 9 which is perpendicular to both the vectors \(4\hat{i} - \hat{j} + 3\hat{k}\) and \(-2\hat{i} + \hat{j} -2\hat{k}\) (3 Marks)
Ans. Let \(\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}\)
\(\vec{b} = -2\hat{i} + \hat{j} -2\hat{k}\)
Then
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix}\)
\(=\hat{i}(2-3) - \hat{j}(-8+6) + \hat{k}(4-2)\)
\(|\vec{a} \times \vec{b}| = \sqrt{-1^2 + 2^2 + 2^2} = \sqrt{9} = 3\)
Required vector = \(9\ \frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}\)
\(= \frac{9}{3} (-\hat{i} + 2\hat{j} + 2\hat{k})\)
\(= (-3\hat{i} + 6\hat{j} + 6\hat{k})\)
Ques. Calculate the cross product between \(\vec{a}\) = (3, −3, 1) and \(\vec{b}\) = (4, 9, 2). (2 marks)
Ans. The cross product is
\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -3 & 1 \\ 4 & 9 & 2 \end{vmatrix}\)
\(=\hat{i}[(-3 \times 2) - (1 \times 9)] - \hat{j}[(3 \times 2) - (1 \times 4)] + \hat{k}[(3 \times 9) - (-3 \times 4)]\)
\(= -15\hat{i}-2\hat{j}+39\hat{k}\)
Ques. Calculate the area of the parallelogram spanned by the vectors \(\vec{a}\) = (3, −3, 1) and \(\vec{c}\) = (−12, 12, −4). (2 marks)
Ans.
\(\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -3 & 1 \\ -12 & 12 & -4 \end{vmatrix}\)
= (0, 0, 0)
Ques. If \(\vec{w}\) = (3, −1, 5) and \(\vec{v}\) = (0, 4, −2) compute \(\vec{v} \times \vec{w}\) (3 marks)
Ans.
\(\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & -2 \\ 3 & -1 & 5 \end{vmatrix}\)
\(=\hat{i}[(4 \times 5) - (-2 \times -1)] - \hat{j}[(0 \times 5) - (-2 \times 3)] + \hat{k}[(0 \times -1) - (4 \times 3)]\)
\(= 18\hat{i}-6\hat{j}-12\hat{k}\)
Ques. Find a vector that is orthogonal to the plane containing the points P = (3,0,1), Q = (4,-2,1) and R = (5,3,-1). (4 marks)
Ans. We first need two vectors that are both parallel to the plane. Using the points that we are given (all in the plane) we can quickly get quite a few vectors that are parallel to the plane. We’ll use the following two vectors.
\(\vec{PQ} = (1, -2, 0)\) and \(\vec{PR} = (2, 3, -2)\)
Now we know that the cross product of any two vectors will be orthogonal to the two original vectors. Since the two vectors from Step 1 are parallel to the plane (they actually lie in the plane in this case) we know that the cross product must then also be orthogonal, or normal, to the plane.
\(\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ 2 & 3 & -2 \end{vmatrix}\)
\(=\hat{i}[(-2 \times -2) - (0 \times 3)] - \hat{j}[(1 \times -2) - (0 \times 2)] + \hat{k}[(1 \times 3) - (-2 \times 2)]\)
\(= 4\hat{i}+2\hat{j}+7\hat{k}\)
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