The cross product is a binary operation on two vectors in three-dimensional space. It again results in a vector which is perpendicular to both vectors. The cross product of two vectors is calculated by the right-hand rule.
The right-hand rule is the resultant of any two vectors perpendicular to the other two vectors. Using the cross product, we can also find the magnitude of the resulting vector.
Definition
Vector product or cross product of two vectors
\(\begin{array}{l}\vec{A}\ \text{and}\ \vec{B}\ \text{is denoted by}\ \vec{A}\times \vec{B}\end{array} \)
and its resultant vector is perpendicular to the vectors \(\begin{array}{l}\vec{A}\ \text{and}\ \vec{B}.\end{array} \)
Key points to remember:
- The cross product of two vectors is always a vector quantity.
- In vector product, the resulting vector contains a negative sign if the order of vectors is changed.
- The direction of
\(\begin{array}{l}\vec A \times \vec B\end{array} \)
is always perpendicular to the plane containing \(\begin{array}{l}\vec{A}\end{array} \)
and \(\begin{array}{l}\vec{B}\end{array} \)
.
- The cross product of any two linear vectors is always a null vector.
Cross Product Formula
If θ is the angle between the given vectors, then the formula is given by
\(\begin{array}{l}A \times B = AB\ sin \theta\end{array} \)
\(\begin{array}{l}\text{Where}\ \hat n\ \text{is the unit vector.}\end{array} \)
Cross Product of Two Vectors
The cross product of two vectors is indicated like,
\(\begin{array}{l}\vec X \times \vec Y = |\vec X|. |\vec Y| \sin \theta\end{array} \)
Let us take any two vectors
\(\begin{array}{l}\vec X = x \vec i + y \vec j + z \vec k\end{array} \)
and \(\begin{array}{l}\vec y = a \vec i + b \vec j + c \vec k\end{array} \)
So, the cross product of these two vectors can be defined by matrices form, also called determinant form.
\(\begin{array}{l}\vec X \times \vec Y = \vec i (yc – zb) – \vec j (xc – za) + \vec k (xb – ya)\end{array} \)
Proof
Let A = ai + bj + ck and B = di + ej + fk are the two given vectors, then
\(\begin{array}{l}\left \| A\times B \right \|= \sqrt{(bf-ec)^{2}+ (dc-fa)^{2}+ (ae-bd)^{2}}\end{array} \)
\(\begin{array}{l}=\sqrt{b^{2}f^{2}+e^{2}c^{2}-2bcef+d^{2}c^{2}+ f^{2}a^{2}-2acdf + a^{2}e^{2}+ b^{2}d^{2}-2abde}\end{array} \)
\(\begin{array}{l}=\sqrt{(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})\left \{ 1-\frac{\left ( (ai+bj+ck).(di+ej+fk) \right )^{2}}{(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})} \right \}}\end{array} \)
\(\begin{array}{l}=\left \| A \right \|.\left \| B \right \|\sqrt{1- cos ^{2}\theta}\end{array} \)
\(\begin{array}{l}=\left \| A \right \|.\left \| B \right \|sin \theta\end{array} \)
This shows that the magnitude of the cross product is the area of the parallelogram, which is formed by the use of the two given vectors.
Direction of cross product
The direction of the cross product of two non-zero parallel vectors, a and b, is given by the right-hand thumb rule. In your right hand, point your index finger along vector a and point your middle finger along vector b, then the thumb gives the direction of the cross product.Â
Triple Cross Product
The product of three vectors is called the triple product; in other words, the cross product of one vector with the cross product of another two vectors.
\(\begin{array}{l}\text{If we have three vectors},\ \vec A, \vec B,\ \text{and}\ \vec {C},\end{array} \)
then the vector triple product is denoted as follows:
A × (B × C) = (A . C) B − (A . B) C
(A × B) × C = −C × (A × B) = −(C . B) A + (C . A) B
Cross Product Rules
Anti-Commutative Property
The cross product is an anti-commutative property
It means
\(\begin{array}{l}\vec A \times \vec B = – \vec B \times \vec A\end{array} \)
\(\begin{array}{l}\text{Let}\ \vec A = a_1 \vec i + a_2 \vec j + a_3 \vec k\ \text{and}\ \vec B = b_1 \vec i + b_2 \vec j + b_3 \vec k,\end{array} \)
\(\begin{array}{l}\vec A \times \vec B = \begin{vmatrix} i& j& k&\\ a_1& a_2& a_3& \\ b_1& b_2& b_3& \end{vmatrix}\end{array} \)
\(\begin{array}{l}=(a_2b_3 – b_2a_3)\vec i – (a_1b_3 – b_1a_3 ) \vec j + (a_1b_2 – a_2b_1) \vec k\end{array} \)
\(\begin{array}{l}\vec B \times \vec A = \begin{vmatrix} i& j& k&\\ b_1& b_2& b_3& \\ a_1& a_2& a_3& \end{vmatrix}\end{array} \)
\(\begin{array}{l}=(b_2a_3 – a_2b_3) \vec i – (b_1a_3 – a_1b_3) \vec j + (b_1a_2 – b_2a_1) \vec k\end{array} \)
\(\begin{array}{l}=-[(a_2b_3 – b_2a_3) \vec i – (a_1b_3 – b_1a_3) \vec j + (a_1b_2 – a_2b_1) \vec k]\end{array} \)
\(\begin{array}{l}=- \vec A \times \vec B\end{array} \)
Therefore,
\(\begin{array}{l}\vec A \times \vec B = – \vec B \times \vec A\end{array} \)
Distributive Property
The cross product has the distributive property over addition.
\(\begin{array}{l}\vec A \times (\vec B + \vec C) = \vec A \times \vec B + \vec A \times \vec C\end{array} \)
\(\begin{array}{l}\text{Let}\ \vec A = a_1 \vec i + a_2 \vec j + a_3 \vec k, \ \vec B = b_1 \vec i + b_2 \vec j + b_3 \vec k\ \text{and}\ \vec C = c_1 \vec i + c_2 \vec j + c_3 \vec k.\end{array} \)
\(\begin{array}{l}\vec A \times \vec B + \vec A \times \vec C = \begin{vmatrix} i& j& k&\\ a_1& a_2& a_3& \\ b_1& b_2& b_3& \end{vmatrix}+\begin{vmatrix} i& j& k&\\ a_1& a_2& a_3& \\ c_1& c_2& c_3& \end{vmatrix}\end{array} \)
\(\begin{array}{l}=(a_2b_3 – b_2a_3) \vec i – (a_1b_3 – b_1a_3 ) \vec j + (a_1b_2 – a_2b_1) \vec k + (a_2c_3 – c_2a_3) \vec i – (a_1c_3 – c_1a_3) \vec j + (a_1c_2 – a_2c_1) \vec k\end{array} \)
\(\begin{array}{l}=(a_2a_3 – b_2a_3 + a_2a_3 – c_2a_3) \vec i – (a_1b_3 – b_1a_3 – a_1c_3 + c_1a_3 ) \vec j + (a_1b_2 – a_2b_1 + a_1c_2 – a_2c_1) \vec k\end{array} \)
\(\begin{array}{l}=\vec A \times (\vec B + \vec C)\end{array} \)
Jacobi Property
Cross-product will satisfy the Jacobi property.
\(\begin{array}{l}\vec A \times (\vec B \times \vec C) + \vec B \times (\vec C \times \vec A) + \vec C \times (\vec A \times \vec B) = 0\end{array} \)
Zero Vector Property
\(\begin{array}{l}a \times b = 0\ if\ a = 0\ or\ b = 0\end{array} \)
\(\begin{array}{l}\text{Let}\ \vec a = 0 \vec i + 0 \vec j + 0 \vec k = 0\ \text{and}\ \vec b = b_1 \vec i + b_2 \vec j + b_3 \vec k.\end{array} \)
Then,
\(\begin{array}{l}a \times b = \begin{vmatrix} i& j& k&\\ 0& 0& 0& \\ b_1& b_2& b_3& \end{vmatrix}\end{array} \)
\(\begin{array}{l} = (0 – 0) \vec i – (0 – 0) \vec j + (0 – 0) \vec k\end{array} \)
= 0
Important results
1. The area of a parallelogram with adjacent sides is \(\begin{array}{l}\vec{a}\ \text{and}\ \vec{b}\ \text{is}\ \left | \vec{a} \times \vec{b}\right |\end{array} \)
Â
2. The area of a triangle whose adjacent sides are \(\begin{array}{l}\vec{a}\ \text{and}\ \vec{b}\ \text{is}\ \frac{1}{2}\left | \vec{a} \times \vec{b}\right |\end{array} \)
3. The area of a triangle ABC is \(\begin{array}{l}\frac{1}{2}\left | \vec{AB} \times \vec{AC}\right |\ \text{or}\ \frac{1}{2}\left | \vec{BC} \times \vec{BA}\right |\ \text{or}\ \frac{1}{2}\left | \vec{CB} \times \vec{CA}\right |\end{array} \)
4. The area of a parallelogram with diagonals is \(\begin{array}{l}\vec{a}\ \text{and}\ \vec{b}\ \text{is}\ \frac{1}{2}\left | \vec{a} \times \vec{b}\right |\end{array} \)
Â
5. The area of a plane quadrilateral ABCD is \(\begin{array}{l}\frac{1}{2}\left | \vec{AC} \times \vec{BD}\right |\end{array} \)
, AC and BD are the diagonals of it.
Cross Product in Spherical Coordinates
The resultant vector of the cross product of two vectors is perpendicular to both vectors, and it is normal to the plane in which they lie. We can use spherical coordinates in a 3-dimensional system to represent the same. We can define any vector in a 3-dimension system as, first, radical distance r, i.e. the distance of a fixed point to the origin, and second, the polar angle θ and third, azimuth angle ϕ.
We know that the transformation from Cartesian to spherical is
X = r sin θ cos ϕ
Y = r sin θ sin ϕ
Z = r cos θ
Also, Read:
Dot Product
Scalar Triple Product
Solved Examples
Example 1: \(\begin{array}{l}\text{Find}\ \vec{a}\times \vec{b}\ \text{if}\ \vec{a} = 2\hat{i}+\hat{k}\ \text{and}\ \vec{b} = \hat{i}+\hat{k}+\hat{k}.\end{array} \)
Solution:
Given \(\begin{array}{l}\vec{a} = 2\hat{i} + 0\hat{j}+\hat{k}\end{array} \)
and \(\begin{array}{l}\vec{b} = \hat{i}+\hat{k}+\hat{k}\end{array} \)
So \(\begin{array}{l}\vec{a}\times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2& 0&1 \\ 1 & 1& 1 \end{vmatrix}\end{array} \)
Â
= i(0 – 1) – j(2 – 1) + k(2 – 0)
\(\begin{array}{l}=-\hat{i}-\hat{j}+2\hat{k}\end{array} \)
Example 2: Given \(\begin{array}{l}\vec{a} = (\hat{i}+3\hat{j}-2\hat{k})\times (-\hat{i}+3\hat{k})\end{array} \)
\(\begin{array}{l}\text{Find the magnitude of}\ \vec{a}.\end{array} \)
Solution:
\(\begin{array}{l}\vec{a} = (\hat{i}+3\hat{j}-2\hat{k})\times (-\hat{i}+3\hat{k})\end{array} \)
\(\begin{array}{l}\vec{a} = \begin{vmatrix} \hat{i}& \hat{j} &\hat{k} \\ 1& 3 &-2 \\ -1& 0 & 3 \end{vmatrix}\end{array} \)
= i(9-0) -j(3-2)+k(0+3)
= 9i-j+3k
\(\begin{array}{l}\left | \vec{a} \right |= \sqrt{9^{2}+1^{2}+3^{2}}\end{array} \)
\(\begin{array}{l}=\sqrt{91}\end{array} \)
Example 3: Find a vector of magnitude 9 which is perpendicular to both the vectors 4i – j + 3k and -2i + j – 2k.
Solution:
Let \(\begin{array}{l}\vec{a} = 4\hat{i} – \hat{j} + 3\hat{k}\end{array} \)
Â
\(\begin{array}{l}\vec{b}=-2\hat{i} + \hat{j} – 2\hat{k}\end{array} \)
Â
Then \(\begin{array}{l}\vec{a}\times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j}& \hat{k}\\ 4& -1 &3 \\ -2 & 1& -2 \end{vmatrix}\end{array} \)
= i(2-3)-j(-8+6)+k(4-2)
\(\begin{array}{l}\left | \vec{a}\times \vec{b} \right | = \sqrt{-1^{2}+2^{2}+2^{2}} = \sqrt{9} = 3\end{array} \)
Required vector \(\begin{array}{l} = 9\frac{(\vec{a}\times \vec{b})}{\left | \vec{a}\times \vec{b}\right |}\end{array} \)
\(\begin{array}{l}=\frac{9}{3}(-\hat{i}+2\hat{j}+2\hat{k})\end{array} \)
\(\begin{array}{l}=(-3\hat{i}+6\hat{j}+6\hat{k})\end{array} \)
Example 4: A = i + 2j + 3k, B = 2i + 3j − 2k, C = i + j + k.
Find A × (B × C), if A × (B × C) = (A . C) B − (A . B) C
Solution:
A.C = (i + 2j + 3k) . (i + j + k) = 6
A.B = (i + 2j + 3k) . (2i + 3j − 2k) = 2
Then, A × (B × C) = 6(2i + 3j − 2k) − 2(i + j + k)
= 10i − 16j − 14k
Example 5: Find the unit vector to the two vectors:
\(\begin{array}{l}\vec{A} = 2\hat{i} + (-1)\hat{j} + 3\hat{k},\ \vec{B} = \hat{i} – 2\hat{j} + 2\hat{k}\end{array} \)
Solution:
\(\begin{array}{l}\text{Let}\ \vec{C}\ \text{be the unit vector.}\end{array} \)
Cross-product of A and B,
\(\begin{array}{l}\vec{A} \times \vec{B} = 4\hat{i} – \hat{j} – 3\hat{k}\end{array} \)
\(\begin{array}{l}\text{Now unit vector in the direction of}\ \vec{C}\ \text{is}\ \frac{\vec{C}}{|\vec{C}|}.\end{array} \)
\(\begin{array}{l}|\vec{C}|=\sqrt{26}\end{array} \)
Therefore desired unit vector is (1/√26) (4i – j – 3k)
Example 6: What is the number of vectors of unit length perpendicular to vectors a = (1, 1, 0) and b = (0, 1, 1)?
Solution:
The vector perpendicular to a and b is
\(\begin{array}{l}\mathbf{a}\times \mathbf{b}=\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{matrix} \right|=\mathbf{i}-\mathbf{j}+\mathbf{k}\\\end{array} \)
Since the length of this vector is √3, the unit vector perpendicular to a and b is:
\(\begin{array}{l} \pm \frac{a\times b}{|a\times b|}\\=\pm \frac{1}{\sqrt{3}}(\mathbf{i}-\mathbf{j}+\mathbf{k})\\\end{array} \)
Hence, there are two such vectors.
Frequently Asked Questions
Q1
What do you mean by the cross product of two vectors?
The cross product of two vectors, vector A and vector B, is denoted by A×B = |A||B| sin θ n, where n is the unit vector.
Q2
Is the cross product commutative?
No, the cross product is anticommutative. A×B = -B×A.
Q3
What do you mean by dot product of 2 vectors?
Dot product is equal to the product of the magnitudes of the two vectors and the cosine of the angle between the two vectors.
Comments