Vector product is also known as cross product. Cross product of two vectors will give the resultant as a vector. Cross product of two vectors is calculated by right hand rule. Right hand rule is nothing but the resultant of any two vectors is perpendicular to the other two vectors. Using cross product, we can find the magnitude of the resultant vector. The usage of right hand rule is to find the direction of the resultant.

Cross product is the binary operation on two vectors in three dimensional space. It again results in a vector which is perpendicular to both the vectors. Also called cross product of two vectors, vector product is zero, if both vector are parallel. Vector product of two vectors a and b is denoted by a x b. It's resultant vector is perpendicular to a and b.

Related Calculators | |

Cross Product Calculator | Cross Multiplication Calculator |

Calculating Dot Product | Product Rule Calculator |

Vector product or cross product of two vectors $\vec{A}$ and $\vec{B}$ is denoted by $A \times B$ and its resultant vector is perpendicular to the vectors $\vec{A}$ and $\vec{B}$.

### Cross Product Formula:

If $\theta$ is the angle between the given vectors, then the formula for cross product of vectors is given by

$A \times B = AB \sin \theta$

### Cross Product of Two Vectors

Cross product of two vectors is indicated like the following $\vec X \times \vec Y = |\vec X| . |\vec Y| \sin \theta$

Let us take any two vectors $\vec X = x \vec i + y \vec j + z \vec k$ and $\vec y = a \vec i + b \vec j + c \vec k$

So, cross product of these two vectors can be defined by matrices form. This is called determinant form.

$\vec X \times \vec Y = \vec i (yc - zb) - \vec j (xc - za) + \vec k (xb - ya)$. To find the cross product of two matrices, the necessary condition is that the given matrices is in the form of column matrix. If it is not, then the cross product is not defined.

Let A = (a_{1}, a_{2}, a_{3})^{T} and B = (b_{1}, b_{2}, b_{3})^{T} are two vectors in three dimension vector matrices, then

A x B = (a_{2}b_{3} - b_{2}a_{3}, b_{1}a_{3} - a_{1}b_{3}, a_{1}b_{2} - a_{2}b_{1})^{T}

A x B = $\begin{pmatrix}

a_{2}b_{3}-b_{2}a_{3}

\\b_{1}a_{3}-a_{1}b_{3}

\\a_{1}b_{2}-a_{2}b_{1}

\end{pmatrix}$

Let $A = ai + bj + ck$ and $B = di + ej + fk$ are the two given vectors, then

$\left \| A\times B \right \|= \sqrt{(bf-ec)^{2}+ (dc-fa)^{2}+ (ae-bd)^{2}}$

= $\sqrt{b^{2}f^{2}+e^{2}c^{2}-2bcef+d^{2}c^{2}+ f^{2}a^{2}-2acdf + a^{2}e^{2}+ b^{2}d^{2}-2abde}$

= $\sqrt{(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})\left \{ 1-\frac{\left ( (ai+bj+ck).(di+ej+fk) \right )^{2}}{(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})} \right \}}$

= $\left \| A \right \|.\left \| B \right \|\sqrt{1- cos ^{2}\theta}$

= $\left \| A \right \|.\left \| B \right \|sin \theta$

This shows that the magnitude of the cross product is the area of the parallelogram which is formed by the use of given two vectors.

The product of three vectors is called the triple product. If we calculate the cross product of three vectors, then it is known as the triple cross product. In other words, the cross product of one vector with the cross product of another two vectors is called the vector triple product.

If we have three vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ then the vector triple product is denoted as follows:

$A \times (B \times C)= (A.C)B -(A.B) C$

$(A \times B)\times C = -C\times (A\times B) = -(C.B)A + (C.A)B$### Solved Example

**Question: **$A = i + 2j + 3k$, $B = 2i + 3j - 2k$, $C = i + j + k$. Find $A \times (B \times C) = (A . C)B - (A . B)C$

** Solution: **
Given below are some of the properties of cross product.

### Anti-Commutative Property

The Cross product is an anti commutative property

It means $\vec A \times \vec B = - \vec B \times \vec A$

Let $\vec A = a_1 \vec i + a_2 \vec j + a_3 \vec k$ and $\vec B = b_1 \vec i + b_2 \vec j + b_3 \vec k$

$\vec A \times \vec B$ = $\begin{vmatrix}

i& j& k&\\

a_1& a_2& a_3& \\

b_1& b_2& b_3&

\end{vmatrix}$

= $(a_2b_3 - b_2a_3)\vec i - (a_1b_3 - b_1a_3 ) \vec j + (a_1b_2 - a_2b_1) \vec k$

$\vec B \times \vec A$ = $\begin{vmatrix}

i& j& k&\\

b_1& b_2& b_3& \\

a_1& a_2& a_3&

\end{vmatrix}$

= $(b_2a_3 - a_2b_3) \vec i - (b_1a_3 - a_1b_3) \vec j + (b_1a_2 - b_2a_1) \vec k$

= $-[(a_2b_3 - b_2a_3) \vec i - (a_1b_3 - b_1a_3) \vec j + (a_1b_2 - a_2b_1) \vec k]$

= $- \vec A \times \vec B$

$\vec A \times \vec B = - \vec B \times \vec A$

### Distributive Property

The cross product is having the distributive property over addition.

$\vec A \times (\vec B + \vec C) = \vec A \times \vec B + \vec A \times \vec C$

Let $\vec A = a_1 \vec i + a_2 \vec j + a_3 \vec k, \vec B = b_1 \vec i + b_2 \vec j + b_3 \vec k$ and $\vec C = c_1 \vec i + c_2 \vec j + c_3 \vec k$

$\vec A \times \vec B + \vec A \times \vec C$ = $\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}$

= $(a_2b_3 - b_2a_3) \vec i - (a_1b_3 - b_1a_3 ) \vec j + (a_1b_2 - a_2b_1) \vec k + (a_2c_3 - c_2a_3) \vec i - (a_1c_3 - c_1a_3) \vec j + (a_1c_2 - a_2c_1) \vec k$

= $(a_2a_3 - b_2a_3 + a_2a_3 - c_2a_3) \vec i - (a_1b_3 - b_1a_3 - a_1c_3 + c_1a_3 ) \vec j + (a_1b_2 - a_2b_1 + a_1c_2 - a_2c_1) \vec k$

= $\vec A \times (\vec B + \vec C)$

### Jacobi Property

Cross product will satisfy the Jacobi property.

$\vec A \times (\vec B \times \vec C) + \vec B \times (\vec C \times \vec A) + \vec C \times (\vec A \times \vec B) = 0$

### Zero Vector Property

$a \times b = 0$ if $a = 0$ or $b = 0$

Let $\vec a = 0 \vec i + 0 \vec j + 0 \vec k = 0$ and $\vec b = b_1 \vec i + b_2 \vec j + b_3 \vec k$

Then, $a \times b$ = $\begin{vmatrix}

i& j& k&\\

0& 0& 0& \\

b_1& b_2& b_3&

\end{vmatrix}$

= $(0 - 0) \vec i - (0 - 0) \vec j + (0 - 0) \vec k$

= $0$ We can find the direction of the unit vector n with the help of right hand rule. In this rule, we can stretch our right hand in a way that the index finger of the right hand in the direction of first vector and the middle finger is in the direction of second vector. Then, the thumb of the right hand indicate the direction or unit vector n. With the help of the right hand rule, we can easily show that the cross product of vectors is not commutative. If we have two vectors A and B, then the diagram for the right hand rule is as follows:

Cross product of two vectors is equal to the product of their magnitude which also represents the area of rectangle with sides X and Y. If vectors are**perpendicular** to
each other, then

** **

$\Theta $ = 0 degree

**$\Theta $ = **$\pi$

$\vec X \times \vec Y = |\vec X| . |\vec Y| \times 0$

$\vec X \times \vec Y$ = 0 vector If we have two vectors $\vec{A}$ and $\vec{B}$ in such a way that $\vec{A} = A_x + A_y + A_z$ and $\vec{B} = B_x + B_y + B_z$, then the magnitude of these two vectors are given by the formula $A = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}$ and $B = \sqrt{B_{x}^{2} + B_{y}^{2} + B_{z}^{2}}$

Then, the magnitude of the cross product of these two vectors is given by $\left | A x B \right |= \left | A \right |\left | B \right | \sin \theta$

where $\theta$ is the angle between the given two vectors and it gives the positive area of the parallelogram formed by these two vectors.

In mathematics, we can discuss the cross product on two vectors in three dimension coordinate system. The resultant vectors of cross product of two vectors is perpendicular to both of them and it is normal to the plane in which they lie. We can use spherical coordinate in 3-dimension system. We can define any vector in 3-dimension system as, first radical distance r i.e. the distance of a fixed point to the origin, second the polar angle $\theta$ and third azimuth angle $\phi$.

We know that the transformation from Cartesian to spherical is $x = r \sin \theta \cos \phi$, $y = r \sin \theta \sin \phi$, $z = r \cos \theta$

If we have A(a, b, c) and B(d, e, f) are two coordinates in 3-dimension system, then we want to apply the cross product on these two A and B. Then, to find the coordinate of the cross product AxB,

Here, we assume that the vectors A and B are the function of x. We can also show this in the determinant form as follows:

$\frac{d}{dx}\begin{vmatrix}

i &j &k \\

A_{x} &A_{y} &A_{z} \\

B_{x} &B_{y} &B_{z}

\end{vmatrix}

= \begin{vmatrix}

i &j &k \\

\frac{dA_{x}}{dx} &\frac{dA_{y}}{dx} &\frac{dA_{y}}{dx} \\

B_{x} &B_{y} &B_{z}

\end{vmatrix} + \begin{vmatrix}

i &j &k \\

A_{x} &A_{y} &A_{z} \\

\frac{dB_{x}}{dx} &\frac{dB_{x}}{dx} &\frac{dB_{x}}{dx}

\end{vmatrix}$

The value of vector product or cross product is zero at $\theta$ = 0, 180^{0} since the values of the sin 0 = 0 and sin 180^{0} = 0. This property is very useful to find the product of unit vectors. If i, j and k are the unit vectors of two given vectors, then we have

$i \times i = j \times j = k \times k = 0$

Now, the product of two different unit vectors are

$i \times j = k$, $ j\times k = i$, $k \times i = j$

$j \times i = -k$, $k\times j = -i$, $i \times k = -j$

To know about the sign of product of unit vectors, we form a pair sequence of i, j and k as we move from left to right or clockwise like i x j, j x k and k x i and get the sign of these pairs is positive. And, the other pairs have the negative sign.

With the help of the determinant method of vectors, we can calculate the cross product of two given vectors. For this, if we have two vectors

$A = a_{1}i + a_{2}j + a_{3}k, B = b_{1}i + b_{2}j + b_{3}k$

Then,

$A \times B = \begin{vmatrix}

i &j & k\\

a_{1} &a_{2} &a_{3} \\

b_{1} &b_{2} &b_{3}

\end{vmatrix}$

= $ i\begin{vmatrix}

a_{2} &a_{3} \\

b_{2} &b_{3}

\end{vmatrix}-j\begin{vmatrix}

a_{1} &a_{3} \\

b_{1} &b_{3}

\end{vmatrix}+k\begin{vmatrix}

a_{1} &a_{2} \\

b_{1} &b_{2}

\end{vmatrix}$

= $i(a_{2}b_{3} - a_{3}b_{2}) - j(a_{1}b_{3} - a_{3}b_{1}) + k(a_{1}b_{2} - a_{2}b_{1})$

Cross product of vectors is not associative. To prove this let us assume A = a_{1}i + a_{2}j + a_{3}k, B = b_{1}i + b_{2}j + b_{3}k and C = c_{1}i + c_{2}j + c_{3}k are three vectors.

Now, we have to show that

$(A \times B) \times C \neq A \times (B \times C)$

Then, $A \times B$ = $\begin{vmatrix}

i &j &k \\

a_{1} &a_{2} &a_{3} \\

b_{1} &b_{2} &b_{3}

\end{vmatrix}$

= $i(a_{2}b_{3} - a_{3}b_{2}) + j(a_{3}b_{1} - a_{1}b_{3}) + k(a_{1}b_{2} - a_{2}b_{1})$

Now, we calculate

$(A \times B) \times C = \begin{vmatrix}

i &j &k \\

(a_{2}b_{3}-a_{3}b_{2}) &(a_{3}b_{1}-a_{1}b_{3}) &(a_{1}b_{2}-a_{2}b_{1}) \\

c_{1} &c_{2} &c_{3}

\end{vmatrix}$

Again, $B \times C$ = $\begin{vmatrix}

i &j &k \\

b_{1} &b_{2} &b_{3} \\

c_{1} &c_{2} &c_{3}

\end{vmatrix}$

= $i(b_{2}c_{3} - b_{3}c_{2}) + j(c_{1}b_{3} - b_{1}c_{3}) + k(b_{1}c_{2} - b_{2}c_{1})$

Again, we find

$A \times (B \times C) = \begin{vmatrix}

i &j &k \\

a_{1} &a_{2} &a_{3} \\

(b_{2}c_{3}-b_{3}c_{2}) &(c_{1}b_{3}-b_{1}c_{3}) &(b_{1}c_{2}-b_{2}c_{1})

\end{vmatrix}$

So, it is clear that

$\begin{vmatrix}

i &j &k \\

b_{1} &b_{2} &b_{3} \\

c_{1} &c_{2} &c_{3}

\end{vmatrix}

= i(b_{2}c_{3} - b_{3}c_{2}) + j(c_{1}b_{3} - b_{1}c_{3}) + k(b_{1}c_{2} - b_{2}c_{1})$

$A \times (B \times C) = \begin{vmatrix}

i &j &k \\

a_{1} &a_{2} &a_{3} \\

(b_{2}c_{3}-b_{3}c_{2}) &(c_{1}b_{3}-b_{1}c_{3}) &(b_{1}c_{2}-b_{2}c_{1})

\end{vmatrix}$

So, it is clear that $(A \times B) \times C \neq A \times (B \times C)$

Given below are some of the examples on cross product of two vectors.### Solved Examples

**Question 1: **Find the cross product of the following vectors. $\vec X = 5 \vec i + 6 \vec j + 2 \vec k$ and $\vec Y = \vec i + \vec j + \vec k$

** Solution: **
**Question 2: **Find the cross product of the following vectors. $\vec X = 4 \vec i - 5 \vec j + 2 \vec k$ and $\vec Y = 2 \vec i + 4 \vec j + 5 \vec k$

** Solution: **
**Question 3: **Find the cross product of the following vectors. $\vec X = 6 \vec j + 2 \vec k$ and $\vec Y = \vec i - 7 \vec j + \vec k$

** Solution: **

If $\theta$ is the angle between the given vectors, then the formula for cross product of vectors is given by

$A \times B = AB \sin \theta$

Cross product of two vectors is indicated like the following $\vec X \times \vec Y = |\vec X| . |\vec Y| \sin \theta$

Let us take any two vectors $\vec X = x \vec i + y \vec j + z \vec k$ and $\vec y = a \vec i + b \vec j + c \vec k$

So, cross product of these two vectors can be defined by matrices form. This is called determinant form.

$\vec X \times \vec Y = \vec i (yc - zb) - \vec j (xc - za) + \vec k (xb - ya)$. To find the cross product of two matrices, the necessary condition is that the given matrices is in the form of column matrix. If it is not, then the cross product is not defined.

Let A = (a

A x B = (a

A x B = $\begin{pmatrix}

a_{2}b_{3}-b_{2}a_{3}

\\b_{1}a_{3}-a_{1}b_{3}

\\a_{1}b_{2}-a_{2}b_{1}

\end{pmatrix}$

Let $A = ai + bj + ck$ and $B = di + ej + fk$ are the two given vectors, then

$\left \| A\times B \right \|= \sqrt{(bf-ec)^{2}+ (dc-fa)^{2}+ (ae-bd)^{2}}$

= $\sqrt{b^{2}f^{2}+e^{2}c^{2}-2bcef+d^{2}c^{2}+ f^{2}a^{2}-2acdf + a^{2}e^{2}+ b^{2}d^{2}-2abde}$

= $\sqrt{(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})\left \{ 1-\frac{\left ( (ai+bj+ck).(di+ej+fk) \right )^{2}}{(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})} \right \}}$

= $\left \| A \right \|.\left \| B \right \|\sqrt{1- cos ^{2}\theta}$

= $\left \| A \right \|.\left \| B \right \|sin \theta$

This shows that the magnitude of the cross product is the area of the parallelogram which is formed by the use of given two vectors.

The product of three vectors is called the triple product. If we calculate the cross product of three vectors, then it is known as the triple cross product. In other words, the cross product of one vector with the cross product of another two vectors is called the vector triple product.

If we have three vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ then the vector triple product is denoted as follows:

$A \times (B \times C)= (A.C)B -(A.B) C$

$(A \times B)\times C = -C\times (A\times B) = -(C.B)A + (C.A)B$

$A . C = (i + 2j + 3k).(i + j + k) = 6$

$A . B = (i + 2j + 3k).(2i + 3j - 2k) = 2$

Then, $A \times (B \times C) = 6(2i + 3j - 2k) - 2(i + j + k)$

= $10i - 16j - 14k$

$A . B = (i + 2j + 3k).(2i + 3j - 2k) = 2$

Then, $A \times (B \times C) = 6(2i + 3j - 2k) - 2(i + j + k)$

= $10i - 16j - 14k$

The Cross product is an anti commutative property

It means $\vec A \times \vec B = - \vec B \times \vec A$

Let $\vec A = a_1 \vec i + a_2 \vec j + a_3 \vec k$ and $\vec B = b_1 \vec i + b_2 \vec j + b_3 \vec k$

$\vec A \times \vec B$ = $\begin{vmatrix}

i& j& k&\\

a_1& a_2& a_3& \\

b_1& b_2& b_3&

\end{vmatrix}$

= $(a_2b_3 - b_2a_3)\vec i - (a_1b_3 - b_1a_3 ) \vec j + (a_1b_2 - a_2b_1) \vec k$

$\vec B \times \vec A$ = $\begin{vmatrix}

i& j& k&\\

b_1& b_2& b_3& \\

a_1& a_2& a_3&

\end{vmatrix}$

= $(b_2a_3 - a_2b_3) \vec i - (b_1a_3 - a_1b_3) \vec j + (b_1a_2 - b_2a_1) \vec k$

= $-[(a_2b_3 - b_2a_3) \vec i - (a_1b_3 - b_1a_3) \vec j + (a_1b_2 - a_2b_1) \vec k]$

= $- \vec A \times \vec B$

$\vec A \times \vec B = - \vec B \times \vec A$

The cross product is having the distributive property over addition.

$\vec A \times (\vec B + \vec C) = \vec A \times \vec B + \vec A \times \vec C$

Let $\vec A = a_1 \vec i + a_2 \vec j + a_3 \vec k, \vec B = b_1 \vec i + b_2 \vec j + b_3 \vec k$ and $\vec C = c_1 \vec i + c_2 \vec j + c_3 \vec k$

$\vec A \times \vec B + \vec A \times \vec C$ = $\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}$

= $(a_2b_3 - b_2a_3) \vec i - (a_1b_3 - b_1a_3 ) \vec j + (a_1b_2 - a_2b_1) \vec k + (a_2c_3 - c_2a_3) \vec i - (a_1c_3 - c_1a_3) \vec j + (a_1c_2 - a_2c_1) \vec k$

= $(a_2a_3 - b_2a_3 + a_2a_3 - c_2a_3) \vec i - (a_1b_3 - b_1a_3 - a_1c_3 + c_1a_3 ) \vec j + (a_1b_2 - a_2b_1 + a_1c_2 - a_2c_1) \vec k$

= $\vec A \times (\vec B + \vec C)$

Cross product will satisfy the Jacobi property.

$\vec A \times (\vec B \times \vec C) + \vec B \times (\vec C \times \vec A) + \vec C \times (\vec A \times \vec B) = 0$

$a \times b = 0$ if $a = 0$ or $b = 0$

Let $\vec a = 0 \vec i + 0 \vec j + 0 \vec k = 0$ and $\vec b = b_1 \vec i + b_2 \vec j + b_3 \vec k$

Then, $a \times b$ = $\begin{vmatrix}

i& j& k&\\

0& 0& 0& \\

b_1& b_2& b_3&

\end{vmatrix}$

= $(0 - 0) \vec i - (0 - 0) \vec j + (0 - 0) \vec k$

= $0$ We can find the direction of the unit vector n with the help of right hand rule. In this rule, we can stretch our right hand in a way that the index finger of the right hand in the direction of first vector and the middle finger is in the direction of second vector. Then, the thumb of the right hand indicate the direction or unit vector n. With the help of the right hand rule, we can easily show that the cross product of vectors is not commutative. If we have two vectors A and B, then the diagram for the right hand rule is as follows:

Cross product of two vectors is equal to the product of their magnitude which also represents the area of rectangle with sides X and Y. If vectors are

** $\Theta $ = 90 degree **

$\sin 90^{0} = 1$

$\vec X \times \vec Y = |\vec X| . |\vec Y| \sin \theta$

$\vec X \times \vec Y = |\vec X| . |\vec Y| \sin 90^{0}$

= $|X| |Y|$

= Area of rectangle with sides X and Y.

$\vec X \times \vec Y = |X| |Y|$

$\Theta $ = 0 degree

** **** **$\sin 0^{0} = 0 = \sin \pi$

$\vec X \times \vec Y = |\vec X| . |\vec Y| \times 0$

$\vec X \times \vec Y$ = 0 vector If we have two vectors $\vec{A}$ and $\vec{B}$ in such a way that $\vec{A} = A_x + A_y + A_z$ and $\vec{B} = B_x + B_y + B_z$, then the magnitude of these two vectors are given by the formula $A = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}$ and $B = \sqrt{B_{x}^{2} + B_{y}^{2} + B_{z}^{2}}$

Then, the magnitude of the cross product of these two vectors is given by $\left | A x B \right |= \left | A \right |\left | B \right | \sin \theta$

where $\theta$ is the angle between the given two vectors and it gives the positive area of the parallelogram formed by these two vectors.

In mathematics, we can discuss the cross product on two vectors in three dimension coordinate system. The resultant vectors of cross product of two vectors is perpendicular to both of them and it is normal to the plane in which they lie. We can use spherical coordinate in 3-dimension system. We can define any vector in 3-dimension system as, first radical distance r i.e. the distance of a fixed point to the origin, second the polar angle $\theta$ and third azimuth angle $\phi$.

We know that the transformation from Cartesian to spherical is $x = r \sin \theta \cos \phi$, $y = r \sin \theta \sin \phi$, $z = r \cos \theta$

If we have A(a, b, c) and B(d, e, f) are two coordinates in 3-dimension system, then we want to apply the cross product on these two A and B. Then, to find the coordinate of the cross product AxB,

- Avoid the first coordinate of A and B, multiply b with f and c with e and get bf and ce. Then, subtract ce from bf i.e. bf - ce.
- Now, avoid the second coordinate of A and B, multiply d with c and a with f and get dc and af. Then, subtract af from dc i.e. dc - af.
- Lastly, avoid the third coordinate of A and B, multiply a with e and b with d and get ae and bd. Then, subtract bd from ae i.e. ae - bd.
- After adding all these, we get the resultant coordinate as (bf - ce, dc - af, ae - bd).

- Find the cross product of the given vectors and after that differentiate it.
- Use the product rule of differentiation to find the derivative of the cross product

Here, we assume that the vectors A and B are the function of x. We can also show this in the determinant form as follows:

$\frac{d}{dx}\begin{vmatrix}

i &j &k \\

A_{x} &A_{y} &A_{z} \\

B_{x} &B_{y} &B_{z}

\end{vmatrix}

= \begin{vmatrix}

i &j &k \\

\frac{dA_{x}}{dx} &\frac{dA_{y}}{dx} &\frac{dA_{y}}{dx} \\

B_{x} &B_{y} &B_{z}

\end{vmatrix} + \begin{vmatrix}

i &j &k \\

A_{x} &A_{y} &A_{z} \\

\frac{dB_{x}}{dx} &\frac{dB_{x}}{dx} &\frac{dB_{x}}{dx}

\end{vmatrix}$

The value of vector product or cross product is zero at $\theta$ = 0, 180

$i \times i = j \times j = k \times k = 0$

Now, the product of two different unit vectors are

$i \times j = k$, $ j\times k = i$, $k \times i = j$

$j \times i = -k$, $k\times j = -i$, $i \times k = -j$

To know about the sign of product of unit vectors, we form a pair sequence of i, j and k as we move from left to right or clockwise like i x j, j x k and k x i and get the sign of these pairs is positive. And, the other pairs have the negative sign.

With the help of the determinant method of vectors, we can calculate the cross product of two given vectors. For this, if we have two vectors

$A = a_{1}i + a_{2}j + a_{3}k, B = b_{1}i + b_{2}j + b_{3}k$

Then,

$A \times B = \begin{vmatrix}

i &j & k\\

a_{1} &a_{2} &a_{3} \\

b_{1} &b_{2} &b_{3}

\end{vmatrix}$

= $ i\begin{vmatrix}

a_{2} &a_{3} \\

b_{2} &b_{3}

\end{vmatrix}-j\begin{vmatrix}

a_{1} &a_{3} \\

b_{1} &b_{3}

\end{vmatrix}+k\begin{vmatrix}

a_{1} &a_{2} \\

b_{1} &b_{2}

\end{vmatrix}$

= $i(a_{2}b_{3} - a_{3}b_{2}) - j(a_{1}b_{3} - a_{3}b_{1}) + k(a_{1}b_{2} - a_{2}b_{1})$

Cross product of vectors is not associative. To prove this let us assume A = a

Now, we have to show that

$(A \times B) \times C \neq A \times (B \times C)$

Then, $A \times B$ = $\begin{vmatrix}

i &j &k \\

a_{1} &a_{2} &a_{3} \\

b_{1} &b_{2} &b_{3}

\end{vmatrix}$

= $i(a_{2}b_{3} - a_{3}b_{2}) + j(a_{3}b_{1} - a_{1}b_{3}) + k(a_{1}b_{2} - a_{2}b_{1})$

Now, we calculate

$(A \times B) \times C = \begin{vmatrix}

i &j &k \\

(a_{2}b_{3}-a_{3}b_{2}) &(a_{3}b_{1}-a_{1}b_{3}) &(a_{1}b_{2}-a_{2}b_{1}) \\

c_{1} &c_{2} &c_{3}

\end{vmatrix}$

Again, $B \times C$ = $\begin{vmatrix}

i &j &k \\

b_{1} &b_{2} &b_{3} \\

c_{1} &c_{2} &c_{3}

\end{vmatrix}$

= $i(b_{2}c_{3} - b_{3}c_{2}) + j(c_{1}b_{3} - b_{1}c_{3}) + k(b_{1}c_{2} - b_{2}c_{1})$

Again, we find

$A \times (B \times C) = \begin{vmatrix}

i &j &k \\

a_{1} &a_{2} &a_{3} \\

(b_{2}c_{3}-b_{3}c_{2}) &(c_{1}b_{3}-b_{1}c_{3}) &(b_{1}c_{2}-b_{2}c_{1})

\end{vmatrix}$

So, it is clear that

$\begin{vmatrix}

i &j &k \\

b_{1} &b_{2} &b_{3} \\

c_{1} &c_{2} &c_{3}

\end{vmatrix}

= i(b_{2}c_{3} - b_{3}c_{2}) + j(c_{1}b_{3} - b_{1}c_{3}) + k(b_{1}c_{2} - b_{2}c_{1})$

$A \times (B \times C) = \begin{vmatrix}

i &j &k \\

a_{1} &a_{2} &a_{3} \\

(b_{2}c_{3}-b_{3}c_{2}) &(c_{1}b_{3}-b_{1}c_{3}) &(b_{1}c_{2}-b_{2}c_{1})

\end{vmatrix}$

So, it is clear that $(A \times B) \times C \neq A \times (B \times C)$

Given below are some of the examples on cross product of two vectors.

To find the cross product of two vectors, we have to write the given vectors in determinant form. Using the determinant form, we can find the Cross product of two vectors.

$\vec X = 5 \vec i + 6 \vec j + 2 \vec k$ and $\vec Y = \vec i + \vec j + \vec k$

$\vec X \times \vec Y = (6 - 2)\vec i - (5 - 2) \vec j + (5 - 6) \vec k$

= $4 \vec i - 3 \vec j - \vec k$

$\vec X = 5 \vec i + 6 \vec j + 2 \vec k$ and $\vec Y = \vec i + \vec j + \vec k$

$\vec X \times \vec Y = (6 - 2)\vec i - (5 - 2) \vec j + (5 - 6) \vec k$

= $4 \vec i - 3 \vec j - \vec k$

To find the cross product of two vectors, we have to write the given vectors in determinant form. Using the determinant form, we can find the cross product of two vectors.

$\vec X = 4 \vec i - 5 \vec j + 2 \vec k$ and $\vec Y = 2 \vec i + 4 \vec j + 5 \vec k$

$\vec X \times \vec Y = (-25 - 8) \vec i - (20 - 4 ) \vec j + (16 + 10) \vec k$

= $- 33 \vec i - 16 \vec j + 26 \vec k$

$\vec X = 4 \vec i - 5 \vec j + 2 \vec k$ and $\vec Y = 2 \vec i + 4 \vec j + 5 \vec k$

$\vec X \times \vec Y = (-25 - 8) \vec i - (20 - 4 ) \vec j + (16 + 10) \vec k$

= $- 33 \vec i - 16 \vec j + 26 \vec k$

To find the cross product of two vectors, we have to write the given vectors in determinant form. Using the determinant form, we can find the cross product of two vectors.

$\vec X = 0 \vec i + 6 \vec j + 2 \vec k$ and $\vec Y = \vec i + - 7 \vec j + \vec k$

$\vec X \times \vec Y = (6 + 14)\vec i - (0 - 2) \vec j + (0 - 6) \vec k$

= $20 \vec i + 2 \vec j - 6 \vec k$

$\vec X = 0 \vec i + 6 \vec j + 2 \vec k$ and $\vec Y = \vec i + - 7 \vec j + \vec k$

$\vec X \times \vec Y = (6 + 14)\vec i - (0 - 2) \vec j + (0 - 6) \vec k$

= $20 \vec i + 2 \vec j - 6 \vec k$

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