Product of Two Vectors - Vector (Or Cross) Product of Two Vectors

The vector product of two nonzero vectors , is denoted by and defined as  ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\left|\overrightarrow{a}\right| \left|\overrightarrow{b}\right|\sin \theta \hat{n}}$,
where ,θ is the angle between , ${\displaystyle \overrightarrow{a}}$ and ${\displaystyle \overrightarrow{b}}$ = 0 ≤ θ ≤ π and ${\displaystyle \hat{n}}$ is a unit vector perpendicular to both ${\displaystyle \overrightarrow{a}}$ and ${\displaystyle \overrightarrow{b}}$, such that ${\displaystyle \overrightarrow{a}}$, ${\displaystyle \overrightarrow{b}}$ and ${\displaystyle \hat{n}}$ form a right handed system in following fig . 

 i.e., the right handed system rotated from ${\displaystyle \overrightarrow{a}}$ to ${\displaystyle \overrightarrow{b}}$ moves in the direction ${\displaystyle \hat{n}}$.
If either ${\displaystyle \overrightarrow{a}=\overrightarrow{0}}$ or ${\displaystyle \overrightarrow{b}=\overrightarrow{0}}$ , then θ is not defined and in this case, we define ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{0}}$.

Observations: 
1) ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}}$ is a vector. 

2) Let ${\displaystyle \overrightarrow{a}}$ and ${\displaystyle \overrightarrow{b}}$ be two  nonzero vectors. Then ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{0}}$ if and only if ${\displaystyle \overrightarrow{a}}$ and ${\displaystyle \overrightarrow{b}}$ are parallel (or collinear) to each other, i.e.,
${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{0}\iff \overrightarrow{a}\mid \mid \overrightarrow{b}}$
In particular , ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{0}}$ and  ${\displaystyle \overrightarrow{a}\times (-\overrightarrow{a})=\overrightarrow{0}}$, since in the first situation, θ = 0 and in the second one, θ = π, making the value of sinθ to be 0.

3) If  , θ = ${\displaystyle \frac{\pi }{2}}$ then ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|.}$

4)  In view of the Observations 2 and 3, for mutually perpendicular unit vectors ${\displaystyle \hat{i},\hat{j}}$ and${\displaystyle \hat{k}}$ fig.

${\displaystyle \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=\overrightarrow{0}}$
${\displaystyle \hat{i}\times \hat{j}=\hat{k}, \hat{j}\times \hat{k}=\hat{i},\hat{k}\times \hat{i}=\hat{j}}$

5) In terms of vector product, the angle between two vectors ${\displaystyle \overrightarrow{a}}$ and ${\displaystyle \overrightarrow{b}}$ may be given as
${\displaystyle \sin \theta =\frac{|\overrightarrow{a}\times \overrightarrow{b}|}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}$

6)  It is always true that the vector product is not commutative, as ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}.}$
Indeed ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right| \sin \theta {\hat{n}}_1}$, where ${\displaystyle \overrightarrow{b},\overrightarrow{a} \text{and} {\hat{n}}_1}$ form a right handed system, i.e., θ is traversed from ${\displaystyle \overrightarrow{b} \text{to} \overrightarrow{a}}$  in following fig. 

While , ${\displaystyle \overrightarrow{b}\times \overrightarrow{a}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right| \sin \theta {\hat{n}}_1}$ , where ${\displaystyle \overrightarrow{b},\overrightarrow{a} \text{and} {\hat{n}}_1}$ form a right handed system i.e. θ is traversed from ${\displaystyle \overrightarrow{b}\text{to} \overrightarrow{a}}$,
Fig.

Thus, if we assume ${\displaystyle \overrightarrow{a}}$ and ${\displaystyle \overrightarrow{b}}$ to lie  in the plane of the paper, then ${\displaystyle \hat{n}}$ and ${\displaystyle {\hat{n}}_1}$ both will be perpendicular to the plane of the paper. But, ${\displaystyle \hat{n}}$  being directed above the paper while ${\displaystyle {\hat{n}}_1}$ directed below the paper i.e. ${\displaystyle {\hat{n}}_1=-\hat{n}.}$
Hence ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right| \sin \theta \hat{n}}$
=${\displaystyle -\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right| =\sin \theta {\hat{n}}_1}$ 
= -${\displaystyle \overrightarrow{b}\times \overrightarrow{a}=}$

7) In view of the Observations 4 and 6, we have 
${\displaystyle \hat{j}\times \hat{i}=-\hat{k},\hat{k}\times \hat{j}=-\hat{i}}$ and  ${\displaystyle \hat{i}\times \hat{k}=-\hat{j}.}$

 8) If ${\displaystyle \overrightarrow{a}}$ and ${\displaystyle \overrightarrow{b}}$ represent the adjacent sides of a triangle then its area is given as ${\displaystyle \frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|}$.
By definition of the area of a triangle, we have from fig. 

Area of triangle ABC =  ${\displaystyle \frac{1}{2}}$ AB .CD
But AB = ${\displaystyle \left|\overrightarrow{b}\right|}$ (as given ), and  CD = ${\displaystyle \left|\overrightarrow{a}\right| \sin \theta .}$

Thus,  Area of triangle ABC =  ${\displaystyle \frac{1}{2}\left|\overrightarrow{b}\right|\left|\overrightarrow{a}\right|\sin \theta =\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|}$. 

9)  If ${\displaystyle \overrightarrow{a} \text{and} \overrightarrow{b}}$ represent the adjacent sides of a  parallelogram, then its area is given by |vec a xx vec b|.From the following Fig. we have