Explain cross product of two vectors.
Defination : The vector product or cross product of two vector $\vec{a}$ and $\vec{b}$ is another vector $\vec{c}$, whose magnitude is equal to product of magnitude of the two vectors and sine of the smaller angle between them. $OR$
If the product of two vector gives resultant vector quantity then this product is vector product. Suppose two vectors $\vec{a}$ and $\vec{b}$ and angle between them is $\theta$
$\therefore \text { Vector product } \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}$ $=a b \sin \theta \hat{n}$
where $|\vec{a}|=a$ and $|\vec{b}|=b$
and $\hat{n}$ is a unit vector perpendicular to the plane form by $\vec{a}$ and $\vec{b}$
The product is known as cross ${~ }\times$ product also.
Suppose $\vec{a} \times \vec{b}$ is denoted by $\vec{c}$ then
$\vec{c}=a b \sin \theta \hat{n}$
and magnitude of $c=a b \sin \theta$
Direction of $\vec{c}$ is perpendicular to the plane form by $\vec{a}$ and $\vec{b}$ and its direction is given by right hand screw rule.
If $\overrightarrow{ A }=(2 \hat{ i }+3 \hat{ j }-\hat{ k }) \;m$ and $\overrightarrow{ B }=(\hat{ i }+2 \hat{ j }+2 \hat{ k })\; m$. The magnitude of component of vector $\overrightarrow{ A }$ along vector $\vec{B}$ will be $......m$.
The area of the parallelogram represented by the vectors $\overrightarrow A = 2\hat i + 3\hat j$ and $\overrightarrow B = \hat i + 4\hat j$ is.......$units$
Find the angle between two vectors $\vec A = 2\hat i + \hat j - \hat k$ and $\vec B = \hat i - \hat k$ ....... $^o$
If $\overrightarrow{ F }=2 \hat{ i }+\hat{ j }-\hat{ k }$ and $\overrightarrow{ r }=3 \hat{ i }+2 \hat{ j }-2 \hat{ k }$, then the scalar and vector products of $\overrightarrow{ F }$ and $\overrightarrow{ r }$ have the magnitudes respectively as
If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation $\vec{A} . \vec{B}=[\vec{A} \times \vec{B}]$. Then the value of $[\vec{A}-\vec{B}]$. will be :