If $\vec A,\vec B$ and $\vec C$ are vectors having a unit magnitude. If $\vec A + \vec B + \vec C = \vec 0$ then $\vec A.\vec B + \vec B.\vec C + \vec C.\vec A$ will be 

A vector $\overrightarrow A $ points vertically upward and $\overrightarrow B $points towards north. The vector product $\overrightarrow A \times \overrightarrow B $ is

Let $\left| {{{\vec A}_1}} \right| = 3,\,\left| {\vec A_2} \right| = 5$, and $\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = 5$. The value of $\left( {2{{\vec A}_1} + 3{{\vec A}_2}} \right)\cdot \left( {3{{\vec A}_1} - 2{{\vec A}_2}} \right)$ is

For three vectors $\vec{A}=(-x \hat{i}-6 \hat{j}-2 \hat{k})$, $\vec{B}=(-\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{C}=(-8 \hat{i}-\hat{j}+3 \hat{k})$, if $\overrightarrow{\mathrm{A}} \cdot(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}})=0$, them value of $\mathrm{x}$ is. . . . . .. 

The two vectors have magnitudes $3$ and $5$. If angle between them is $60^o$, then the dot product of two vectors will be

Find the scalar and vector products of two vectors. $a =(3 \hat{ i }-4 \hat{ j }+5 \hat{ k })$ and $b =(- 2 \hat{ i }+\hat{ j }- 3 \hat { k } )$