The position vectors of points $A, B, C$ and $D$ are $A = 3\hat i + 4\hat j + 5\hat k,\,\,B = 4\hat i + 5\hat j + 6\hat k,\,\,C = 7\hat i + 9\hat j + 3\hat k$ and $D = 4\hat i + 6\hat j$ then the displacement vectors $AB$ and $CD $ are

Obtain scalar product in terms of Cartesian component of vectors.

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. . . . . .. 

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

The vector $\overrightarrow P = a\hat i + a\hat j + 3\hat k$ and $\overrightarrow Q = a\hat i - 2\hat j - \hat k$ are perpendicular to each other. The positive value of $a$ is

The angle between the vectors $\overrightarrow A $ and $\overrightarrow B $ is $\theta .$ The value of the triple product $\overrightarrow A \,.\,(\overrightarrow B \times \overrightarrow A \,)$ is