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. . . . . ..
For any two vectors $\overrightarrow A $ and $\overrightarrow B $, if $\overrightarrow A \,.\,\overrightarrow B = \,\,|\overrightarrow A \times \overrightarrow B |,$ the magnitude of $\overrightarrow C = \overrightarrow A + \overrightarrow B $ is equal to
If $\left| {\vec A } \right|\, = \,2$ and $\left| {\vec B } \right|\, = \,4$ then match the relation in Column $-I$ with the angle $\theta $ between $\vec A$ and $\vec B$ in Column $-II$.
Column $-I$ | Column $-II$ |
$(a)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,0$ | $(i)$ $\theta = \,{30^o}$ |
$(b)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,8$ | $(ii)$ $\theta = \,{45^o}$ |
$(c)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4$ | $(iii)$ $\theta = \,{90^o}$ |
$(d)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4\sqrt 2$ | $(iv)$ $\theta = \,{0^o}$ |
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
Given $\left| {{\vec A_1}} \right| = 2,\,\left| {{\vec A_2}} \right| = 3$ and $\left| {{{\vec A}_1} + {{\vec A}_2}} \right| = 3$. Find the value or $\left| {\left( {{{\vec A}_1} + 2{{\vec A}_2}} \right) \times \left( {3{{\vec A}_1} - 4{{\vec A}_2}} \right)} \right|$