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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. . . . . ..
$2$
$4$
$6$
$8$
Solution
$\vec{B} \times \vec{C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3\end{array}\right|=15 \hat{i}-21 \hat{j}+33 \hat{k}$
$\vec{A} \cdot(\vec{B} \times \vec{C})=(-x \hat{i}-6 \hat{j}-2 \hat{k}) \cdot(15 \hat{i}-21 \hat{j}+33 \hat{k})$
$0=-15 x+126-66$
$15 x=60$
$x=4$
Similar Questions
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 $-II$ |
| $(a)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,0$ | $(i)$ $\theta = \,{30^o}$ |
| $(b)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,8$ | $(ii)$ $\theta = \,{45^o}$ |
| $(c)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4$ | $(iii)$ $\theta = \,{90^o}$ |
| $(d)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4\sqrt 2$ | $(iv)$ $\theta = \,{0^o}$ |
