- Home
- Standard 11
- Physics
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
${A^2}B$
Zero
${A^2}B\sin \theta $
${A^2}B\cos \theta $
Solution
(b) Let $\overrightarrow A \,.(\overrightarrow B \, \times \overrightarrow A ) = \overrightarrow A \,.\,\overrightarrow C \,$
Here $\overrightarrow C = \overrightarrow B \times \overrightarrow A $ Which is perpendicular to both vector
$\overrightarrow A $ and $\overrightarrow B $
$\therefore\,\overrightarrow A \,.\overrightarrow {\,C} = 0$
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 $-II$ |
| $(a)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,0$ | $(i)$ $\theta = \,{30^o}$ |
| $(b)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,8$ | $(ii)$ $\theta = \,{45^o}$ |
| $(c)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4$ | $(iii)$ $\theta = \,{90^o}$ |
| $(d)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4\sqrt 2$ | $(iv)$ $\theta = \,{0^o}$ |
