A vector $\overrightarrow A $ points vertically upward and $\overrightarrow B $points towards north. The vector product $\overrightarrow A \times \overrightarrow B $ is
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 $-II$ |
$(a)$ $\vec A \,.\,\,\vec B \, = \,\,0$ | $(i)$ $\theta = \,{0^o}$ |
$(b)$ $\vec A \,.\,\,\vec B \, = \,\,+8$ | $(ii)$ $\theta = \,{90^o}$ |
$(c)$ $\vec A \,.\,\,\vec B \, = \,\,4$ | $(iii)$ $\theta = \,{180^o}$ |
$(d)$ $\vec A \,.\,\,\vec B \, = \,\,-8$ | $(iv)$ $\theta = \,{60^o}$ |
Which of the following is not true ? If $\overrightarrow A = 3\hat i + 4\hat j$ and $\overrightarrow B = 6\hat i + 8\hat j$ where $ A$ and $B$ are the magnitudes of $\overrightarrow A $ and $\overrightarrow B $
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