Obtain the scalar product of two mutually perpendicular vectors.
Obtain the scalar product of two mutually perpendicular vectors.
$\text { If } \vec{A} \perp \vec{B}, \text { then } \theta=90^{\circ}$
$\therefore \quad \vec{A} \cdot \vec{B}=\mathrm{ABcos} 90^{\circ}$
$=0$
$\because \cos 90^{\circ}=0$
This is the condition for mutually perpendicular of two non-zero vectors.
Similar Questions
If $\overrightarrow A \times \overrightarrow B = \overrightarrow C + \overrightarrow D,$ then select the correct alternative-
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The two vectors $\vec A$ and $\vec B$ that are parallel to each other are
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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 $-II$
$(a)$ $\vec A \,.\,\,\vec B \, = \,\,0$
$(i)$ $\theta = \,{0^o}$
$(b)$ $\vec A \,.\,\,\vec B \, = \,\,+8$
$(ii)$ $\theta = \,{90^o}$
$(c)$ $\vec A \,.\,\,\vec B \, = \,\,4$
$(iii)$ $\theta = \,{180^o}$
$(d)$ $\vec A \,.\,\,\vec B \, = \,\,-8$
$(iv)$ $\theta = \,{60^o}$
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 $-II$ |
| $(a)$ $\vec A \,.\,\,\vec B \, = \,\,0$ | $(i)$ $\theta = \,{0^o}$ |
| $(b)$ $\vec A \,.\,\,\vec B \, = \,\,+8$ | $(ii)$ $\theta = \,{90^o}$ |
| $(c)$ $\vec A \,.\,\,\vec B \, = \,\,4$ | $(iii)$ $\theta = \,{180^o}$ |
| $(d)$ $\vec A \,.\,\,\vec B \, = \,\,-8$ | $(iv)$ $\theta = \,{60^o}$ |
${\vec A }$, ${\vec B }$ and ${\vec C }$ are three non-collinear, non co-planar vectors. What can you say about directin of $\vec A \, \times \,\left( {\vec B \, \times \vec {\,C} } \right)$ ?
${\vec A }$, ${\vec B }$ and ${\vec C }$ are three non-collinear, non co-planar vectors. What can you say about directin of $\vec A \, \times \,\left( {\vec B \, \times \vec {\,C} } \right)$ ?
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
- [AIIMS 2012]