Show that $a \cdot( b \times c )$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors, $a, b$ and $c$.
Volume of the given parallelepiped $=a b c$
$\overrightarrow{ OC }=\vec{a}$
$\overrightarrow{ OB }=\vec{b}$
$\overrightarrow{ OC }=\vec{c}$
Let $\hat{ n }$ be a unit vector perpendicular to both $b$ and $c .$ Hence, $\quad \hat{ n }$ and $a$ have the same direction. $\therefore \vec{b} \times \vec{c}=b c \sin \theta \hat{ n }$
$=b c \sin 90^{\circ} \hat{ n }$
$=b c \hat{n}$
$\vec{a} \cdot(\vec{b} \times \vec{c})$
$=a \cdot(b c \hat{ n })$
$=a b c \cos \theta \hat{ n }$
$=a b c \cos 0^{\circ}$
$=a b c$
$=$ Volume of the parallelepiped
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 $-II$ |
$(a)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,0$ | $(i)$ $\theta = \,{30^o}$ |
$(b)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,8$ | $(ii)$ $\theta = \,{45^o}$ |
$(c)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4$ | $(iii)$ $\theta = \,{90^o}$ |
$(d)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4\sqrt 2$ | $(iv)$ $\theta = \,{0^o}$ |
Explain the kinds of multiplication operations for vectors.
Dot product of two mutual perpendicular vector is
The area of the triangle formed by $2\hat i + \hat j - \hat k$ and $\hat i + \hat j + \hat k$ is