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 $c$.

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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

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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 $-II$
$(a)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,0$ $(i)$ $\theta = \,{30^o}$
$(b)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,8$ $(ii)$ $\theta = \,{45^o}$
$(c)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4$ $(iii)$ $\theta = \,{90^o}$
$(d)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4\sqrt 2$ $(iv)$ $\theta = \,{0^o}$

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