First slide

Scalar triple product of vectors

Question

$If \vec{a}, \vec{b}, \vec{c} are three non-coplanar and \vec{p}, \vec{q}, \vec{r} are reciprocal vectors to \vec{a}, \vec{b}$ $and \vec{c} respectively, then (l \vec{a} + m \vec{b} + n \vec{c}) \cdot (l \vec{p} + m \vec{q} + n \vec{r}) is equal to : (where$ $1, m, \underline{n} are scalars)$

Moderate

A

$l^{2} + m^{2} + n^{2}$
B

$lm + mn + nl$
C

0
D

None of these

Solution

$\vec{p} = \frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]}; \vec{q} = \frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]}; \vec{r} = \frac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]}$
$\vec{a} . \vec{p} = \vec{a} \cdot \frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]} = \frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]} =1$ and $\vec{a} . \vec{q} = \vec{a} \cdot \frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]} = \frac{[\vec{a} \vec{c} \vec{a}]}{[\vec{a} \vec{b} \vec{c}]} =0$
$s i m i l a r l y \vec{b} \cdot \vec{q} = 1 \vec{c} \cdot \vec{r} = 1$
$(l \vec{a} + m \vec{b} + n \vec{c}) \cdot (l \vec{p} + m \vec{q} + n \vec{r}) = l^{2} \vec{a} \cdot \vec{p} + m^{2} \vec{b} . \vec{q} + n^{2} \vec{c} \cdot \vec{r}$
$= l^{2} + m^{2} + n^{2}$

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