Question

If $\overset{\rightarrow}{a}$, $\overset{\rightarrow}{b}$, $\overset{\rightarrow}{c}$ are three non-coplanar vector $\overset{\rightarrow}{p},\overset{\rightarrow}{q},\overset{\rightarrow}{r}$ and are reciprocal vectors, then $\left( \overset{\rightarrow}{\mathcal{l}a} + \overset{\rightarrow}{mb} + \overset{\rightarrow}{nc} \right)$. $\left( \overset{\rightarrow}{\mathcal{l}p} + \overset{\rightarrow}{mq} + \overset{\rightarrow}{nr} \right)$ is equal to-

• A. l2 + m2 + n2
• B. lm + mn + nl
• C. 0
• D. None of these

Answer 1

Answer: (A)

l2 + m2 + n2

Answer 2

The vectors reciprocal to $\overset{\rightarrow}{a}$, $\overset{\rightarrow}{b}$, $\overset{\rightarrow}{c}$are given by

$\overset{\rightarrow}{p}$ = $\frac{\overset{\rightarrow}{b} \times \overset{\rightarrow}{c}}{(\overset{\rightarrow}{a}\overset{\rightarrow}{b}\overset{\rightarrow}{c})}$, $\frac{\overset{\rightarrow}{c} \times \overset{\rightarrow}{a}}{(\overset{\rightarrow}{a}\overset{\rightarrow}{b}\overset{\rightarrow}{c})}$, $\frac{\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}}{(\overset{\rightarrow}{a}\overset{\rightarrow}{b}\overset{\rightarrow}{c})}$= so that

$\overset{\rightarrow}{a}$. $\overset{\rightarrow}{p}$= 1, $\overset{\rightarrow}{a}$ . $\overset{\rightarrow}{q}$= $\overset{\rightarrow}{a}$ . $\overset{\rightarrow}{r}$ = 0, $\overline{b}$. $\overline{q}$= 1, $\overset{\rightarrow}{c}$ . $\overline{q}$= $\overset{\rightarrow}{a}$. $\overset{\rightarrow}{q}$

= 0, $\overline{c}.\overline{r}$ = 1, $\overline{c}.\overline{p}$ = $\overline{c}.\overline{r}$ = 0

This gives ($\overset{\rightarrow}{a}$ + m$\overset{\rightarrow}{b}$+ n$\overset{\rightarrow}{c}$). (l$\overset{\rightarrow}{p}$+ m$\overset{\rightarrow}{q}$ + n$\overset{\rightarrow}{r}$)

= l2 + m2 + n2.