Question

If$\mathit{a}\mathbf{,}\mathbf{}\mathit{b}\mathbf{}and\mathit{c}$ are three non-coplanar vectors and$\mathit{p}\mathbf{,}\mathbf{}\mathit{q}and\mathit{r}$ reciprocal vectors, then$(\mathit{l}\mathit{a}+\mathit{m}\mathit{b}+\mathit{n}\mathit{c}).(\mathit{l}\mathit{p}+\mathit{m}\mathit{q}+\mathit{n}\mathit{r})$ is equal to?

• A. $\mathit{l}+\mathit{m}+\mathit{n}$
• B. ${\mathit{l}}^{\mathbf{3}}+{\mathit{m}}^{\mathbf{3}}+{\mathit{n}}^{\mathbf{3}}$
• C. ${\mathit{l}}^{\mathbf{2}}+{\mathit{m}}^{\mathbf{2}}+{\mathit{n}}^{\mathbf{2}}$
• D. $noneofthese$

Answer 1

The correct option is C

${\mathit{l}}^{\mathbf{2}}+{\mathit{m}}^{\mathbf{2}}+{\mathit{n}}^{\mathbf{2}}$

Explanation for the correct option:

Find out: $(la+mb+nc).(lp+mq+nr)$:

Given: $a,bandc$ are three non-coplanar vectors and $p,qandr$ reciprocal vectors

As $p,qandr$ are the reciprocal vectors of $a,b\mathbf{}andc$ vectors respectively,

So,

$$\begin{gathered} p.a=1,p.b=0\mathbf{}andp.c=0and \\ q.a=0,q.b=1andq.c=0 \\ r.a=0,r.b=0\mathbf{}andr.c=1 \end{gathered}$$

now,

$$\begin{gathered} (la+mb+nc).(lp+mq+nr)=l^{2}a.p+m^{2}b.q+n^{2}c.r \\ \Rightarrow l^2+m^2+n^2 \end{gathered}$$

Hence option $\mathbf{\left(}\mathit{C}\mathbf{\right)}$is the correct answer.