Question

If $a,b,c$ are three non-coplanar vectors and $p,q,r$ reciprocal vectors, then $(la+mb+nc).(lp+mq+nr)$ is equal to?
(1) $l+m+n$
(2) $l^3+m^3+n^3$
(3) $l^2+m^2+n^2$
(4) None of these

Answer 1

Hint:Here in this question we have been asked to value of $(la+mb+nc).(lp+mq+nr)$ given that $a,b,c$ are three non-coplanar vectors and $p,q,r$ reciprocal vectors. As it is given that $p,q,r$ are reciprocal vectors of $a,b,c$ are three non-coplanar vectors we can say that $p.a=1$ , $q.b=1$ and $r.c=1$.

Complete step-by-step solution:
Now considering from the question we have been asked to value of $(la+mb+nc).(lp+mq+nr)$ given that $a,b,c$ are three non-coplanar vectors and $p,q,r$ reciprocal vectors.
From the basic concepts we know that the product of a vector and its reciprocal vector is 1.
Since it is given that $a,b,c$ are three non-coplanar vectors and $p,q,r$ reciprocal vectors we can say that
$p.a=1$ , $p.b=0$ and $p.c=0$ .
$q.b=1$ , $q.a=0$ and $q.c=0$ .
$r.c=1$ , $r.a=0$ and $r.b=0$ .
Now we can evaluate the value of the given expression $(la+mb+nc).(lp+mq+nr)$ using the values we have got, by doing that we will have
 $\begin{array}{rl}& (la+mb+nc).(lp+mq+nr)\\ & \Rightarrow l^{2}a.p+m^{2}b.q+n^{2}c.r\end{array}$ .
Now by substituting $p.a=q.b=c.r=1$ we can say that the value of the expression is $\Rightarrow l^2+m^2+n^2$ .
Therefore we can conclude that the value of the given expression $(la+mb+nc).(lp+mq+nr)$ is $l^2+m^2+n^2$ when it is given that $a,b,c$ are three non-coplanar vectors and $p,q,r$ reciprocal vectors.
Hence we will mark the option “3” as correct.

Note:This is a very easy and simple question and can be answered in a short span of time. If someone had a misconception and considered that $p.a=0$ ,$q.b=0$ and $r.c=0$ then they will end up having the value of the expression as $(la+mb+nc).(lp+mq+nr)=0$ which is a wrong answer.