
If $a,b,c$ are three non-coplanar vectors and $p,q,r$ reciprocal vectors, then $\left( la+mb+nc \right).\left( lp+mq+nr \right)$ 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
510.3k+ views
Hint:Here in this question we have been asked to value of $\left( la+mb+nc \right).\left( lp+mq+nr \right)$ 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 $\left( la+mb+nc \right).\left( lp+mq+nr \right)$ 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 $\left( la+mb+nc \right).\left( lp+mq+nr \right)$ using the values we have got, by doing that we will have
$\begin{align}
& \left( la+mb+nc \right).\left( lp+mq+nr \right) \\
& \Rightarrow {{l}^{2}}a.p+{{m}^{2}}b.q+{{n}^{2}}c.r \\
\end{align}$ .
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 $\left( la+mb+nc \right).\left( lp+mq+nr \right)$ 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 $\left( la+mb+nc \right).\left( lp+mq+nr \right)=0$ which is a wrong answer.
Complete step-by-step solution:
Now considering from the question we have been asked to value of $\left( la+mb+nc \right).\left( lp+mq+nr \right)$ 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 $\left( la+mb+nc \right).\left( lp+mq+nr \right)$ using the values we have got, by doing that we will have
$\begin{align}
& \left( la+mb+nc \right).\left( lp+mq+nr \right) \\
& \Rightarrow {{l}^{2}}a.p+{{m}^{2}}b.q+{{n}^{2}}c.r \\
\end{align}$ .
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 $\left( la+mb+nc \right).\left( lp+mq+nr \right)$ 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 $\left( la+mb+nc \right).\left( lp+mq+nr \right)=0$ which is a wrong answer.
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