Answer:
The
greatest resultant of given two vectors = \[\left( \text{P}+\text{Q} \right)\]
and
the least resultant of given two vectors = \[\left(
\text{P}\,\text{-}\,\text{Q} \right)\]
According
to questions ; \[\left( \text{P}+\text{Q} \right)=\text{n}\left(
\text{P}-\text{Q} \right)\]
or
\[Q=\frac{\left(
n-1 \right)}{\left( n+1 \right)}P\]
If
angle between the vectors is \[\text{ }\!\!\theta\!\!\text{ }\], then the
resultant is given by
\[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos
\text{ }\!\!\theta\!\!\text{ }\] ...(i)
Given
\[R=\frac{P+Q}{2}=\frac{1}{2}\left[ P+\left( \frac{n-1}{n+1} \right)P
\right]=\frac{nP}{\left( n+1 \right)}\]
Putting
the values in (i), we get
\[\frac{{{n}^{2}}{{P}^{2}}}{{{\left(
n+1 \right)}^{2}}}={{P}^{2}}+\frac{{{\left( n-1 \right)}^{2}}}{{{\left( n+1
\right)}^{2}}}{{P}^{2}}+2P\frac{\left( n-1 \right)}{\left( n+1 \right)}P\cos
\text{ }\!\!\theta\!\!\text{ }\]
On
solving we get,
\[\cos \text{ }\!\!\theta\!\!\text{
}\,\text{=}\,\text{-}\,\frac{\left( {{n}^{2}}+2 \right)}{\left( {{n}^{2}}-1
\right)}\]
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