question_answer

The greatest resultant of two vectors \[\overset{\to }{\mathop{P}}\,\] and \[\overset{\to }{\mathop{Q}}\,\] is n times their least resultant. Given \[|\overset{\to }{\mathop{P}}\,|\,\,>\,|\overset{\to }{\mathop{Q}}\,|\]. When \[\text{ }\!\!\theta\!\!\text{ }\] is the angle between the two vectors, their resultant is half the sum of the two vectors. Show that, \[\text{cos }\!\!\theta\!\!\text{ }=-\left( {{\text{n}}^{\text{2}}}+\text{2 } \right)/\left( {{\text{n}}^{\text{2}}}-\text{1} \right)\]

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, \[\cos \text{ }\!\!\theta\!\!\text{ }\,\text{=}\,\text{-}\,\frac{\left( {{n}^{2}}+2 \right)}{\left( {{n}^{2}}-1 \right)}\]