The cosine of the angle between vectors p and q such that 2p+q=i+j , p+2q=i-j is

Dear Student,

$$\begin{gathered} Given : 2\stackrel{\rightharpoonup }{ p }+\stackrel{\rightharpoonup }{ q }= \hat{i} + \hat{j} and \stackrel{\rightharpoonup }{ p }+2\stackrel{\rightharpoonup }{ q }= \hat{i} - \hat{j} \\ To calculate : \cos ine of angle between \stackrel{\rightharpoonup }{ p }and\stackrel{\rightharpoonup }{ q }. \\ solution\hspace{0.17em}: \\ 2\stackrel{\rightharpoonup }{ p }+\stackrel{\rightharpoonup }{ q }= \hat{i} + \hat{j} ..................................\overline{)1} \\ \stackrel{\rightharpoonup }{ p }+2\stackrel{\rightharpoonup }{ q }= \hat{i} - \hat{j} ..................................\overline{)2} \\ Multiplying \overline{)2} with 2 subtract from \overline{)1} \\ \Rightarrow (2 \stackrel{\rightharpoonup }{ p }+\stackrel{\rightharpoonup }{ q })-2( \stackrel{\rightharpoonup }{ p }+2\stackrel{\rightharpoonup }{ q })=(\hat{i} + \hat{j})-2( \hat{i} - \hat{j}) \\ \Rightarrow \cancel{2 \stackrel{\rightharpoonup }{ p }}-\cancel{2 \stackrel{\rightharpoonup }{ p }}+ \stackrel{\rightharpoonup }{ q }-4 \stackrel{\rightharpoonup }{ q }= \hat{i} - 2 \hat{i} + \hat{j}+2\widehat{ j} \\ \Rightarrow -3 \stackrel{\rightharpoonup }{ q }=- \hat{i}+3 \hat{j} \\ \Rightarrow \stackrel{\rightharpoonup }{ q }=\frac{-1}{-3} \hat{i}+\frac{3}{-3} \hat{j} \\ \Rightarrow \stackrel{\rightharpoonup }{ q }=\frac{1}{3} \hat{i}- \hat{j}......................................\overline{)3} \\ putting the value of \overline{)3} into \overline{)2} \\ \stackrel{\rightharpoonup }{ p }+2\stackrel{\rightharpoonup }{ q }= \hat{i} - \hat{j} \\ \Rightarrow \stackrel{\rightharpoonup }{ p }+2(\frac{1}{3} \hat{i}- \hat{j})= \hat{i} - \hat{j} \\ \Rightarrow \stackrel{\rightharpoonup }{ p }+\frac{2}{3} \hat{i}-2 \hat{j}= \hat{i} - \hat{j} \\ \Rightarrow \stackrel{\rightharpoonup }{ p }=-\frac{2}{3} \hat{i}+2 \hat{j}+ \hat{i} - \hat{j} \\ \Rightarrow \stackrel{\rightharpoonup }{ p }=\frac{1}{3} \hat{i}+ \hat{j}.......................................\overline{)4} \\ taking dot product of \overline{)3} and \overline{)4} \\ \Rightarrow \stackrel{\rightharpoonup }{ p }.\stackrel{\rightharpoonup }{ q }=\left|\stackrel{\rightharpoonup }{ p }\right| \left|\stackrel{\rightharpoonup }{ q }\right| \cos \theta \\ \Rightarrow \cos \theta =\frac{\stackrel{\rightharpoonup }{ p }.\stackrel{\rightharpoonup }{ q }}{\left|\stackrel{\rightharpoonup }{ p }\right| \left|\stackrel{\rightharpoonup }{ q }\right|} \\ \Rightarrow \cos \theta =\frac{(\frac{1}{3} \hat{i}+ \hat{j}).(\frac{1}{3} \hat{i}- \hat{j})}{\left(\sqrt{({\displaystyle \frac{1}{3}}{)}^2+1^2}\right) \left(\sqrt{(\frac{1}{3}{)}^2+1^2}\right)} \\ \Rightarrow \cos \theta =\frac{(\frac{1}{3}{)}^2 -1}{(\sqrt{({\displaystyle \frac{1}{3}}{)}^2+1^2}{)}^2} \\ \Rightarrow \cos \theta =\frac{-{\displaystyle \frac{8}{9}}}{{\displaystyle \frac{10}{9}}} \\ \Rightarrow \cos \theta =\frac{-{\displaystyle 4}}{{\displaystyle 5}} \end{gathered}$$
Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Keep posting!!

Regards