question_answer

What is the angle made by vector \[\vec{A}=2\hat{i}+2\hat{j}\]\[x\]-axis?

Answer:

                 The angle \[\theta \] between \[\vec{A}\] and \[x\]-axis is given by \[\cos \theta =\frac{\vec{A}.\hat{i}}{|\vec{A}\text{ }\!\!|\!\!\text{   }\!\!|\!\!\text{ }\hat{i}|}=\frac{(2\hat{i}+2\hat{j})}{|2\hat{i}+2\hat{j}\text{ }\!\!|\!\!\text{   }\!\!|\!\!\text{ \hat{i} }\!\!|\!\!\text{ }}=\frac{2\times 1+2\times 0}{\sqrt{{{2}^{2}}+{{2}^{2}}}\sqrt{{{1}^{2}}}}=\frac{2}{2\sqrt{2}}\]or        \[\cos \theta =\frac{1}{\sqrt{2}}\] \[\therefore \]  \[\theta ={{45}^{\circ }}\].