question_answer

What is the difference between the following data? (i) 3 (5\[~\text{km}{{\text{h}}^{-\text{1}}}\], west) (ii) 3 hour (\[~\text{km}{{\text{h}}^{-\text{1}}}\], west).

Answer:

                Draw \[(\overrightarrow{PQ})=\vec{A}\]and from the arrow head of\[\vec{A}\], draw \[(\overrightarrow{QS})=\vec{B}\], of the same length (i.e., QS = PQ) and perpendicular to\[\vec{A}\] Fig. 2(c).53. Now \[(\overrightarrow{PS})\]will represent \[(\vec{A}+\,\vec{B})\]. Here,                     \[\text{tan}{{\theta }_{\text{1}}}=\frac{QS}{PQ}=\text{1}\]or \[\theta =\text{45}{}^\circ \] Now draw \[(\overrightarrow{QT})=-\vec{B}\]where\[\text{QT}=\text{QS}\]. Now \[(\overrightarrow{PT})\] will represent \[(\vec{A}-\vec{B})\]. Here, \[\text{tan}{{\theta }_{\text{2}}}=\frac{QT}{PQ}=1\]or \[{{\theta }_{\text{2}}}=\text{45}{}^\circ \]. On measuring, the lengths of \[(\vec{A}+\vec{B})\] and \[(\vec{A}-\vec{B})\]come out to be the same and angle between them \[({{\theta }_{\text{1}}}+{{\theta }_{\text{2}}})=\text{45}{}^\circ +\text{45}{}^\circ =\text{9}0{}^\circ \].