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
\[({{\theta }_{\text{1}}}+{{\theta }_{\text{2}}})=\text{45}{}^\circ
+\text{45}{}^\circ =\text{9}0{}^\circ \].
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