A) \[90{}^\circ \]
B) \[45{}^\circ \]
C) Zero
D) \[60{}^\circ \]
Correct Answer: C
Solution :
The scalar product of two vectors is defined as a scalar quantity equal to the product of their magnitude and cosine of the angle between them. \[\overrightarrow{A}.\overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|\cos \theta \] \[\Rightarrow \]\[\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}=\frac{(5\hat{i}-5\hat{j}).(5\hat{i}-5\hat{j})}{\sqrt{{{5}^{2}}+{{5}^{2}}}\sqrt{{{5}^{2}}+{{5}^{2}}}}\] \[=\frac{25+25}{25\times 2}=\frac{50}{50}=1\] \[\Rightarrow \] \[\theta ={{0}^{o}}\] Alternative: As given,\[\overrightarrow{A}\]and\[\overrightarrow{B}\]are parallel vectors and hence, angle between parallel vectors is zero.
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