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Question

(C) The angle $	heta$ between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by the dot product formula: $\overrightarrow{A} \cdo

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The angle $	heta$ between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by the dot product formula: $\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos 	heta$.First,calculate the dot product: $\overrightarrow{A} \cdot \overrightarrow{B} = (5\hat{i} + 5\hat{j}) \cdot (5\hat{i} - 5\hat{j}) = (5 	imes 5) + (5 	imes -5) = 25 - 25 = 0$.Since the dot product is $0$,we have $0 = |\overrightarrow{A}| |\overrightarrow{B}| \cos 	heta$.This implies $\cos 	heta = 0$,which means $	heta = 90^\circ$.

The angle between the two vectors $\overrightarrow{A} = 5\hat{i} + 5\hat{j}$ and $\overrightarrow{B} = 5\hat{i} - 5\hat{j}$ will be ....... $^\circ$.

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