Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
(N/A) Let $\overrightarrow{A}$ be a vector. The scalar product (dot product) of the vector $\overrightarrow{A}$ with itself is defined as $\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}| |\overrightarrow{A}| \cos \theta$.
Since the angle $\theta$ between a vector and itself is $0^{\circ}$,we have $\cos 0^{\circ} = 1$.
Therefore,$\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}| |\overrightarrow{A}| (1) = |\overrightarrow{A}|^2$.
Taking the square root on both sides,we get $|\overrightarrow{A}| = \sqrt{\overrightarrow{A} \cdot \overrightarrow{A}}$.
Thus,the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
Since the angle $\theta$ between a vector and itself is $0^{\circ}$,we have $\cos 0^{\circ} = 1$.
Therefore,$\overrightarrow{A} \cdot \overrightarrow{A} = |\overrightarrow{A}| |\overrightarrow{A}| (1) = |\overrightarrow{A}|^2$.
Taking the square root on both sides,we get $|\overrightarrow{A}| = \sqrt{\overrightarrow{A} \cdot \overrightarrow{A}}$.
Thus,the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
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