Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
If $\overrightarrow{\mathrm{A}} \| \overrightarrow{\mathrm{B}}$, then $\theta=0^{\circ}$
$\therefore \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=\mathrm{AB} \text { and } \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}=|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{A}}|=\mathrm{A}^{2}$
$\therefore|\overrightarrow{\mathrm{A}}|=\sqrt{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}}$
hence the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
Magnitude of vector is equal to the square root of the scalar product of the vector with itself.
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- [AIPMT 2003]