3-1.Vectors
medium

Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.

Option A
Option B
Option C
Option D

Solution

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.

Standard 11
Physics

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