Question

Draw diagrams and prove (4.1).

Short answer

The diagram is verified using (4.1) theorem.

Step-by-step solution

Formula for the magnitude of the components of the vector:

If ${\mathbf{A}}_{\mathbf{x}}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{,}\mathbf{}$and ${\mathbf{A}}_{\mathbf{z}}$are the magnitudes of components of vector$\overrightarrow{\mathbf{A}}$along the coordinate axis, then $\overrightarrow{\mathbf{A}}\mathbf{=}{\mathbf{A}}_{\mathbf{x}}\hat{\mathbf{i}}\mathbf{+}{\mathbf{A}}_{\mathbf{y}}\hat{\mathbf{j}}\mathbf{+}{\mathbf{A}}_{\mathbf{z}}\mathbf{}\hat{\mathbf{k}}\mathbf{}$, where$\hat{\mathbf{i}}\mathbf{,}\hat{\mathbf{j}}\mathbf{,}\hat{\mathbf{k}}\mathbf{}$respectively are the unit vectors along $\mathbf{x}\mathbf{,}\mathbf{y}$and zaxis.

The diagram of vector A→  in two dimensions:

Using Pythagoras theorem, the length of A , named $A=\left|A\right|$is given by,

$$\begin{gathered} A^2=A_x^2+A_y^2 \\ A=\sqrt{A_x^2+A_y^2} \end{gathered}$$

Where the dotted lines represent the components of the vector $A_x$ and $A_y$

The diagram of vector A→  in three dimensions:

Using Pythagoras theorem in $3D$, the length of A , named $A=\left|A\right|$is given by,

$$\begin{gathered} {\mathrm{A}}^2={\mathrm{A}}_{\mathrm{x}}^2+{\mathrm{A}}_{\mathrm{y}}^2+{\mathrm{A}}_{\mathrm{z}}^2 \\ \mathrm{A}=\sqrt{{\mathrm{A}}_{\mathrm{x}}^2+{\mathrm{A}}_{\mathrm{y}}^2+{\mathrm{A}}_{\mathrm{z}}^2} \end{gathered}$$

Where the dotted lines represent the components of the vector$A_x,A_y$and $A_z$.

Hence, the condition is proved.