Question

Add vectors $\overrightarrow{\mathrm{A}},\overrightarrow{\mathrm{B}}\mathrm{and}\overrightarrow{\mathrm{C}}$ each having magnitude of 100 unit and inclined to the X-axis at angles 45°, 135° and 315° respectively.

Answer 1

First, we will find the components of the vector along the x-axis and y-axis. Then we will find the resultant x and y-components.
x-component of $\overrightarrow{\mathrm{A}}=\mathrm{Acos}45°=100\cos 45°=\frac{100}{\sqrt{2}}\mathrm{unit}$
x-component of $\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}\cos 135°=-\frac{100}{\sqrt{2}}$
x-component of $\overrightarrow{\mathrm{C}}$ = $\overrightarrow{\mathrm{C}}$cos315$°$ = 100 cos 315°
$=100\cos 45°=\frac{100}{\sqrt{2}}$
Resultant x-component $=\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}$



Now, y-component of $\overrightarrow{\mathrm{A}}=100\sin 45°=\frac{100}{\sqrt{2}}$
y-component of $\overrightarrow{\mathrm{B}}=100\sin 135°=\frac{100}{\sqrt{2}}$
y-component of $\overrightarrow{\mathrm{C}}=100\sin 315°=-\frac{100}{\sqrt{2}}$
Resultant y-component$=\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}$

Magnitude of the resultant$=\sqrt{{\left(\frac{100}{\sqrt{2}}\right)}^2+{\left(\frac{100}{\sqrt{2}}\right)}^2}$
$=\sqrt{10000}=100$
Angle made by the resultant vector with the x-axis is given by
$$\begin{gathered} \tan \mathrm{\alpha }=\frac{y\mathrm{comp}}{x\mathrm{comp}} \\ =\frac{100\sqrt{2}}{100\sqrt{2}}=1 \end{gathered}$$
⇒ α = tan⁻¹ (1) = 45°

∴ The magnitude of the resultant vector is 100 units and it makes an angle of 45° with the x-axis.