Question

Add vectors \[\vec{A} , \vec{B} \text { and } \vec{C}\]  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 \[\vec{A} = \ A\ cos \ 45^\circ =100 \cos 45^\circ = \frac{100}{\sqrt{2}} \text { unit }\]

x-component of \[\vec{B} = \vec{B} \cos 135^\circ = - \frac{100}{\sqrt{2}}\]

x-component of \[\vec{C}\] = \[\vec{C}\] cos 315\[^\circ\]

= 100 cos 315°

\[= 100 \cos 45^\circ = \frac{100}{\sqrt{2}}\]

Resultant x-component \[= \frac{100}{\sqrt{2}} - \frac{100}{\sqrt{2}} + \frac{100}{\sqrt{2}} = \frac{100}{\sqrt{2}}\]

Now, y-component of \[\vec{A} = 100 \sin 45^\circ = \frac{100}{\sqrt{2}}\]

y-component of \[\vec{B} = 100 \sin 135^\circ = \frac{100}{\sqrt{2}}\] 

y-component of \[\vec{C} = 100 \sin 315^\circ = - \frac{100}{\sqrt{2}}\]

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

\[\tan \alpha = \frac{\text { y comp}}{\text { x comp }}\]

\[ = \frac{100\sqrt{2}}{100\sqrt{2}} = 1\]

⇒ 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.