Let \[\vec{A} \text { and } \vec{B}\] be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angle 30° and 60° respectively, find the resultant.
Solution
Angle between \[\vec{A} \text { and } \vec{B}\],θ = 60° − 30° = 30°
\[\left| \vec{A} \right| = \left| \vec{B} \right| = 10 \text { units }\]
The magnitude of the resultant vector is given by
\[R = \sqrt{A^2 + A^2 + 2AA\cos\theta}\]
\[ = \sqrt{{10}^2 + {10}^2 + 2 \times 10 \times 10 \times \cos 30^\circ}\]
\[ = \sqrt{200 + 200 \cos 30^\circ}\]
\[ = \sqrt{200 + 100}\]
\[ = \sqrt{300} = 17 . 3 \text { units }\]
Let β be the angle between
\[\vec{R} \text { and } \vec{A}\].
\[\therefore \beta = \tan^{- 1} \left( \frac{A \sin 30^\circ}{A + A \cos 30^\circ} \right)\]
\[ \Rightarrow \beta = \tan^{- 1} \left( \frac{10 \sin 30^\circ}{10 + 10 \cos 30^\circ} \right)\]
\[ \Rightarrow \beta = \tan^{- 1} \left( \frac{1}{2 + \sqrt{3}} \right) = \tan^{- 1} \left( \frac{1}{3 . 372} \right)\]
\[ \Rightarrow \beta = \tan^{- 1} \left( 0 . 26795 \right) = 15^\circ\]
Angle made by the resultant vector with the X-axis = 15° + 30° = 45°
∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the X-axis.
