Question

The magnitude of vectors \(\vec{OA}\), \(\vec{OB}\) and \(\vec{OC}\) in the given figure are equal. The direction of \(\vec{OA} + \vec{OB} - \vec{OC}\) with x-axis will be : 

• A. \(\tan^{-1} \left( \frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}} \right)\)
• B. \(\tan^{-1} \left( \frac{\sqrt{3} - 1 + \sqrt{2}}{1 + \sqrt{3} - \sqrt{2}} \right)\)
• C. \(\tan^{-1} \left( \frac{1 + \sqrt{3} - \sqrt{2}}{1 - \sqrt{3} - \sqrt{2}} \right)\)
• D. \(\tan^{-1} \left( \frac{\sqrt{3} - 1 + \sqrt{2}}{1 - \sqrt{3} + \sqrt{2}} \right)\)

Hint:

Be careful with the signs of trigonometric components based on the quadrant where the vector lies. Subtracting a vector is equivalent to adding its negative.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Vector addition and subtraction are performed by breaking the vectors into their horizontal (x) and vertical (y) components and then finding the angle of the resultant using \(\tan \theta = \frac{R_y}{R_x}\).
Step 2: Key Formula or Approach:
1. \(\vec{A} = A_x \hat{i} + A_y \hat{j}\)
2. \(\theta = \tan^{-1} \left( \frac{\sum Y}{\sum X} \right)\)
Step 3: Detailed Explanation:
Let the magnitude of each vector be \(V\). From the diagram:
- \(\vec{OA}\) makes \(30^\circ\) with x-axis: \(V(\cos 30^\circ \hat{i} + \sin 30^\circ \hat{j})\).
- \(\vec{OB}\) makes \(60^\circ\) below positive x-axis: \(V(\cos(-60^\circ) \hat{i} + \sin(-60^\circ) \hat{j}) = V(\cos 60^\circ \hat{i} - \sin 60^\circ \hat{j})\).
- \(\vec{OC}\) makes \(45^\circ\) with negative x-axis: \(V(\cos 135^\circ \hat{i} + \sin 135^\circ \hat{j}) = V(-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})\).
Calculate Resultant \(\vec{R} = \vec{OA} + \vec{OB} - \vec{OC}\):
\[ R_x = V[\cos 30^\circ + \cos 60^\circ - \cos 135^\circ] = V[\frac{\sqrt{3}}{2} + \frac{1}{2} - (-\frac{1}{\sqrt{2}})] = \frac{V}{2} [1 + \sqrt{3} + \sqrt{2}] \]
\[ R_y = V[\sin 30^\circ - \sin 60^\circ - \sin 135^\circ] = V[\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}] = \frac{V}{2} [1 - \sqrt{3} - \sqrt{2}] \]
The direction \(\theta\) with x-axis is:
\[ \tan \theta = \frac{R_y}{R_x} = \frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}} \]
Step 4: Final Answer:
The direction is \(\tan^{-1} \left( \frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}} \right)\).