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

To determine the direction of the vector \(\vec{OA} + \vec{OB} - \vec{OC}\) with respect to the x-axis, we need to calculate the components of each vector and then find the resultant vector. 

1. Vector Components: Given that the magnitudes of vectors \(\vec{OA}\), \(\vec{OB}\), and \(\vec{OC}\) are equal, we denote their magnitude by \(r\).
   • Vector \(\vec{OA}\): \(OA_x = r \cos 30^\circ = \frac{\sqrt{3}}{2} r\), \(OA_y = r \sin 30^\circ = \frac{1}{2} r\)
   • Vector \(\vec{OB}\): \(OB_x = r \cos 60^\circ = \frac{1}{2} r\), \(OB_y = -r \sin 60^\circ = -\frac{\sqrt{3}}{2} r\)
   • Vector \(\vec{OC}\): \(OC_x = -r \cos 45^\circ = -\frac{1}{\sqrt{2}} r\), \(OC_y = r \sin 45^\circ = \frac{1}{\sqrt{2}} r\)
2. Resultant Vector: The vector \(\vec{R} = \vec{OA} + \vec{OB} - \vec{OC}\) has components:
   • \(R_x = OA_x + OB_x - OC_x\)
     • \(\(R_x = \frac{\sqrt{3}}{2}r + \frac{1}{2}r + \frac{1}{\sqrt{2}}r\)\)
   • \(R_y = OA_y + OB_y - OC_y\)
     • \(\(R_y = \frac{1}{2}r - \frac{\sqrt{3}}{2}r - \frac{1}{\sqrt{2}}r\)\)
3. Direction with x-axis: The direction is given by \(\theta = \tan^{-1} \left(\frac{R_y}{R_x}\right)\).
   • \(\theta = \tan^{-1} \left(\frac{\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}}\right)\)
   • This simplifies to: \(\theta = \tan^{-1} \left(\frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}}\right)\)

Therefore, the direction of \(\vec{OA} + \vec{OB} - \vec{OC}\) with the x-axis is given by: \(\tan^{-1} \left( \frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}} \right)\).