1. **Problem Statement:** We have three displacement vectors \(\vec{A}, \vec{B}, \vec{C}\) with magnitudes \(A=10\) cm, \(B=7\) cm, \(C=8\) cm and direction angles \(\alpha=53^\circ, \beta=120^\circ, \gamma=210^\circ\) respectively. We want to find the magnitude and direction of: (a) \(\vec{R} = \vec{A} + \vec{B}\) (b) \(\vec{P} = \vec{A} - \vec{B}\) (c) \(\vec{Q} = \vec{A} - 2\vec{B} + \vec{C}\) 2. **Formula and Rules:** - Vector components: \(x = r \cos \theta\), \(y = r \sin \theta\) - Vector addition/subtraction is done component-wise. - Magnitude of vector \(\vec{V} = (V_x, V_y)\) is \(\sqrt{V_x^2 + V_y^2}\). - Direction angle \(\phi = \tan^{-1}(\frac{V_y}{V_x})\), adjusted for quadrant. 3. **Calculate components:** \(A_x = 10 \cos 53^\circ = 10 \times 0.6018 = 6.018\) cm \(A_y = 10 \sin 53^\circ = 10 \times 0.7986 = 7.986\) cm \(B_x = 7 \cos 120^\circ = 7 \times (-0.5) = -3.5\) cm \(B_y = 7 \sin 120^\circ = 7 \times 0.8660 = 6.062\) cm \(C_x = 8 \cos 210^\circ = 8 \times (-0.8660) = -6.928\) cm \(C_y = 8 \sin 210^\circ = 8 \times (-0.5) = -4.0\) cm 4. **(a) Vector \(\vec{R} = \vec{A} + \vec{B}\):** \(R_x = A_x + B_x = 6.018 - 3.5 = 2.518\) cm \(R_y = A_y + B_y = 7.986 + 6.062 = 14.048\) cm Magnitude: $$R = \sqrt{2.518^2 + 14.048^2} = \sqrt{6.34 + 197.35} = \sqrt{203.69} = 14.27\text{ cm}$$ Direction: $$\theta_R = \tan^{-1}\left(\frac{14.048}{2.518}\right) = \tan^{-1}(5.58) = 80.2^\circ$$ 5. **(b) Vector \(\vec{P} = \vec{A} - \vec{B}\):** \(P_x = A_x - B_x = 6.018 - (-3.5) = 9.518\) cm \(P_y = A_y - B_y = 7.986 - 6.062 = 1.924\) cm Magnitude: $$P = \sqrt{9.518^2 + 1.924^2} = \sqrt{90.61 + 3.70} = \sqrt{94.31} = 9.71\text{ cm}$$ Direction: $$\theta_P = \tan^{-1}\left(\frac{1.924}{9.518}\right) = \tan^{-1}(0.202) = 11.5^\circ$$ 6. **(c) Vector \(\vec{Q} = \vec{A} - 2\vec{B} + \vec{C}\):** Calculate \(-2\vec{B}\): \((-2B)_x = -2 \times (-3.5) = 7.0\) cm \((-2B)_y = -2 \times 6.062 = -12.124\) cm Sum components: \(Q_x = A_x + (-2B)_x + C_x = 6.018 + 7.0 - 6.928 = 6.09\) cm \(Q_y = A_y + (-2B)_y + C_y = 7.986 - 12.124 - 4.0 = -8.138\) cm Magnitude: $$Q = \sqrt{6.09^2 + (-8.138)^2} = \sqrt{37.09 + 66.23} = \sqrt{103.32} = 10.17\text{ cm}$$ Direction: $$\theta_Q = \tan^{-1}\left(\frac{-8.138}{6.09}\right) = \tan^{-1}(-1.337) = -53.3^\circ$$ Since \(Q_y < 0\) and \(Q_x > 0\), angle is in 4th quadrant, so $$\theta_Q = 360^\circ - 53.3^\circ = 306.7^\circ$$ **Final answers:** - (a) \(R = 14.27\) cm, \(\theta_R = 80.2^\circ\) - (b) \(P = 9.71\) cm, \(\theta_P = 11.5^\circ\) - (c) \(Q = 10.17\) cm, \(\theta_Q = 306.7^\circ\)