Vector \(\overrightarrow{\mathbf{B}}\) has magnitude 7.1 and direction \(14^{\circ}\) below the \(+x\) -axis. Vector \(\quad \overrightarrow{\mathbf{C}}\) has \(x\) -component \(C_{x}=-1.8\) and \(y\) -component \(C_{y}=-6.7 .\) Compute \((a)\) the \(x-\) and \(y\) -components of \(\overrightarrow{\mathbf{B}} ;\) (b) the magnitude and direction of \(\overrightarrow{\mathbf{C}} ;\) (c) the magnitude and direction of \(\overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{B}} ;\) (d) the magnitude and direction of \(\overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}} ;\) (e) the \(x\) - and \(y\) -components of \(\overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}\)

To find the \(x\) and \(y\) components of vector \(\overrightarrow{\mathbf{B}}\), we can use the magnitude and direction of the vector and trigonometric functions. We have: \(B_x = B \cos\theta\) \(B_y = B \sin\theta\) Where \(B\) = 7.1 (magnitude of \(\overrightarrow{\mathbf{B}}\)) and \(\theta = 14^{\circ}\) (angle below the \(+x\)-axis). \(B_x = 7.1 \cos(14^{\circ})\) \(B_y = -7.1 \sin(14^{\circ})\) After calculating these values, we get: \(B_x \approx 6.83\) \(B_y \approx -1.70\)

We can find the magnitude of vector \(\overrightarrow{\mathbf{C}}\) using the Pythagorean theorem: \(C = \sqrt{(-1.8)^2 + (-6.7)^2}\) After calculating this value, we get: \(C\approx7.22\) To find the angle, we can use the inverse tangent function: \(\phi = \tan^{-1}\left(\frac{C_y}{C_x}\right)\) \(\phi = \tan^{-1}\left(\frac{-6.7}{-1.8}\right)\) After calculating this value, we get: \(\phi = 104.70^{\circ}\) (measured counter-clockwise from the \(x\)-axis)

To find the sum of two vectors, we first add their corresponding components: \(D_x = C_x + B_x\) \(D_y = C_y + B_y\) \(D_x = -1.8 + 6.83\) \(D_y = -6.7 + (-1.70)\) After calculating these values, we get: \(D_x \approx 5.03\) \(D_y \approx -8.40\) Now, to find the magnitude of vector \(\overrightarrow{\mathbf{D}} = \overrightarrow{\mathbf{C}} + \overrightarrow{\mathbf{B}}\), we can use the Pythagorean theorem: \(D = \sqrt{(5.03)^2 + (-8.4)^2}\) After calculating this value, we get: \(D\approx9.74\) To find the angle, we can use the inverse tangent function: \(\psi = \tan^{-1}\left(\frac{D_y}{D_x}\right)\) \(\psi = \tan^{-1}\left(\frac{-8.4}{5.03}\right)\) After calculating this value, we get: \(\psi\approx59.37^{\circ}\) (measured counter-clockwise from the \(x\)-axis)

To find the difference of two vectors, we subtract their corresponding components: \(E_x = C_x - B_x\) \(E_y = C_y - B_y\) \(E_x = -1.8 - 6.83\) \(E_y = -6.7 - (-1.7)\) After calculating these values, we get: \(E_x \approx -8.63\) \(E_y \approx -5.00\) Now, to find the magnitude of vector \(\overrightarrow{\mathbf{E}} = \overrightarrow{\mathbf{C}} - \overrightarrow{\mathbf{B}}\), we can use the Pythagorean theorem: \(E = \sqrt{(-8.63)^2 + (-5.00)^2}\) After calculating this value, we get: \(E\approx10.03\) To find the angle, we can use the inverse tangent function: \(\chi = \tan^{-1}\left(\frac{E_y}{E_x}\right)\) \(\chi = \tan^{-1}\left(\frac{-5.00}{-8.63}\right)\) After calculating this value, we get: \(\chi\approx30.04^{\circ}\) (measured counter-clockwise from the \(x\)-axis)

Since we have already found the components of \(\overrightarrow{\mathbf{E}} = \overrightarrow{\mathbf{C}} - \overrightarrow{\mathbf{B}}\) in Part (d), we can simply list them: \(E_x \approx -8.63\) \(E_y \approx -5.00\)