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Given the vector \( \mathbf{u} \) equal to \( 6\left\langle\cos 205^{\circ}, \sin 205^{\circ}\right\rangle \) and vector \( \mathbf{v} \) equal to \( 5\left\langle\cos 120^{\circ}, \sin 120^{\circ}\right\rangle \), find the \( \operatorname{sum} \mathbf{u}+\mathbf{v} \) and write your answer in magnitude and direction form with the magnitude rounded to the nearest tenth and the direction rounded to the nearest degree, \( 0^{\circ} \leq \theta<360^{\circ} \).

Ask by Ortiz Kelly. in the United States
Jan 24,2025

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The sum of vectors \( \mathbf{u} \) and \( \mathbf{v} \) is approximately 8.1 units at 167 degrees.

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To find the sum \(\mathbf{u} + \mathbf{v}\), we first need to compute the components of \(\mathbf{u}\) and \(\mathbf{v}\). The vector \(\mathbf{u} = 6\left\langle\cos 205^{\circ}, \sin 205^{\circ}\right\rangle\): - \(\cos 205^{\circ} = -\cos 25^{\circ} \approx -0.9063\) - \(\sin 205^{\circ} = -\sin 25^{\circ} \approx -0.4226\) Calculating the components of \(\mathbf{u}\): \[ \mathbf{u} = 6\left\langle -0.9063, -0.4226 \right\rangle \approx \left\langle -5.4378, -2.5356 \right\rangle \] Now for the vector \(\mathbf{v} = 5\left\langle\cos 120^{\circ}, \sin 120^{\circ}\right\rangle\): - \(\cos 120^{\circ} = -\frac{1}{2} = -0.5\) - \(\sin 120^{\circ} = \frac{\sqrt{3}}{2} \approx 0.8660\) Calculating the components of \(\mathbf{v}\): \[ \mathbf{v} = 5\left\langle -0.5, 0.8660 \right\rangle \approx \left\langle -2.5, 4.33 \right\rangle \] Now, we sum \(\mathbf{u}\) and \(\mathbf{v}\): \[ \mathbf{u} + \mathbf{v} = \left\langle -5.4378 - 2.5, -2.5356 + 4.33 \right\rangle \approx \left\langle -7.9378, 1.7944 \right\rangle \] Next, we find the magnitude of \(\mathbf{u} + \mathbf{v}\): \[ \text{Magnitude} = \sqrt{(-7.9378)^2 + (1.7944)^2} \approx \sqrt{63.0270 + 3.2195} \approx \sqrt{66.2465} \approx 8.1 \] Now to find the direction \(\theta\): \[ \theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{1.7944}{-7.9378}\right) \] Calculating the angle: - Since \(\mathbf{x}\) is negative and \(\mathbf{y}\) is positive, we are in the second quadrant. Calculating: \[ \theta = \arctan\left(-0.2260\right) \approx -12.6^{\circ} \] However, to convert it to the correct range for angle in the second quadrant: \[ \theta = 180^{\circ} - 12.6^{\circ} \approx 167.4^{\circ} \approx 167^{\circ} \] Thus, the final answer is: \[ \text{Magnitude} = 8.1, \quad \text{Direction} = 167^\circ \] In summary, the sum \(\mathbf{u} + \mathbf{v}\) in magnitude and direction form is approximately: \[ \boxed{8.1 \text{ at } 167^{\circ}} \]

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