Problem 42 - Contemporary Precalculus

Question

Find the magnitude and direction angle of the vector \(\boldsymbol{v}\). $$\mathbf{v}=-15 \mathbf{i}-10 \mathbf{j}$$

Step-by-Step Solution

Answer

Answer: The magnitude of the vector \(\mathbf{v}\) is \(\sqrt{325}\), and its direction angle is \(213.69^{\circ}\).

Step 1: Find the Magnitude of the Vector

To find the magnitude of the vector \(\mathbf{v} = -15\mathbf{i} - 10\mathbf{j}\), we use the formula: $$\|\mathbf{v}\| = \sqrt{(-15)^2 + (-10)^2}$$ Calculate the magnitude: $$\|\mathbf{v}\| = \sqrt{225 + 100} = \sqrt{325}$$ So, the magnitude of the vector \(\mathbf{v}\) is \(\sqrt{325}\).

Step 2: Find the Direction Angle of the Vector

To find the direction angle \(\theta\) of the vector \(\mathbf{v} = -15\mathbf{i} - 10\mathbf{j}\), we can use the tangent function: $$\tan\theta = \frac{-10}{-15}$$ Simplify the fraction: $$\tan\theta = \frac{2}{3}$$ Now, we find the angle using the inverse tangent function: $$\theta = \arctan{\frac{2}{3}}$$ Calculating the angle, we get: $$\theta \approx 33.69^{\circ}$$ Since the vector is in the third quadrant (both components are negative), we need to add \(180^{\circ}\) to the angle to get the correct direction angle: $$\theta = 33.69^{\circ} + 180^{\circ} = 213.69^{\circ}$$

Step 3: Write the Final Answer

The magnitude and direction angle of the vector \(\mathbf{v} = -15\mathbf{i} - 10\mathbf{j}\) are: Magnitude: \(\|\mathbf{v}\| = \sqrt{325}\) Direction angle: \(\theta = 213.69^{\circ}\)

Key Concepts

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