Find the magnitude and direction angle for each vector. Round angle measures to the nearest tenth, as necessary.
〈15, -8〉

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Identify the components of the vector. Here, the vector is given as \( \langle 15, -8 \rangle \), where 15 is the x-component and -8 is the y-component.

Calculate the magnitude of the vector using the formula for the length of a vector: \( \text{magnitude} = \sqrt{x^2 + y^2} \). Substitute the values: \( \sqrt{15^2 + (-8)^2} \).

Find the direction angle \( \theta \) of the vector relative to the positive x-axis using the inverse tangent function: \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \). Substitute the values: \( \tan^{-1} \left( \frac{-8}{15} \right) \).

Determine the correct quadrant for the angle. Since the x-component is positive and the y-component is negative, the vector lies in the fourth quadrant. Adjust the angle accordingly if necessary to express it as a positive angle measured counterclockwise from the positive x-axis.

Round the direction angle to the nearest tenth of a degree as required.

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Vector Magnitude

The magnitude of a vector represents its length and is calculated using the Pythagorean theorem. For a vector with components (x, y), the magnitude is √(x² + y²). This gives a non-negative scalar value indicating the vector's size.

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04:44

Finding Magnitude of a Vector

Direction Angle of a Vector

The direction angle of a vector is the angle it makes with the positive x-axis, measured counterclockwise. It can be found using the inverse tangent function: θ = arctan(y/x). Adjustments may be needed based on the vector's quadrant to get the correct angle.

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Finding Direction of a Vector

Quadrant Considerations in Angle Calculation

Since arctan(y/x) only returns values between -90° and 90°, the vector's quadrant must be considered to determine the correct direction angle. For vectors in quadrants II and III, add 180°; for quadrant IV, add 360° if needed to express the angle between 0° and 360°.

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