Problem 49 - Precalculus Mathematics for Calculus

Question

Find the magnitude and direction (in degrees) of the vector. $$\mathbf{v}=\langle- 12,5\rangle$$

Step-by-Step Solution

Answer

Magnitude: 13, Direction: 157.38°

Step 1: Understanding the Components

The vector \( \mathbf{v} = \langle -12, 5 \rangle \) has two components: -12 along the x-axis and 5 along the y-axis. We will use these components to find the magnitude and direction of the vector.

Step 2: Calculating the Magnitude

To find the magnitude of \( \mathbf{v} \), we use the formula for the magnitude of a vector: \[ |\mathbf{v} | = \sqrt{x^2 + y^2} \]Substitute \( x = -12 \) and \( y = 5 \) into the formula:\[ |\mathbf{v} | = \sqrt{(-12)^2 + (5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]Thus, the magnitude of \( \mathbf{v} \) is 13.

Step 3: Calculating the Direction in Degrees

To find the direction, we use the formula: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]where \( x = -12 \) and \( y = 5 \). Substitute the values into the equation:\[ \theta = \tan^{-1}\left(\frac{5}{-12}\right) \approx -22.62^\circ \]This angle is measured from the negative x-axis toward the positive y-axis. Since it's in the second quadrant, we add 180° to find the standard position angle:\[ \theta = -22.62^\circ + 180^\circ = 157.38^\circ \]So, the direction of the vector is approximately 157.38°.

Key Concepts

Vector Components

Vectors consist of components, which represent their values in specific directions. For any vector \( \mathbf{v} = \langle x, y \rangle \), these components have significant importance because they determine the vector's position and movement.

• The first value (-12 in our example) is the x-component. It shows how far the vector moves to the left or right on a standard graph.
• The second value (5 in the example) is the y-component. It represents the movement up or down.

These components provide a way to break down complex movements into simpler, one-dimensional movements. This simplification makes calculations involving vectors more manageable.

Magnitude of a Vector

The magnitude of a vector is like its length or size, quantifying how much of something the vector represents. To find this, we use the formula:\[|\mathbf{v}| = \sqrt{x^2 + y^2}\]This formula is derived from the Pythagorean theorem, which we use to find distances in geometry. Consider

• Squaring each component (x and y) ensures we're dealing with positive values.
• Adding the squared values to find the sum represents a complete vector size.
• Taking the square root gives us a real number, indicating the total length.

Using the given vector \( \mathbf{v} = \langle -12, 5 \rangle \), applying the formula gives us a magnitude of 13. This tells us the vector's distance from the origin on a graph.

Direction of a Vector

The direction of a vector plays a crucial role in understanding where it's headed on a graph. This direction is defined as the angle that the vector forms from a reference line, typically from the positive x-axis. The formula we use is:\[\theta = \tan^{-1}\left(\frac{y}{x}\right)\]Here:

• "\( y \)" and "\( x \)" are the vector components.
• "\( \tan^{-1} \)" or "arctan" helps us find the angle of the vector compared to the x-axis.

For our vector \( \mathbf{v} = \langle -12, 5 \rangle \):

• Calculate the initial angle using inverse tangent, which gives us \( -22.62^\circ \).
• Add 180° to shift this to a standard position, since our vector resides in the second quadrant, resulting in approximately \(157.38^\circ \).

This angle denotes how the vector points relative to the positive x-axis, which is essential for understanding its trajectory or direction.