Problem 75 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the magnitude and direction angle of the vector v. $$\mathbf{v}=12 \mathbf{i}+15 \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude of the vector is 19.2093727 and the direction angle is 51.34 degrees.

Step 1: Calculate the magnitude of the vector

Use the formula for the magnitude (\(M\)) of a 2D vector; \(M = \sqrt{x^{2} + y^{2}}\). Here \(x = 12\) and \(y = 15\). So substitute the values to get \(M = \sqrt{12^{2} + 15^{2}} = \sqrt{144 + 225} = \sqrt{369} = 19.2093727\). Therefore, the magnitude is 19.2093727.

Step 2: Calculate the direction angle of the vector

Use the formula \(tan(\theta) = \frac{y}{x}\) to find the direction angle. Here \(x = 12\) and \(y = 15\). So substitute the values to get \(tan(\theta) = \frac{15}{12} = 1.25\). Therefore, \(\theta = arctan(1.25)\) which gives the result in radians. To convert to degrees, multiply by \(\frac{180}{\pi}\). So \(\theta = arctan(1.25) * \frac{180}{\pi} = 51.34\) degrees. Therefore, the direction angle is 51.34 degrees.

Key Concepts

Magnitude of Vector

To find the magnitude, imagine the length of a vector as the straight-line distance from its starting point to its end point in a 2D space. For the vector \( \mathbf{v} = 12 \mathbf{i} + 15 \mathbf{j} \), this calculation helps you understand "how long" the vector is.You use the Pythagorean theorem, which is usually written as \( \sqrt{x^2 + y^2} \), where:

• \( x \) is the horizontal component (in this case, 12).
• \( y \) is the vertical component (here, 15).

Plug these into the formula: \( M = \sqrt{12^2 + 15^2} = \sqrt{369} \). The value comes out to approximately \( 19.21 \).
This number tells you how far the vector stretches, showing its "size" in the 2D plane.

Direction Angle

The direction angle gives you a sense of "where" the vector is pointing relative to a baseline, typically the positive x-axis in 2D space. It's like asking where a compass needle is pointing.To calculate it, use the tangent function: \( \tan(\theta) = \frac{y}{x} \). In our vector, \( x = 12 \) and \( y = 15 \), so \( \tan(\theta) = \frac{15}{12} = 1.25 \).The angle \( \theta \) is then found by taking the inverse tangent, or \( \arctan(1.25) \). This gives us a value in degrees to better match everyday use, by converting radians to degrees using \( \frac{180}{\pi} \). After conversion, \( \theta \approx 51.34^\circ \).Thus, this angle suggests that our vector points diagonally upward and rightward in the 2D plane.

2D Vector Analysis

2D vector analysis pulls these elements together to give a full picture of a vector's positioning in a plane.

• Components: Vectors in 2D have two parts: an i-component (horizontal) and a j-component (vertical).
• Magnitude: Calculated by \( \sqrt{x^2 + y^2} \), it shows the vector's length.
• Direction Angle: Using \( \tan^{-1} \), it offers insight into the orientation of the vector.

Breaking vectors into these components simplifies complex problems, much like breaking down complicated tasks into easier steps. This approach allows for clearer understanding and application in various fields, such as physics, engineering, and computer graphics.
By dissecting vectors into magnitudes and direction angles, we unlock their full potential in describing motion, force, and more.