Problem 33 - Algebra and Trigonometry

Question

Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\langle 3,4\rangle $$

Step-by-Step Solution

Answer

Magnitude: 5, Direction: 53.13°

Step 1: Understanding Vector Components

The vector \( \mathbf{v} = \langle 3, 4 \rangle \) is given in component form, where \( 3 \) is the horizontal (\( x \)) component and \( 4 \) is the vertical (\( y \)) component. We need to calculate the magnitude and direction of this vector.

Step 2: Calculating the Magnitude

The magnitude of a vector \( \mathbf{v} = \langle x, y \rangle \) is given by the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \). Substitute \( x = 3 \) and \( y = 4 \) into the formula:\[ \| \mathbf{v} \| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \].The magnitude of the vector is 5.

Step 3: Finding the Direction in Degrees

The direction of a vector is given by the angle \( \theta \) with respect to the positive \( x \)-axis. Use the tangent function: \( \tan \theta = \frac{y}{x} \). Here, \( y = 4 \) and \( x = 3 \), so:\[ \tan \theta = \frac{4}{3} \].To find \( \theta \), take the arctangent (inverse tangent):\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^{\circ} \].Thus, the direction of the vector is approximately 53.13 degrees.

Key Concepts

Vector Components

Vectors are mathematical objects that have both a magnitude and a direction. To help understand their position in a plane, they can be broken down into two important parts called components. These components are simply the projections of the vector along the horizontal and vertical axes.

• The horizontal component is often referred to as the "x-component." This indicates how far the vector stretches in the left or right direction.
• The vertical component, or the "y-component," shows how tall the vector is in the upward or downward direction.

For the vector given in the exercise, \( \mathbf{v} = \langle 3, 4 \rangle \), the number 3 represents the x-component, while 4 represents the y-component. Think of these components as building blocks that come together to form the original vector.

Magnitude Formula

The magnitude of a vector gives the length or size of the vector. It's a measure of how long the vector is, visually speaking.To calculate the magnitude, you take the square root of the sum of the squares of its components. This is often referred to as "finding the hypotenuse," as it follows the formula for the length of the hypotenuse in a right-angled triangle.The formula to find the magnitude \( \| \mathbf{v} \| \) of a vector \( \mathbf{v} = \langle x, y \rangle \) is:\[ \| \mathbf{v} \| = \sqrt{x^2 + y^2} \]Plugging in the example values, for \( \mathbf{v} = \langle 3, 4 \rangle \):

• First, square the components: \( 3^2 = 9 \) and \( 4^2 = 16 \).
• Then, add these squares: \( 9 + 16 = 25 \).
• Finally, take the square root of this sum: \( \sqrt{25} = 5 \).

Hence, the magnitude of the vector is 5.

Direction Angle Calculation

Once you have the components of a vector, you can also find the direction it points in. This direction is usually specified by the angle it makes with the positive x-axis.The formula to find this angle \( \theta \) is based on the tangent function which relates the opposite and adjacent sides in a right triangle:\[ \tan \theta = \frac{y}{x} \]Here, you simply divide the y-component by the x-component to find the tangent of the angle.For the vector \( \mathbf{v} = \langle 3, 4 \rangle \):

• Calculate the tangent: \( \tan \theta = \frac{4}{3} \).

This result tells us that the vector's angle is dependent on this ratio, setting up the stage for finding the exact angle using the next step.

Inverse Tangent Function

The inverse tangent function, often denoted as \( \tan^{-1} \) or "arctan", is used to find an angle given its tangent value. Basically, it helps us reverse what the tangent did earlier to find the exact direction angle \( \theta \) of the vector.To apply this, you use the value obtained from the tangent calculation:\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]For our vector \( \mathbf{v} = \langle 3, 4 \rangle \), where we had \( \tan \theta = \frac{4}{3} \):

• Use the inverse tangent: \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).
• Calculate \( \theta \approx 53.13^{\circ} \).

Now, we know the direction of the vector is approximately 53.13 degrees from the positive x-axis. This angle provides crucial directional information, defining exactly where the vector "points" in a 2D space.