Problem 76 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the magnitude and direction angle of the vector v.$$\mathbf{v}=8 \mathbf{i}-3 \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude of the vector \(\mathbf{v} = 8\mathbf{i} - 3\mathbf{j}\) is \( |\mathbf{v}| = \sqrt{73} \) and the direction angle is \(339.44^{\circ}\), measured anti-clockwise from the positive x-axis.

Step 1: Calculate the Magnitude of the Vector

Use the formula: \( \|v\| = \sqrt{(v_1)^2 + (v_2)^2} \). Substituting \( v_1 = 8 \) and \( v_2 = -3 \), the magnitude is \( \|v\| = \sqrt{(8)^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73} \).

Step 2: Find the Direction Angle of the Vector

Use the formula \(\tan\theta = \frac{v_2}{v_1}\). Substituting \( v_1 = 8 \) and \( v_2 = -3 \), we get \( \tan\theta = \frac{-3}{8} \). Calculating the atan (inverse of tan) of \( \frac{-3}{8} \), it is found that \(\theta \approx -20.56^{\circ}\). The negative sign indicates the angle is measured clockwise from the positive x-axis.

Step 3: Making the Angle Positive

In the standard convention, angles are measured anti-clockwise from the positive x-axis and are kept within the range [0, 360). So, adding 360° to the clockwise angle generates a positive anti-clockwise angle. \(\theta \approx -20.56^{\circ} + 360^{\circ} = 339.44^{\circ}\)

Key Concepts

Magnitude of a Vector

The magnitude of a vector provides a mathematical measure of its length or size. You can think of it as the distance from the vector's starting point to its endpoint in a coordinate system.
To calculate the magnitude (often represented as \( \| \mathbf{v} \| \)), use the formula:
\[\| \mathbf{v} \| = \sqrt{(v_1)^2 + (v_2)^2}\]
This formula applies the Pythagorean theorem. For example, given a vector \( \mathbf{v} = 8 \mathbf{i} - 3 \mathbf{j} \), where \( v_1 = 8 \) and \( v_2 = -3 \), the magnitude is calculated as:
\[ \| \mathbf{v} \| = \sqrt{(8)^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73} \] This defines the length of the vector in a 2D coordinate plane.

Direction Angle

The direction angle of a vector describes the angle it makes with the positive x-axis. It essentially shows the vector's orientation in the plane.
To find the direction angle \( \theta \), we use the tangent function which relates to the vector's components:
\( \tan\theta = \frac{v_2}{v_1} \).
In our example, with \( v_1 = 8 \) and \( v_2 = -3 \), you find that:
\( \tan\theta = \frac{-3}{8} \).
The direction angle \( \theta \approx -20.56^{\circ} \), which is negative and suggests the angle is measured clockwise from the x-axis.
Ensuring your angle is positive for most applications involves adding \( 360^{\circ} \) to adjust to the standard convention, resulting in an angle of \( \theta \approx 339.44^{\circ} \). This position is now correctly interpreted on a circular unit.

Vector Components

Vector components help in breaking down a vector into its foundational parts along the coordinate axes.
Any vector, like \( \mathbf{v} = 8 \mathbf{i} - 3 \mathbf{j} \) in 2D, consists of two components:

• The horizontal component \( 8 \mathbf{i} \) (along the x-axis)
• The vertical component \( -3 \mathbf{j} \) (along the y-axis)

These components follow logically from how vectors are used to represent quantities with both magnitude and direction.
The decomposition into components is a helpful way to understand complex vector interactions and the resulting calculations, such as finding the magnitude or direction angle.

Tangent Function

The tangent function primarily connects an angle from the trigonometric circle to the ratio of two sides of a right triangle.
In a vector context, it helps find the direction angle using the formula \( \tan\theta = \frac{v_2}{v_1} \).
Here, \( v_1 \) is the adjacent side, and \( v_2 \) is the opposite side of the angle in the right triangle formed by the vector.
The value \( \frac{v_2}{v_1} \) gives a specific tangent value, but remember, tangent values do not give the angle directly due to the multiple possibilities across the circular angles.

Inverse Tangent

The inverse tangent function, also known as \( \arctan \), is used to calculate the angle when you know the tangent's value.
This function "undoes" what tangent does—taking a ratio from a right triangle's sides and giving back an angle.
For a given tangent result from a vector, like \( \frac{-3}{8} \), the inverse tangent \( \theta = \arctan\left(\frac{-3}{8}\right) \approx -20.56^{\circ} \) finds the angle.
Keep in mind that this gives us an angle in the range \([-90^{\circ}, 90^{\circ}]\).
Using inverse tangent helps convert the numerical relationship between vector components into a directional angle, a key detail for understanding vector direction and orientation.