Problem 37 - Algebra and Trigonometry

Question

Find the magnitude and direction (in degrees) of the vector. $$ \mathbf{v}=\mathbf{i}+\sqrt{3} \mathbf{j} $$

Step-by-Step Solution

Answer

Magnitude is 2; direction is 60 degrees.

Step 1: Identify the Components of the Vector

The given vector \( \mathbf{v} \) can be written as: \( \mathbf{v} = 1\mathbf{i} + \sqrt{3} \mathbf{j} \). Hence, the components of the vector are 1 in the \( x \)-direction and \( \sqrt{3} \) in the \( y \)-direction.

Step 2: Calculate the Magnitude of the Vector

To find the magnitude of a vector \( \mathbf{v} = x \mathbf{i} + y \mathbf{j} \), use the formula: \[ \| \mathbf{v} \| = \sqrt{x^2 + y^2} \] Substituting the values, \( x = 1 \) and \( y = \sqrt{3} \):\[ \| \mathbf{v} \| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

Step 3: Find the Direction of the Vector

The direction of the vector is the angle \( \theta \) that \( \mathbf{v} \) makes with the positive \( x \)-axis. Calculate \( \theta \) using: \[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \] Substituting, \( x = 1 \) and \( y = \sqrt{3} \):\[ \theta = \tan^{-1} \left( \frac{\sqrt{3}}{1} \right) = \tan^{-1}(\sqrt{3}) = 60^{\circ} \]

Step 4: Summarize the Magnitude and Direction

The magnitude of the vector \( \mathbf{v} \) is 2 and the direction angle is \( 60^{\circ} \).

Key Concepts

Vector Components

When dealing with vectors, one of the first steps is to break the vector into its components. These components describe how far the vector stretches in each direction (usually along the x and y axes). For example, in the vector \( \mathbf{v} = 1\mathbf{i} + \sqrt{3} \mathbf{j} \), we have two components:

• 1 in the x-direction (represented by \( \mathbf{i} \))
• \( \sqrt{3} \) in the y-direction (represented by \( \mathbf{j} \))

Understanding vector components is crucial because they tell us the vector's influence in each direction. This foundational step helps in further calculations like determining magnitude and direction.

Magnitude Calculation

The magnitude of a vector represents its length. It's a measure of how much ground a vector covers, regardless of direction. To calculate it, apply the Pythagorean theorem using the components of the vector. For a vector \( \mathbf{v} = x\mathbf{i} + y\mathbf{j} \), the magnitude \( \| \mathbf{v} \| \) is calculated as follows:

\[ \| \mathbf{v} \| = \sqrt{x^2 + y^2} \]

In our example, the components are \( x = 1 \) and \( y = \sqrt{3} \). Plug these into the formula:

\[ \| \mathbf{v} \| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

This tells us that the vector is 2 units long. Magnitude gives us a sense of scale, making it essential for knowing how strong or intense a vector is over its given direction.

Direction Angle

Vectors have not only magnitude but also direction, defined by the angle they make with a reference axis, usually the positive x-axis. This angle is called the direction angle. It is essential for understanding where a vector points in the plane.

To find this angle, use the tangent function from trigonometry, which relates the sides of a right triangle. Given a vector \( \mathbf{v} = x\mathbf{i} + y\mathbf{j} \), the direction angle \( \theta \) can be determined using:

\[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \]

For our vector, where \( x = 1 \) and \( y = \sqrt{3} \):

\[ \theta = \tan^{-1} \left( \frac{\sqrt{3}}{1} \right) = \tan^{-1}(\sqrt{3}) = 60^{\circ} \]

Thus, the vector forms a \( 60^{\circ} \) angle with the positive x-axis, directing us to imagine it pointing diagonally upwards on a graph.

Trigonometry

Trigonometry is a branch of mathematics featuring prominently in vector analysis. It deals with the relationships between the angles and sides of triangles, especially right triangles. Understanding trigonometry is vital when calculating direction angles of vectors.

In our context, trigonometry helps us find angles using the tangent function as applied with:

\[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \]

Here, \( y \) and \( x \) are the lengths of the opposite and adjacent sides of a right triangle formed by the vector projection on the x and y axes. The inverse tangent (\( \tan^{-1} \)) function returns the angle whose tangent is the quotient of these sides.

Grasping these trigonometric concepts allows you to determine angles effectively, thus depicting the exact orientation of vectors in a plane.