Problem 46 - Algebra and Trigonometry

Question

\(41-46\) . Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors \(\mathbf{i}\) and \(\mathbf{j}\) . $$ |\mathbf{v}|=\sqrt{3}, \quad \theta=300^{\circ} $$

Step-by-Step Solution

Answer

The vector is \( \frac{\sqrt{3}}{2} \mathbf{i} - \frac{3}{2} \mathbf{j} \).

Step 1: Understand Vector Components

The vector can be expressed in terms of its magnitude, \(|\mathbf{v}|\), and angle \(\theta\) using trigonometry. The horizontal (x) component can be found using the cosine of the angle, and the vertical (y) component using the sine of the angle.

Step 2: Convert Angle to Radians

Since angles in trigonometric calculations often use radians, but here degrees are given, convert \(300^{\circ}\) to radians: \( \theta = 300^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{5\pi}{3} \) radians.

Step 3: Calculate Horizontal Component

The horizontal component (\( v_x \)) of the vector is given by \( v_x = |\mathbf{v}| \cos(\theta) \). Substituting the values: \( v_x = \sqrt{3} \cos(300^{\circ}) = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} \).

Step 4: Calculate Vertical Component

The vertical component (\( v_y \)) of the vector is given by \( v_y = |\mathbf{v}| \sin(\theta) \). Substituting the values: \( v_y = \sqrt{3} \sin(300^{\circ}) = \sqrt{3} \times \left(-\frac{\sqrt{3}}{2}\right) = -\frac{3}{2} \).

Step 5: Express Vector in terms of \(\mathbf{i}\) and \(\mathbf{j}\)

The vector can now be written as \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} = \frac{\sqrt{3}}{2} \mathbf{i} - \frac{3}{2} \mathbf{j} \). This combines both horizontal and vertical components in terms of unit vectors \(\mathbf{i}\) and \(\mathbf{j}\).

Key Concepts

Vector Magnitude

A vector's magnitude is essentially its length or size. It's a crucial part of understanding how vectors work. In this case, the magnitude of the vector is given as \(\sqrt{3}\). This tells us how 'long' the vector is, which will directly influence the size of its components. The magnitude is essential because it scales the horizontal and vertical components of a vector, providing a scalar multiplier to whatever trigonometric projections we calculate from the given angle. Think of magnitude as the overall force that gets "sliced" into both the x and y components.To understand why this is important, imagine the vector as a hypotenuse of a right triangle, where the length is \(\sqrt{3}\). The magnitudes of its horizontal and vertical components are essentially the adjacent and opposite sides of this triangle.

Angle Conversion to Radians

Angles can be measured in degrees or radians, and it's common to switch between these two forms in trigonometry. Radians are often more convenient for calculations involving periodic functions like sine and cosine. For instance, the given angle of \(300^\circ\) needs to be converted to radians for these calculations:

• To convert degrees to radians, use the formula: \( \text{radians} = \text{degrees} \times \frac{\pi}{180^\circ} \).
• Applying this to the given angle: \( \theta = 300^\circ \times \frac{\pi}{180^\circ} = \frac{5\pi}{3} \) radians.

This conversion is critical as it allows us to use trigonometric functions effectively, ensuring that our calculations for the vector components are accurate. Remember that radians provide a natural way to measure angles that directly relate the arc length to the radius of a circle.

Trigonometric Functions: Sine and Cosine

Understanding sine and cosine is key to breaking down a vector into its components. These functions help us project the magnitude of the vector onto the horizontal (x-axis) and vertical (y-axis) planes respectively.

• The cosine of an angle in a right triangle describes the ratio of the adjacent side (horizontal component) to the hypotenuse (magnitude).
• The sine of an angle describes the ratio of the opposite side (vertical component) to the hypotenuse.

For example, let's consider \(300^\circ\):

• The horizontal component \(v_x = |\mathbf{v}| \cos(\theta)\), with \(\cos(300^\circ) = \frac{1}{2}\), results in \(v_x = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}\).
• The vertical component \(v_y = |\mathbf{v}| \sin(\theta)\), with \(\sin(300^\circ) = -\frac{\sqrt{3}}{2}\), results in \(v_y = \sqrt{3} \times \left(-\frac{\sqrt{3}}{2}\right) = -\frac{3}{2}\).

These calculations show how the trigonometric functions, sine and cosine, are handy for understanding the components of a vector relative to its orientation in a plane.

Unit Vectors i and j

Unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are the building blocks for vectors in a 2D plane.

• \(\mathbf{i}\) represents the unit vector along the x-axis, typically having a magnitude of 1 in that direction.
• \(\mathbf{j}\) is the unit vector along the y-axis, with a magnitude of 1 in the vertical direction.

These vectors allow us to describe any 2D vector in terms of its components as a linear combination. The original problem asks us to express the vector in terms of \(\mathbf{i}\) and \(\mathbf{j}\):Here, the vector is: \[\mathbf{v} = \frac{\sqrt{3}}{2} \mathbf{i} - \frac{3}{2} \mathbf{j}\]This means that the vector's horizontal component, \(\frac{\sqrt{3}}{2}\), is expressed in terms of \(\mathbf{i}\), and its vertical component, \(-\frac{3}{2}\), is expressed in terms of \(\mathbf{j}\). Using \(\mathbf{i}\) and \(\mathbf{j}\) in this way provides a clear and standardized way to represent vector quantities in terms of their direction and magnitude.