Problem 70 - Precalculus with Limits

Question

Finding the Component Form of a Vector In Exercises \(67-74\) , find the component form of \(v\) given its magnitude and the angle it makes with the positive \(x\) -axis. Sketch y. $$\begin{array}{ll}{\text { Magnitude }} & {\text { Angle }} \\\ {\|\mathbf{v}\|=\frac{3}{4}} & {\theta=150^{\circ}}\end{array}$$

Step-by-Step Solution

Answer

The component form of vector \(v\) is \(v = - \frac{3\sqrt{3}}{8}\hat{i} + \frac{3}{8}\hat{j}\).

Step 1: Determine the x-component of the vector

Remember the x-component of the vector is given by \( v_x = \|v\| \cdot \cos(\theta) \) where \(\|v\|\) is the magnitude of the vector and \(\theta\) is the angle it makes with the positive x-axis. Substituting \(\|v\| = \frac{3}{4}\) and \(\theta=150^{\circ}\) into the equation yields \( v_x =\frac{3}{4} \cdot \cos(150^{\circ})\).

Step 2: Calculate the x-component

Compute the value of the above equation to get the x-component. Note that \(\cos(150^{\circ}) = - \frac{\sqrt{3}}{2}\) which gives \(v_x = - \frac{3\sqrt{3}}{8}\).

Step 3: Determine the y-component of the vector

Similarly, the y-component of the vector is given by \( v_y = \|v\| \cdot \sin(\theta) \) . Substituting the same values \(\|v\| = \frac{3}{4}\) and \(\theta=150^{\circ}\) into the equation yields \( v_y =\frac{3}{4} \cdot \sin(150^{\circ})\).

Step 4: Calculate the y-component

Compute the value of the above equation to get the y-component. Note that \(\sin(150^{\circ}) = \frac{1}{2}\), which gives \(v_y = \frac{3}{8}\).

Step 5: Write the vector in component form

Combining the x and y components, the vector \(v\) in component form is \(v = v_x \hat{i} + v_y \hat{j}\) which is \(v = - \frac{3\sqrt{3}}{8}\hat{i} + \frac{3}{8}\hat{j}\).

Key Concepts

Magnitude of a Vector

The magnitude of a vector represents its length or size and is often denoted by double vertical bars, such as \( \| \mathbf{v} \| \). Imagine you're holding an arrow; the length of this arrow from its tail to the tip is the vector's magnitude. This is a scalar quantity, meaning it only has magnitude and no direction.

To visualize this better, if you draw a right triangle with a vector as the hypotenuse, the magnitude of the vector would be the length of the hypotenuse. It's calculated with the help of the Pythagorean theorem when dealing with the vector's components, although in our exercise, it's provided directly as \(\|\mathbf{v}\| = \frac{3}{4}\).

Direction Angle

The direction angle of a vector is the angle it makes with the positive x-axis, typically measured in degrees or radians. It's like using a compass to find out which way you're headed, but instead of North, South, East, and West, we're talking about angles from the positive x-axis.

In our example, the vector \( \mathbf{v} \) has a direction angle of 150 degrees. It's important to note that the angle should be measured in a counterclockwise direction from the positive x-axis to the vector to be consistent with standard trigonometric practice.

Trigonometric Functions

Trigonometric functions are like the GPS of mathematics, providing a way to navigate through angles and sides of a triangle. The most common are sine (sin), cosine (cos), and tangent (tan). These functions relate the angles of a triangle to the ratios of its sides.

In the context of vectors, we use \(\cos\) to find the x-component and \(\sin\) to find the y-component. It's like projecting the vector onto the x-axis and y-axis. For our direction angle of 150 degrees, \(\cos(150^\circ) = - \frac{\sqrt{3}}{2}\) and \(\sin(150^\circ) = \frac{1}{2}\), both of which are crucial in calculating the respective components of our vector.

Component Form Calculation

To put a vector into component form, we split it into its x and y parts, which are like the building blocks of the vector's overall direction and magnitude. Think of it as breaking down a lego structure into individual lego pieces.

To do this, we use the equations \(v_x = \|v\| \cdot \cos(\theta)\) for the x-component, and \(v_y = \|v\| \cdot \sin(\theta)\) for the y-component. After calculating these using the provided magnitude and direction angle (as shown in our exercise), we combine them into the vector's component form, which in this case is \(v = -\frac{3\sqrt{3}}{8}\hat{i} + \frac{3}{8}\hat{j}\).