Problem 62 - Algebra and Trigonometry

Question

Write the vector \(\mathbf{v}\) in the form \(\mathbf{ai}+ \mathbf{bj}\), given its magnitude \(\|\mathbf{v}\|\) and the angle \(\alpha\) it makes with the positive \(x\) -axis. \(\mid \mathbf{v} \|=8, \quad \alpha=45^{\circ}\)

Step-by-Step Solution

Answer

\textbf{4\sqrt{2}i} + \textbf{4\sqrt{2}j}

Step 1 - Understand the relationship between magnitude and components

To write a vector in the form \(\textbf{ai} + \textbf{bj}\), recall that \(\textbf{a}\) is the component of the vector along the x-axis, and \(\textbf{b}\) is the component along the y-axis. The relationship between these components, the magnitude \(\textbf{v}\), and the angle \(\textbf{\alpha}\) is given by trigonometric functions: \[ \textbf{a} = \textbf{v} \cdot \cos(\alpha) \] \[ \textbf{b} = \textbf{v} \cdot \sin(\alpha) \]

Step 2 - Calculate the x-component

Using the cosine function, calculate the x-component: \[ \textbf{a} = 8 \cdot \cos(45^{\circ}) \] Since \(\cos(45^{\circ}) = \frac{\textbf{1}}{\textbf{\sqrt{2}}} \), we get: \[ \textbf{a} = 8 \cdot \frac{1}{\textbf{\sqrt{2}}} = \frac{8}{\textbf{\sqrt{2}}} = 4\textbf{\sqrt{2}} \]

Step 3 - Calculate the y-component

Using the sine function, calculate the y-component: \[ \textbf{b} = 8 \cdot \sin(45^{\circ}) \] Since \(\sin(45^{\circ}) = \frac{\textbf{1}}{\textbf{\sqrt{2}}} \), we get: \[ \textbf{b} = 8 \cdot \frac{1}{\textbf{\sqrt{2}}} = \frac{8}{\textbf{\sqrt{2}}} = 4\textbf{\sqrt{2}} \]

Step 4 - Write the vector in the form \(\textbf{ai} + \textbf{bj}\)

Combine the x-component and y-component to form the vector: \[ \textbf{v} = \textbf{4\sqrt{2}i} + \textbf{4\sqrt{2}j} \]

Key Concepts

magnitude

The magnitude of a vector measures its length or size. It is a scalar quantity, meaning it has only magnitude and no direction.
To find the magnitude of a vector, you can use the Pythagorean theorem if you have the components. For a vector \( \textbf{v} = \textbf{ai} + \textbf{bj} \), the magnitude \( \textbf{v} \) is calculated as follows:

\[ \| \textbf{v} \| = \sqrt{ \textbf{a}^2 + \textbf{b}^2} \]
This equation comes from the Pythagorean theorem. In our example, the magnitude of the vector is given as \| \textbf{v} \| = 8.
Understanding the magnitude helps in finding the vector's components and ultimately expressing it in standard form.

trigonometric functions

Trigonometric functions like sine and cosine are crucial in understanding the relationship between a vector's magnitude and its components.
These functions provide a way to link angles with lengths of sides in a right-angled triangle. In our example, we use:

\textbf{cos( \alpha )} to find the x-component
\textbf{sin( \alpha )} to find the y-component

Where \textbf{\alpha} is the angle the vector makes with the positive x-axis. Given the angle \textbf{\alpha = 45^{\circ}}:

Cosine Function: \[ \textbf{cos(45^{\circ})} = \frac{\textbf{1}}{\textbf{\sqrt{2}}} \]
Sine Function: \[ \textbf{sin(45^{\circ})} = \frac{\textbf{1}}{\textbf{\sqrt{2}}} \]

Using these trigonometric functions, we can calculate the specific components of our vector.

x-component

The x-component of a vector represents how much it extends in the x-direction. To find it, you use the cosine function:
\[ \textbf{a} = \| \textbf{v} \| \cdot \textbf{cos}( \alpha ) \] In our exercise:\ \| \textbf{v} \| = 8 and \textbf{ \alpha } = 45^{\circ}
So we calculate:
\[ \textbf{a} = 8 \cdot \textbf{cos}(45^{\circ}) = 8 \cdot \frac{\textbf{1}}{\textbf{\sqrt{2}}} = \frac{8}{\textbf{\sqrt{2}}} = 4 \textbf{ \sqrt{2}} \] This x-component tells us that the vector extends \approx 5.66 units to the right along the x-axis.

y-component

The y-component of a vector represents how much it extends in the y-direction. To find it, you use the sine function:
\[ \textbf{b} = \| \textbf{v} \| \cdot \textbf{sin}( \alpha ) \] In our exercise: \ \| \textbf{v} \| = 8 and \textbf{ \alpha } = 45^{\circ}
So we calculate:
\[ \textbf{b} = 8 \cdot \textbf{sin}(45^{\circ}) = 8 \cdot \frac{\textbf{1}}{\textbf{ \sqrt{2}}} = \frac{8}{\textbf{\sqrt{2}}} = 4 \textbf{ \sqrt{2}} \] This y-component tells us that the vector extends \approx 5.66 units upwards along the y-axis.
Both x and y components are essential for fully describing the vector's direction and length.