91Ó°ÊÓ

Vectors are mathematical quantities that have both magnitude and direction. The magnitude, often referred to as the "length" of a vector, is like measuring the "size" or "extent" of the vector. To calculate the magnitude of a vector, we use the components of the vector, which are essentially the coordinates in the vector notation.

In the original exercise, the given vector is \( \langle 8, -4 \rangle \). Here, the first number (8) is the x-component, and the second number (-4) is the y-component.

The formula for the magnitude of a 2D vector \( \langle a, b \rangle \) is:

• \( \text{Magnitude} = \sqrt{a^2 + b^2} \)

Applying this formula to \( \langle 8, -4 \rangle \):

• Substitute \(a = 8\) and \(b = -4\) into the formula.
• \( \text{Magnitude} = \sqrt{8^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} \)
• Finally, simplify \( \sqrt{80} \) to get \( 4\sqrt{5} \).

Therefore, the magnitude of the vector \( \langle 8, -4 \rangle \) is \( 4\sqrt{5} \). This tells us how long the vector is when represented in a coordinate system.

Understanding vector components is essential for analyzing vectors. Vectors in two dimensions are generally represented by an ordered pair, such as \( \langle 8, -4 \rangle \), where each part describes how far the vector stretches in the x-direction and y-direction.

Each component acts as a building block of the vector:

• **X-component**: This is the horizontal component that shows how far the vector moves along the x-axis. In our exercise, this is 8.
• **Y-component**: This describes the vertical movement along the y-axis. In our exercise, this is -4.

Knowing the components helps in various calculations, such as finding the vector's magnitude and direction. These components can be visualized as forming a right triangle, with the vector itself being the hypotenuse. This makes calculations straightforward using simple trigonometry.

Finding the direction of a vector provides a means to describe its orientation. Direction is often expressed as an angle formed with respect to the positive x-axis.

To find the direction in geometry, we typically use the tangent function and arc tangent operation (\( \tan^{-1} \)) to calculate this angle:

• The formula is \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \), where \(a\) and \(b\) are the x and y components, respectively.
• For our vector \(\langle 8, -4 \rangle\), substituting the values gives us \( \theta = \tan^{-1}\left(\frac{-4}{8}\right) = \tan^{-1}(-0.5) \).
• This results in an angle of approximately \(-26.57^\circ\), meaning the angle is directed downwards.

Since the angle is negative, it indicates a direction clockwise from the positive x-axis. To convert this to a standard position (0 to 360 degrees), simply add 360 degrees:

• Adding 360 degrees yields \( 333.43^\circ \).

This gives us a complete understanding of the vector's direction in standard position, which is a crucial aspect of vector analysis.