91Ó°ÊÓ

Understanding vector components is key to solving problems involving vector addition and subtraction. Vectors are often described by their direction and magnitude. However, to manipulate them mathematically, we use their horizontal and vertical components.

For example, in our exercise, vector \( \overrightarrow{\mathbf{A}} \) points directly west. This means it only has a horizontal component and can be represented as \((-63, 0)\). Meanwhile, vector \( \overrightarrow{\mathbf{B}} \) points south, giving it a vertical component only, represented by \((0, -63)\).

• The first number in the component form represents the horizontal component (x-axis).
• The second number represents the vertical component (y-axis).

Expressing vectors in this component form allows us to easily perform mathematical operations like addition and subtraction by simply dealing with these parts separately.

When dealing with vectors, knowing how to calculate their magnitude is crucial. A vector's magnitude is like its length, representing the distance from its start to its end point in space. Using the Pythagorean theorem is a common approach for calculating magnitude in two-dimensional space.

Let's consider the vector \((-63, -63)\) obtained from adding \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \):

• Square each component: \((-63)^2 + (-63)^2\).
• Sum these squares: \(7938\).
• Take the square root: \( \sqrt{7938} \approx 89.1 \text{ units}\).

This process gives us the magnitude of both \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} \) and \( \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} \), as both expressions involve the same components.

Determining the direction of a vector tells us where it points relative to a baseline, like due west here. We often use trigonometry for this. The tangent function, specifically \( \tan \theta = \frac{y}{x} \), helps in finding the angle \( \theta \) relative to the positive x-axis.

For vector \((-63, -63)\), found by adding \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \), the calculation looks like:

• Use \( \tan \theta = \frac{63}{63} = 1\).
• Thus, \( \theta = 45^\circ \), resulting from the inverse tangent.
• Since both components are negative, this vector is in the third quadrant, needing an addition of \(180^\circ\) to place it correctly, giving us an overall \(225^\circ\).

The subtraction scenario of \( \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}} \) yields \((-63, 63)\), altering the angle to \(135^\circ\) as it sits in the second quadrant after assessment.