91Ó°ÊÓ

The magnitude of a vector is like the length of a line segment you might draw on paper. For a given vector \( \mathbf{u} = \langle a, b \rangle \), where \( a \) and \( b \) are the components along the x-axis and y-axis respectively, the magnitude tells you the distance from the start of the vector to its endpoint. This distance is always a non-negative number.

You can find the magnitude using the formula:

• \( |\mathbf{u}| = \sqrt{a^2 + b^2} \)

Effectively, this formula is derived from the Pythagorean theorem, which you might recall is used for calculating the hypotenuse of a right triangle. Here, \( a \) and \( b \) are analogous to the triangle's legs, and \( |\mathbf{u}| \) is the hypotenuse.

• For \( \mathbf{u} = \langle -4, 1 \rangle \), plug in \( a = -4 \) and \( b = 1\).
• Thus, the magnitude \( |\mathbf{u}| = \sqrt{(-4)^2 + 1^2} = \sqrt{17} \).

This magnitude helps in determining just how "strong" or "long" the vector is, regardless of its direction.

The direction angle of a vector provides the direction in which the vector is pointing. It is measured from the positive x-axis, moving counterclockwise. For vectors expressed as \( \langle a, b \rangle \), the direction angle \( \theta \) can be found using trigonometry. Specifically, it involves the tangent function:

\( \tan \theta = \frac{b}{a} \)

This formula comes from the basic definition of tangent in a right triangle, where tangent represents the ratio of the opposite side to the adjacent side.

In practical terms:

• Find \( \theta = \arctan\left(\frac{b}{a}\right) \).
• For \( \mathbf{u} = \langle -4, 1 \rangle \), that means \( \tan \theta = \frac{1}{-4} \).
• Using a calculator or trigonometry table, find \( \theta \approx -14.04^\circ \).

However, an angle like \(-14.04^\circ\) needs adjustment based on the vector's quadrant in the coordinate system to represent the accurate direction angle.

Understanding quadrants is essential for accurately determining the vector's direction. The coordinate plane is divided into four quadrants, each with unique characteristics.

Here's a brief overview of the quadrants:

• **Quadrant I**: Both \( x \) and \( y \) components are positive.
• **Quadrant II**: \( x \) is negative and \( y \) is positive.
• **Quadrant III**: Both \( x \) and \( y \) components are negative.
• **Quadrant IV**: \( x \) is positive and \( y \) is negative.

When dealing with the vector \( \mathbf{u} = \langle -4, 1 \rangle \), it's located in Quadrant II since its \( x \) component \(-4\) is negative and its \( y \) component \(1\) is positive.

This placement impacts the direction angle because angles are typically adjusted to reflect their actual compass direction, especially when extending past the typical 90-degree quadrant boundaries.

Thus, to adjust the initial \(-14.04^\circ\) (which might indicate a position in Quadrant IV), you must add \(180^\circ\) to move to Quadrant II, resulting in a direction angle of \(165.96^\circ\). This adjustment ensures you've placed the vector in the correct quadrant, reflecting its true direction.