91Ó°ÊÓ

Understanding the components of a vector is crucial in analyzing its behavior in a plane. A vector consists of two elements—one representing the horizontal movement and the other representing the vertical movement. In this scenario, our vector \( \mathbf{u} = \langle -4, 1 \rangle \) breaks down into these components:

• The horizontal component \( a = -4 \)
• The vertical component \( b = 1 \)

This notation means that the vector moves 4 units to the left (since \( a \) is negative) and 1 unit upwards. These components are essential for further calculations, such as finding its magnitude and direction.

The magnitude of a vector represents its length or how much distance it covers from its initial point to its terminal point. For the vector \( \mathbf{u} = \langle a, b \rangle \), the magnitude can be calculated using the formula: \[ ||\mathbf{u}|| = \sqrt{a^2 + b^2} \] This is derived from the Pythagorean theorem, considering the vector as the hypotenuse of a right triangle formed by its components. Now plug in the values \( a = -4 \) and \( b = 1 \):

• Calculate \( (-4)^2 = 16 \)
• Calculate \( 1^2 = 1 \)
• Add: \( 16 + 1 = 17 \)
• Find the square root: \( \sqrt{17} \approx 4.123 \)

So, the magnitude of the vector \( \mathbf{u} \) is approximately 4.123.

The direction angle \( \theta \) of a vector tells you about its inclination with respect to the positive direction of the x-axis. To find this angle, you first use the tangent function, which is the ratio of the vertical component to the horizontal component of the vector. This is given by the formula: \[ \tan \theta = \frac{b}{a} \] Substituting the components \( b = 1 \) and \( a = -4 \), we have: \[ \tan \theta = \frac{1}{-4} = -\frac{1}{4} \] It's important to analyze in which quadrant the angle lies. Given our vector has a negative horizontal and a positive vertical component, it lies in the second quadrant. This affects the angle's final value because standard inverse functions return angles in their principal range (\(-90^\circ \) to \(90^\circ \)). Therefore, adjustments are made.

To find the direction angle precisely, we use the arctangent or inverse tangent function, written as \( \tan^{-1} \). Its role is to give us the angle whose tangent is a given number. Calculating this for our example yields: \[ \theta = \tan^{-1}\left(-\frac{1}{4}\right) \approx -14.04^\circ \] However, because the vector is actually in the second quadrant, you must account for this by adding 180° to the principal value: \[ \theta = 180^\circ - 14.04^\circ \approx 165.96^\circ \] Thus, the adjusted direction angle \( \theta \) is approximately \( 165.96^\circ \). This correction ensures that the angle correctly reflects the vector's true orientation in the Cartesian plane.