Problem 10 - Precalculus

Question

Find the magnitude and direction angle of the given vector. $$\mathbf{u}=\langle-6,-2\rangle$$

Step-by-Step Solution

Answer

Magnitude: \(2\sqrt{10}\); Direction angle: \(198.43^\circ\).

Step 1: Understanding the Vector

The given vector \( \mathbf{u} \) is described by the components \(-6\) and \(-2\). This can be interpreted as moving 6 units left and 2 units down from the origin in a Cartesian coordinate system.

Step 2: Magnitude Calculation

The magnitude of a vector \( \mathbf{u} = \langle a, b \rangle \) is calculated using the formula:\[||\mathbf{u}|| = \sqrt{a^2 + b^2}\]Substituting \( a = -6 \) and \( b = -2 \), we calculate:\[||\mathbf{u}|| = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}\]

Step 3: Direction Angle Calculation

The direction angle \( \theta \) of a vector \( \mathbf{u} = \langle a, b \rangle \) can be found using the arctangent function:\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]For \( a = -6 \) and \( b = -2 \), we calculate:\[\theta = \tan^{-1}\left(\frac{-2}{-6}\right) = \tan^{-1}\left(\frac{1}{3}\right)\]The arctangent value of \( \frac{1}{3} \) is approximately \( 18.43^\circ \). However, since the vector is in the third quadrant, we add \( 180^\circ \), yielding \( \theta \approx 198.43^\circ \).

Key Concepts

Magnitude of a Vector

One of the basic concepts in vector analysis is understanding how the magnitude of a vector is determined. The magnitude measures the size or length of the vector. Think of it as the "distance" from the initial point to the terminal point of the vector when graphed. In formula terms, if you have a vector \( \mathbf{u} = \langle a, b \rangle \), its magnitude is calculated using:
\[||\mathbf{u}|| = \sqrt{a^2 + b^2}\]Let’s consider an example from the exercise. For the vector \( \mathbf{u} = \langle -6, -2 \rangle \):

• First, you square the components: \((-6)^2 = 36\) and \((-2)^2 = 4\).
• Add these squares together: \(36 + 4 = 40\).
• Finally, take the square root: \(\sqrt{40} = 2\sqrt{10}\).

Thus, the magnitude of this vector is \(2\sqrt{10}\). This tells you how "long" the vector is without considering its direction.

Direction Angle

The direction angle of a vector tells you the angle at which the vector is pointing relative to a reference line, usually the positive x-axis. It's similar to pointing out directions with degrees like on a compass.
The direction angle \( \theta \) can be calculated using the formula:
\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]For the vector \( \mathbf{u} = \langle -6, -2 \rangle \):

• Use the components of the vector: \(a = -6\) and \(b = -2\).
• Plug these values into the formula: \(\theta = \tan^{-1}\left(\frac{-2}{-6}\right) = \tan^{-1}\left(\frac{1}{3}\right)\).
• The approximate value of \(\tan^{-1}\left(\frac{1}{3}\right)\) is about \(18.43^\circ\).
• Since the vector lies in the third quadrant (both components are negative), adjust the calculated angle by adding \(180^\circ\).

This gives a final direction angle of approximately \(198.43^\circ\). This adjustment allows us to properly express the vector's true direction in the coordinate plane.

Cartesian Coordinate System

The Cartesian coordinate system is a common way of graphing points and vectors in a plane. Named after René Descartes, it uses two perpendicular axes:

• X-axis (horizontal)
• Y-axis (vertical)

Vectors in this system are represented as ordered pairs, like \( \langle a, b \rangle \). These pairs tell us precisely where to move on the X and Y axes. For instance, consider the vector \( \mathbf{u} = \langle -6, -2 \rangle \):

• The first number, \(-6\), denotes moving 6 units left from the origin along the X-axis.
• The second number, \(-2\), indicates moving 2 units down along the Y-axis.

This system is highly intuitive and is widely used in physics, engineering, and computer graphics for its simplicity in describing positional data and movement.