Problem 30 - Precalculus

Question

Find the vector, given its magnitude and direction angle. $$|\mathbf{u}|=8, \theta=225^{\circ}$$

Step-by-Step Solution

Answer

The vector is \( \mathbf{u} = \langle -4\sqrt{2}, -4\sqrt{2} \rangle \).

Step 1: Understanding the Problem

We are given the magnitude of a vector \( \mathbf{u} \) as 8 and its direction angle \( \theta \) as 225 degrees. We need to find the vector \( \mathbf{u} \).

Step 2: Direction Angle and Components Relationship

Vectors in a 2D plane with a given magnitude \( |\mathbf{u}| \) and direction angle \( \theta \) can be expressed using trigonometric functions. For a vector \( \mathbf{u} = \langle u_x, u_y \rangle \), we have: \[ u_x = |\mathbf{u}| \cos(\theta) \]\[ u_y = |\mathbf{u}| \sin(\theta) \]

Step 3: Applying Magnitude to Components

Substitute the given magnitude \( |\mathbf{u}| = 8 \) into the equations:\[ u_x = 8 \cos(225^{\circ}) \]\[ u_y = 8 \sin(225^{\circ}) \]

Step 4: Calculating Trigonometric Values

Calculate \( \cos(225^{\circ}) \) and \( \sin(225^{\circ}) \). Since 225 degrees is in the third quadrant,\[ \cos(225^{\circ}) = -\frac{\sqrt{2}}{2} \]\[ \sin(225^{\circ}) = -\frac{\sqrt{2}}{2} \]

Step 5: Compute Vector Components

Substitute these trigonometric values back into the component equations:\[ u_x = 8 \left(-\frac{\sqrt{2}}{2}\right) = -4\sqrt{2} \]\[ u_y = 8 \left(-\frac{\sqrt{2}}{2}\right) = -4\sqrt{2} \]

Step 6: Write the Vector

So, the vector \( \mathbf{u} \) in component form is:\[ \mathbf{u} = \langle -4\sqrt{2}, -4\sqrt{2} \rangle \]

Key Concepts

Magnitude

When discussing vectors, the concept of magnitude is crucial. A vector's magnitude is essentially its length. Consider it the distance from the tail of the vector to its tip in a 2D or 3D space.

For the vector \(\mathbf{u}\) with a magnitude of 8, this value tells us the length of the vector lines directly proportional to the number 8, no matter where it points. Magnitude is depicted using absolute value notation, as seen in \(|\mathbf{u}| = 8\). You can almost think of it as the vector's size.

• Magnitude is always non-negative.
• In cartesian coordinates, for a vector \(\langle x,y \rangle\), the magnitude is calculated using the Pythagorean Theorem:\[|\mathbf{u}| = \sqrt{x^2 + y^2}\].
  
• Given directly, it's a starting point to decompose the vector into components with trigonometry.
• Unlike scalar quantities, vectors have both magnitude and direction.

Direction Angle

The direction angle is critical to identifying the direction in which the vector points. It describes the rotational angle from a baseline direction, generally the positive x-axis, clockwise or counter-clockwise.

In our example, the angle is \(225^\circ\). This particular angle places the vector into what is known as the third quadrant of the Cartesian plane (where both cosine and sine values are negative).

• Angles in a 2D space usually range between \(0^\circ\) and \(360^\circ\)
.
• They help convert scalar values like magnitude into x and y components.
• Measured in degrees or radians, angles provide clear direction when combined with magnitude.

Knowing the angle as \(225^\circ\) helps project the vector components onto the x-axis and y-axis correctly using cosine and sine functions respectively.

Trigonometric Functions

Trigonometric functions bridge the gap between scalar and vector by linking the direction angle with the vector components. Trigonometry involves the principles of triangles, and, in this case, we focus on sine and cosine functions.

Vectors typically use these functions to break down the direction and magnitude into understandable parts:

• Cosine (\(\cos\theta\)) indicates how much of the vector's magnitude aligns with the horizontal axis.
• Sine (\(\sin\theta\)) shows the magnitude's alignment with the vertical axis.

For \(\theta = 225^\circ\):

• Since it is in the third quadrant, the values of cosine and sine are both negative:
• \(\cos(225^\circ) = -\frac{\sqrt{2}}{2}\)
• \(\sin(225^\circ) = -\frac{\sqrt{2}}{2}\)

To find the components \(u_x\) and \(u_y\):

• \(u_x = 8 \cos(225^\circ) = -4\sqrt{2}\)
• \(u_y = 8 \sin(225^\circ) = -4\sqrt{2}\)

Trigonometric functions thus make it possible to translate a magnitude and an angle into a fully formed vector.