Problem 18 - College Trigonometry

Question

Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=5 \sqrt{6}\); when drawn in standard position \(\vec{v}\) lies in Quadrant III and makes a \(45^{\circ}\) angle with the negative \(x\) -axis

Step-by-Step Solution

Answer

The component form of \(\vec{v}\) is \((-5 \sqrt{3}, -5 \sqrt{3})\).

Step 1: Determine Unit Vector in Direction

First, we find the unit vector in the direction of \ \(\vec{v}\) \ which lies at a \ \(45^{\circ}\) \ angle with the negative \ \(x\)\ -axis. The direction can be represented using the angle with the positive \ \(x\)\ -axis as \ \(180^{\circ} + 45^{\circ} = 225^{\circ}\). \ The unit vector is given by: \ \[ \vec{u} = (\cos 225^{\circ}, \sin 225^{\circ}) \]

Step 2: Calculate Exact Trigonometric Values

Determine the \ \(\cos\) and \ \(\sin\) for the angle we found in the previous step: \\(\cos 225^{\circ} = -\frac{\sqrt{2}}{2}\) and \ \(\sin 225^{\circ} = -\frac{\sqrt{2}}{2}\). \ So, the unit vector becomes: \ \[ \vec{u} = \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \]

Step 3: Scale Unit Vector by Magnitude

Multiply the unit vector by the magnitude \ \(5 \sqrt{6}\) \ to find the component form of \ \(\vec{v}\): \\[ \vec{v} = 5 \sqrt{6} \times \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \] Simplify each component: \First component: \ \(-5 \sqrt{6} \times \frac{\sqrt{2}}{2} = -\frac{5 \cdot \sqrt{12}}{2} = -\frac{5 \cdot 2 \sqrt{3}}{2} = -5 \sqrt{3}\) \Second component: \( \text{is the same as the first due to symmetry in calculation} \) \Thus, \ \(\vec{v} = (-5 \sqrt{3}, -5 \sqrt{3})\).

Key Concepts

Magnitude

The magnitude of a vector is like its length or size, representing how far it might "reach." For a vector \( \vec{v} \), the magnitude is denoted as \( \|\vec{v}\| \). In our exercise, the magnitude provided is \( 5 \sqrt{6} \). This tells us that the vector extends as far as \( 5 \sqrt{6} \) units from the origin in whatever direction it points. Magnitude is always a positive value because lengths can't be negative. To compute vector components, knowing the magnitude helps set a corresponding scale once we know the vector's direction.

Direction

Understanding the direction of a vector involves knowing the angle it makes with a reference axis. This is usually the positive \( x \)-axis. In the given exercise, the vector \( \vec{v} \) makes an angle of \( 45^{\circ} \) with the negative \( x \)-axis. However, to place this in a standard context, we measure angles from the positive \( x \)-axis, resulting in an angle of \( 225^{\circ} \) for direction in standard position. Vectors in different quadrants can impact their components being positive or negative. In our case, Quadrant III makes both \( x \) and \( y \) components negative.

Trigonometric Values

Trigonometrically, vectors are often broken down into their \( x \) and \( y \) components using cosine and sine values respectively. For our angle of \( 225^{\circ} \), we need these values to find the component form:

• \( \cos 225^{\circ} \): This gives the \( x \) component direction, often negative in Quadrant III, equaling \( -\frac{\sqrt{2}}{2} \).
• \( \sin 225^{\circ} \): Provides the \( y \) component direction, also negative, resulting in \( -\frac{\sqrt{2}}{2} \).

Combining these values gives us the unit vector's direction. These trigonometric values are crucial for determining how much the vector "pushes" in each coordinate direction.

Unit Vector

A unit vector is a vector with a magnitude of 1, often used to describe the direction of a vector without considering its length. For our vector \( \vec{v} \), whose components after direction calculation are \( (-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}) \), this establishes a basic direction in two dimensions. The unit vector helps in scaling vectors to any desired magnitude. Once found, scaling the unit vector by the original vector's magnitude \( 5 \sqrt{6} \) helps find the component form of the vector. Therefore, our result, \( \vec{v} = (-5\sqrt{3}, -5\sqrt{3}) \), represents a vector aligned precisely to its direction with given magnitude.