Problem 33 - Precalculus with Calculus Previews

Question

Express the given vector (a) in trigonometric form and (b) as a linear combination of the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\). $$ \langle-\sqrt{2}, \sqrt{2}\rangle $$

Step-by-Step Solution

Answer

In trigonometric form: \( 2 \operatorname{cis}(\frac{3\pi}{4}) \); As a linear combination: \( -\sqrt{2}\mathbf{i} + \sqrt{2}\mathbf{j} \).

Step 1: Identify Magnitude of the Vector

First, let's calculate the magnitude of the given vector \( \langle -\sqrt{2}, \sqrt{2} \rangle \). The magnitude \( r \) is given by \( r = \sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \).

Step 2: Calculate Angle for Trigonometric Form

Now, let's determine the angle \( \theta \) that the vector makes with the positive x-axis. Since the vector is in the form \( \langle x, y \rangle = \langle -\sqrt{2}, \sqrt{2} \rangle \), the tangent of the angle is \( \tan(\theta) = \frac{\sqrt{2}}{-\sqrt{2}} = -1 \). The angle \( \theta \) for which tangent is -1 is in the second quadrant, thus \( \theta = \frac{3\pi}{4} \).

Step 3: Express Vector in Trigonometric Form

Using the magnitude and the angle, the trigonometric form of the vector is \( 2 \operatorname{cis}\left(\frac{3\pi}{4}\right) \), or equivalently \( 2(\cos(\frac{3\pi}{4}) + i \sin(\frac{3\pi}{4})) \).

Step 4: Express Vector as a Linear Combination

To express the vector as a linear combination of the unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), simply write it as \( -\sqrt{2}\mathbf{i} + \sqrt{2}\mathbf{j} \). This is straightforward as it matches the components of the vector directly.

Key Concepts

Trigonometric Form

The trigonometric form of a vector provides an elegant representation by highlighting its magnitude and direction. To express a vector in this form, you first need to determine two components:

• The magnitude of the vector
• The angle it makes with a reference axis, usually the positive x-axis

To convert the vector \( \langle -\sqrt{2}, \sqrt{2} \rangle \) into trigonometric form, we start by calculating its magnitude, found using the formula \( r = \sqrt{x^2 + y^2} \). Here, substituting our values gives us a magnitude of 2.
Next, we need to find the angle \( \theta \). Using the tangent ratio \( \tan(\theta) = \frac{y}{x} \), we can deduce \( \theta \) based on the ratio and the quadrant the vector is in. For our vector, \( \tan(\theta) = -1 \) places \( \theta \) in the second quadrant, resulting in \( \theta = \frac{3\pi}{4} \).
Finally, the trigonometric form is expressed with the notation \( r \text{cis}(\theta) \), where `cis` stands for \( \cos(\theta) + i \sin(\theta) \). Thus, the vector has a trigonometric form of \( 2 \text{cis} \left( \frac{3\pi}{4} \right) \). This form effectively communicates both the vector's distance from the origin and its direction.

Unit Vectors

Unit vectors are essential in vector representation because they define direction. The standard unit vectors in a 2D plane are \( \mathbf{i} \) and \( \mathbf{j} \).

• \( \mathbf{i} \) is the unit vector in the horizontal (x) direction
• \( \mathbf{j} \) is the unit vector in the vertical (y) direction

Any vector in the plane can be expressed as a linear combination of these unit vectors.
For the vector \( \langle -\sqrt{2}, \sqrt{2} \rangle \), you can easily represent it using \( \mathbf{i} \) and \( \mathbf{j} \). Simply use the components of the vector: \(-\sqrt{2}\mathbf{i} + \sqrt{2}\mathbf{j}\). This breakdown illustrates both direction and magnitude clearly along each axis, helping to understand how each component contributes to the vector's orientation in space.

Magnitude of a Vector

The magnitude, or length, of a vector quantifies how long the vector is from its initial point to its terminal point. To calculate the magnitude of a vector in 2D space, you use the following formula:\[|\mathbf{v}| = \sqrt{x^2 + y^2}\]For the vector \( \langle -\sqrt{2}, \sqrt{2} \rangle \), compute as follows:

• Square each component: \((-\sqrt{2})^2 = 2\) and \((\sqrt{2})^2 = 2\).
• Add the squares: \(2 + 2 = 4\).
• Take the square root of the sum: \(\sqrt{4} = 2\).

Thus, the magnitude is 2. This value tells you the vector's length, which is consistent no matter which direction the vector points. Understanding the magnitude is essential because it provides a scalar quantity that is indifferent to the vector's orientation.