Problem 35 - 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 \(\mathbf{j}\). $$ \langle-3 \sqrt{3}, 3\rangle $$

Step-by-Step Solution

Answer

Vector: 6 (cos 150° i + sin 150° j), Linear: -3√3 i + 3 j.

Step 1: Identify Components of the Vector

The given vector is \( \langle -3 \sqrt{3}, 3 \rangle \). Identify the components: the x-component is \( -3 \sqrt{3} \) and the y-component is 3.

Step 2: Calculate Magnitude of the Vector

The magnitude \( r \) of a vector \( \langle x, y \rangle \) is found using the formula \( r = \sqrt{x^2 + y^2} \). Here, \( r = \sqrt{(-3\sqrt{3})^2 + 3^2} = \sqrt{27 + 9} = \sqrt{36} = 6 \).

Step 3: Find the Angle using Inverse Tangent

The angle \( \theta \) can be found using the inverse tangent: \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \). Using \( y = 3 \) and \( x = -3\sqrt{3} \), we calculate \( \theta = \tan^{-1} \left( \frac{3}{-3\sqrt{3}} \right) = \tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) = -30^\circ \). Adjust to standard position: \(180^\circ - 30^\circ = 150^\circ \).

Step 4: Express in Trigonometric Form

The trigonometric form of the vector is \( r (\cos \theta \mathbf{i} + \sin \theta \mathbf{j}) \). Using magnitude \( r = 6 \) and angle \( \theta = 150^\circ \), the vector becomes \( 6 ( \cos 150^\circ \mathbf{i} + \sin 150^\circ \mathbf{j} ) \).

Step 5: Express as a Linear Combination of Unit Vectors

A vector \( \langle x, y \rangle \) can be written as \( x \mathbf{i} + y \mathbf{j} \). Therefore, \( \langle -3\sqrt{3}, 3 \rangle \) can be expressed as \( -3\sqrt{3} \mathbf{i} + 3 \mathbf{j} \).

Key Concepts

Inverse Tangent

In trigonometry, the inverse tangent function, often denoted as \( \tan^{-1} \) or \( \arctan \), helps us find the angle that corresponds to a given tangent value. Imagine you have a right-angled triangle. The tangent of an angle in that triangle is the ratio between the opposite side and the adjacent side. The inverse tangent function allows us to work backwards from this ratio to find the angle, \( \theta \).
When expressing a vector in trigonometric form, finding this angle is essential. Consider a vector with its initial point at the origin and terminal point at \( \langle x, y \rangle \). Here, \( x \) and \( y \) represent the horizontal and vertical components, respectively. The angle \( \theta \) from the positive x-axis is calculated using:

• \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)

In our example vector \( \langle -3\sqrt{3}, 3 \rangle \), plugging in the values gives us \( \theta = \tan^{-1} \left( \frac{3}{-3\sqrt{3}} \right) = -30^\circ \). The negative angle indicates a clockwise direction, so we adjust it to the standard position in the unit circle by adding 180°, resulting in \( 150^\circ \).
Remember, knowing how to use the inverse tangent function aids in rotating vectors properly and understanding their orientation relative to the coordinate axes.

Magnitude of a Vector

The magnitude of a vector gives us the vector's length or size. When you think about motion or force, this is a measure of how strong or far something goes. For a vector \( \langle x, y \rangle \), its magnitude, usually denoted as \( r \), is computed using the Pythagorean theorem. The formula is:

• \( r = \sqrt{x^2 + y^2} \)

In simple terms, this formula extends the idea of "distance formula" from coordinate geometry to vectors. Applying this to our example vector \( \langle -3\sqrt{3}, 3 \rangle \):

• First, square each component: \( (-3\sqrt{3})^2 = 27 \)
• Next, find the square: \( 3^2 = 9 \)
• Add these squares together: \( 27 + 9 = 36 \)
• Finally, take the square root: \( \sqrt{36} = 6 \)

Thus, the magnitude of our vector is 6. Knowing the magnitude is crucial when you are trying to scale vectors or even when combining vectors mathematically.

Linear Combination of Unit Vectors

Vectors can also be expressed as a linear combination of unit vectors. Typically, the Cartesian plane uses the unit vectors \( \mathbf{i} \) (for the x-direction) and \( \mathbf{j} \) (for the y-direction). Any vector \( \langle x, y \rangle \) can be represented as a combination of these unit vectors:

• \( x \mathbf{i} + y \mathbf{j} \)

In our example, the vector \( \langle -3\sqrt{3}, 3 \rangle \) is expressed as:

• \( -3\sqrt{3} \mathbf{i} + 3 \mathbf{j} \)

This notation is particularly useful in physics and engineering where it helps to break down forces or motions into simpler components along the axes. By understanding how to express vectors in terms of \( \mathbf{i} \) and \( \mathbf{j} \), you can easily analyze and manipulate them within different contexts. This understanding is fundamental when working with vector additions, projections, or even in scenarios involving multiple dimensions.