Problem 18 - College Trigonometry

Question

In Exercises \(1-20\), use the pair of vectors \(\vec{v}\) and \(\vec{w}\) to find the following quantities. $$ \vec{v}=\left\langle\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle \text { and } \vec{w}=\left\langle\frac{1}{2},-\frac{\sqrt{3}}{2}\right\rangle $$

Step-by-Step Solution

Answer

Dot product is \(\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}\); both magnitudes are 1, and use the dot product result to find the angle.

Step 1: Identify the Problem

We need to find specific quantities involving the vectors \( \vec{v} \) and \( \vec{w} \). For most vector exercises, quantities like dot product, cross product, magnitudes, or angles between vectors are usually sought after.

Step 2: Find the Dot Product

The dot product of two vectors \( \vec{v} = \langle a_1, a_2 \rangle \) and \( \vec{w} = \langle b_1, b_2 \rangle \) is calculated as \( a_1b_1 + a_2b_2 \). Thus for \( \vec{v} \) and \( \vec{w} \): \[ \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \].

Step 3: Calculate the Magnitude of the Vectors

The magnitude of a vector \( \vec{a} = \langle x, y \rangle \) is computed by \( \sqrt{x^2 + y^2} \). Calculate magnitudes for both vectors:For \( \vec{v}\): \[ \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1 \]For \( \vec{w}\): \[ \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \].

Step 4: Determine the Angle Between Vectors

The angle \( \theta \) between two vectors can be found using the dot product formula: \[ \vec{v} \cdot \vec{w} = \|\vec{v}\| \|\vec{w}\| \cos(\theta) \].We have \( \vec{v} \cdot \vec{w} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \), and both magnitudes are 1.\[ \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = 1 \times 1 \times \cos(\theta) \]So, \[ \cos(\theta) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \].Finally, use the inverse cosine \( \cos^{-1} \) to find \( \theta \).

Key Concepts

Dot Product

The dot product is a way to multiply two vectors together, and it provides a scalar result, unlike a cross product which results in a vector. To calculate the dot product of two given vectors, let's take \( \vec{v} = \langle a_1, a_2 \rangle \) and \( \vec{w} = \langle b_1, b_2 \rangle \). The dot product is calculated as:

• \( a_1 \times b_1 \)
• plus \( a_2 \times b_2 \)

A geometric interpretation of the dot product is that it represents the product of the magnitudes of the two vectors and the cosine of the angle between them. Understanding the dot product helps in figuring out whether vectors are perpendicular—they are if the dot product is zero—because this implies that the angle between them is 90 degrees.

Magnitude of a Vector

The magnitude of a vector, often referred to as its "length", is essentially the distance from the origin of the coordinate plane to the point where the vector terminates. This gives us an understanding of the vector's scale in the given space.The magnitude of a vector \( \vec{a} = \langle x, y \rangle \) is calculated using the formula:

• \( \sqrt{x^2 + y^2} \)

This results from the Pythagorean theorem, which measures the hypotenuse of a right triangle. The magnitude is always a non-negative number, as it represents a physical distance, and importantly, it remains unchanged if the vector direction is reversed. This property makes vector magnitudes crucial in determining how far an object might move in space, as represented by the vector.

Angle Between Vectors

Finding the angle between two vectors is particularly useful in fields like physics and engineering, as it helps determine how much of one vector goes in the direction of another. The angle \( \theta \) can be discovered using the dot product formula and the magnitudes of the vectors:

• The dot product \( \vec{v} \cdot \vec{w} = \|\vec{v}\| \|\vec{w}\| \cos(\theta) \)

After calculating the dot product and the magnitudes of the vectors, \( \cos(\theta) \) can be found, and subsequently, the angle \( \theta \) itself. When two vectors are perpendicular, \( \theta = 90^{\circ} \); when they are parallel, \( \theta = 0^{\circ} \). The angle between vectors not only influences how the vectors interact but also helps describe their spatial relationship clearly.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles when given the ratios that result from dividing the sides of a triangle. The function \( \cos^{-1} \), or inverse cosine, is particularly useful in vector algebra for determining angles when we know the cosine of the angle.When you have calculated \( \cos(\theta) \) from the dot product:

• Use \( \cos^{-1} \) to derive \( \theta \), the angle.

This is crucial when working with vectors because, often, the direct measurement of angles isn't possible. By interpreting the value of the inverse trigonometric function, we can uncover precise angle measures, shaping a deeper understanding of the vector orientation relative to each other. This ability to translate between linear and angular measurements is indispensable in many technical and scientific computations.