Problem 28 - APEX Calculus

Question

Verify, from Key Idea 10.2.1, that $$ \vec{u}=\langle\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta\rangle $$ is a unit vector for all angles \(\theta\) and \(\varphi\).

Step-by-Step Solution

Answer

The vector \(\vec{u}\) is a unit vector because its magnitude is 1 for all \(\theta\) and \(\varphi\).

Step 1: Understand the Definition of a Unit Vector

A unit vector is a vector that has a magnitude of 1. Therefore, to verify if \( \vec{u} \) is a unit vector, we must ensure that the magnitude of \( \vec{u} \) is equal to 1.

Step 2: Calculate the Magnitude of \( \vec{u} \)

The magnitude of a vector \( \vec{u} = \langle a, b, c \rangle \) is calculated using the formula: \[ \| \vec{u} \| = \sqrt{a^2 + b^2 + c^2}. \] For \( \vec{u} = \langle \sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta \rangle \), the magnitude becomes \[ \| \vec{u} \| = \sqrt{(\sin \theta \cos \varphi)^2 + (\sin \theta \sin \varphi)^2 + (\cos \theta)^2}. \]

Step 3: Simplify the Magnitude Expression

Expanding each term inside the square root, we have:\[ = \sqrt{(\sin^2 \theta \cos^2 \varphi) + (\sin^2 \theta \sin^2 \varphi) + \cos^2 \theta}. \] Factor out \( \sin^2 \theta \) in the first two terms:\[ = \sqrt{\sin^2 \theta (\cos^2 \varphi + \sin^2 \varphi) + \cos^2 \theta}. \] Note that \( \cos^2 \varphi + \sin^2 \varphi = 1 \), due to the Pythagorean identity, thus simplifying:\[ = \sqrt{\sin^2 \theta \cdot 1 + \cos^2 \theta}. \] This reduces further to:\[ = \sqrt{\sin^2 \theta + \cos^2 \theta}. \] Again, using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):\[ = \sqrt{1} = 1. \]

Step 4: Conclusion

Since the magnitude \( \| \vec{u} \| = 1 \), \( \vec{u} \) is indeed a unit vector. This verifies that the given vector \( \langle \sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta \rangle \) is a unit vector for all angles \( \theta \) and \( \varphi \).

Key Concepts

Vector Magnitude

When we talk about the magnitude of a vector, we're referring to its length or size in space. It's like measuring how long a vector is, representing the distance from the point of origin to the vector's endpoint. To make this measurement, we use a formula that is rooted in the essence of geometry.

The magnitude of a vector with components \( \langle a, b, c \rangle \) is calculated using the formula for the Euclidean norm:

• \( \| \vec{u} \| = \sqrt{a^2 + b^2 + c^2} \)

Calculating this, you sum the squares of each component of the vector and then take the square root of that sum.

This formula is perfectly analogous to calculating the hypotenuse of a right triangle. In our problem, we verified that the vector \( \vec{u} = \langle \sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta \rangle \) has a magnitude (length) of 1. Therefore, it's a unit vector since its magnitude equals 1.

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry and describes how certain combinations of trigonometric functions always yield the same value. Specifically, the Pythagorean identity states:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This identity plays a crucial role not only in verifying our vector, but also in many other mathematical contexts. It assures us that regardless of the angle \( \theta \), the sum of the square of the sine of \( \theta \) and the square of the cosine of \( \theta \) will always be one.

In our vector verification, we leveraged this identity multiple times. For example, when simplifying the expression \( \sqrt{\sin^2 \theta + \cos^2 \theta} \), the Pythagorean identity helped us conclude that this expression simplifies to 1.

Trigonometric Functions

Trigonometric functions underpin many aspects of mathematics, particularly when dealing with angles and periodic phenomena. In the context of vectors, these functions often help us describe components in terms of angles.

• \( \sin \theta \)
• \( \cos \theta \)
• \( \tan \theta \)

Each function derives from relationships in a right-angled triangle and the unit circle.

For vectors like our \( \vec{u} \), these functions help form a bridge between angular measurements and linear dimensions. The components of the vector \( \vec{u} \) are products of trigonometric functions, reflecting the angles \( \theta \) and \( \varphi \) in three-dimensional space.

This illustrates how trigonometry enables us to navigate between angles and linear space, providing solutions that are both precise and elegantly structured.