Question

Find a unit vector in \(E^{4}\), with positive components, that forms equal angles with the axes, i.e., with the basic unit vectors (see Problem 7 ).

Short answer

The vector is \(\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\).

Step-by-step solution

Understanding the Problem

We need to find a unit vector in four-dimensional space ( E^4 ) that makes equal angles with each of the coordinate axes. This means the vector has components (x, x, x, x) in E^4 .

Setting Up the Unit Vector Equation

We know that for a vector \(v = (x, x, x, x)\), to be a unit vector, its magnitude must equal 1. Therefore, we must solve the equation: \(\sqrt{x^2 + x^2 + x^2 + x^2} = 1\), or equivalently \(\sqrt{4x^2} = 1\).

Solving for the Vector Component

The equation \(\sqrt{4x^2} = 1\) simplifies to \(2x = 1\), which gives \(x = \frac{1}{2}\). Thus, the vector with equal angles with each axis has components \(\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\).

Verify the Unit Vector

We verify that the vector components satisfy the unit vector condition: compute \(\sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4}} = \sqrt{1} = 1\). This confirms that \(\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\) is indeed a unit vector in E^4.

Key concepts

Four-Dimensional Space

Four-dimensional space, often represented as \( E^4 \), is an extension of the three-dimensional space we are more familiar with. It adds an extra dimension, think of it like adding another direction to move in. Just like how 3D space requires three coordinates \((x, y, z)\) to define a point, points in 4D space require four coordinates, often denoted as \((x_1, x_2, x_3, x_4)\).
This additional dimension allows for more complex geometrical concepts and changes how we visualize mathematical problems. While it's challenging to intuitively visualize 4D shapes, mathematicians can manipulate them using algebraic and geometric methods.

• In 4D space, the concept of distance (or magnitude) extends naturally. The distance from the origin to a point \((x, y, z, w)\) is computed as \( \sqrt{x^2 + y^2 + z^2 + w^2} \).
• Just like in 3D, vectors in 4D are defined by components along these four coordinate directions.
  

Equal Angles

The concept of equal angles in a vector context means that the vector is positioned such that it maintains the same angle with each of the axes. This requires that all the vector's components are identical; thus, the vector \((x, x, x, x)\) has identical interactions with every axis it meets.
Imagine the challenge is to balance a vector so all directions it points to (towards each axis) are equally favored, meaning the vector's angle with each axis is the same. This is achieved when the components of the vector have an equal magnitude.
To find such a vector in 4D that makes equal angles with each axis, first, note down the components and think about how each contributes equally to the vector's overall direction.

Coordinate Axes

Coordinate axes are the backbone of any geometric space. In four-dimensional space, like in 3D, there are four coordinate axes, usually named \( x \), \( y \), \( z \), and \( w \). Each axis is a line that stretches infinitely in both directions and acts as a reference for measuring distance and defining directions.
When discussing vectors in 4D, each component of a vector aligns parallel to one of these coordinate axes. A vector that forms equal angles with each axis will have each of its components equal, representing its equal inclination with respect to each axis. These axes help determine how a vector is oriented in space.
Understanding the role of coordinate axes in higher dimensions is crucial as it translates to comprehending how vectors interact and maintain balance across this multi-directional continuum.

Magnitude of Vector

The magnitude of a vector is a measure of its length or size and is crucial in deciding if a vector is a unit vector. In mathematical terms, for any vector \( \mathbf{v} = (x_1, x_2, x_3, x_4) \), the magnitude is calculated using the formula:

\[ \| \mathbf{v} \| = \sqrt{x_1^2 + x_2^2 + x_3^2 + x_4^2} \]

For a vector to be considered a unit vector, its magnitude should be exactly 1. This condition ensures that the vector is a "standard measure" and can be uniformly applied in computational situations without scaling it further. This is particularly important when determining vectors that evenly distribute influence across all axes.
The unit vector derived in the original solution, \((\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})\), confirms this with a magnitude of 1 when computed, affirming its balanced length across its four dimensions.