Exploring Orthogonality in Three-Dimensional Space: A Deep Dive

In the realm of linear algebra, the concept of orthogonality plays a crucial role, particularly when dealing with vectors in three-dimensional space. This article delves into the specific scenario of whether there exists a non-zero vector v (x, y, z) that is orthogonal to the vectors [1, 0, 1], [0, 2, 1], and [1, 1, -1]. We explore this using various methods, including solving systems of linear equations, analyzing linear independence, and using matrix operations.

Understanding Orthogonal Vectors

Two vectors are orthogonal if their dot product equals zero. The problem at hand asks whether there exists a vector v that is orthogonal to three given vectors. Mathematically, this means finding a vector v such that:

v middot; [1, 0, 1] 0 v middot; [0, 2, 1] 0 v middot; [1, 1, -1] 0

Method 1: Solving the System of Linear Equations

Let's consider the three given vectors in turn and set up the corresponding equations for orthogonality:

1. v middot; [1, 0, 1] 0

This gives us: x z 0

2. v middot; [0, 2, 1] 0

This gives us: 2y z 0

3. v middot; [1, 1, -1] 0

This gives us: x y - z 0

We now have three equations:

x z 0 2y z 0 x y - z 0

From the first equation, x -z. From the second equation, y -z/2. Substituting these into the third equation, we get:

-z - z/2 - z 0 rarr; -5/2z 0 rarr; z 0

As z 0, then x -z 0 and y -z/2 0. Thus, the only solution is (x, y, z) (0, 0, 0), which is the trivial solution. Therefore, there is no non-zero vector orthogonal to all three given vectors.

Method 2: Analyzing Linear Independence

Another approach is to consider whether the given vectors are linearly independent. If the vectors form a basis for R^3, then they are linearly independent and no orthogonal vector exists.

The three vectors [1, 0, 1], [0, 2, 1], and [1, 1, -1] can be used as columns of a matrix and row-reduced to determine linear independence. Row-reducing the matrix:

A(abc)

where

A(1 0 1, 0 2 1, 1 1 -1)

Row-reducing this matrix will show that it has full rank, meaning the vectors are linearly independent.

Method 3: Using Determinants and Matrix Row Reduction

Another way to determine the existence of an orthogonal vector is to consider the determinant of the coefficient matrix formed by the given vectors:

D(1 0 1, 0 2 1, 1 1 -1)

The determinant of this matrix is:

D -5

Since the determinant is non-zero, the system of equations has a trivial solution only, confirming that no non-zero vector orthogonal to the given vectors exists.

Conclusion

Through the above methods, we have shown that no non-zero vector exists that is orthogonal to the vectors [1, 0, 1], [0, 2, 1], and [1, 1, -1]. This conclusion is consistent across different approaches, providing a robust understanding of the problem. Whether through solving systems of equations, analyzing linear independence, or using determinant methods, the outcome remains the same.