Find Two Linearly Independent Vectors Perpendicular To The Vector

Let's explore how to find two linearly independent vectors that are perpendicular to a given vector. This is a fundamental concept in linear algebra with applications in computer graphics, physics, and engineering. Understanding this process allows us to define planes in 3D space, solve systems of equations, and perform various geometric transformations.

Introduction

Finding vectors perpendicular to a given vector, also known as finding orthogonal vectors, is a common task in linear algebra. When working in three-dimensional space (R³), given a single vector, we can always find two linearly independent vectors that are perpendicular to it. This is possible because a single vector in R³ constrains one dimension, leaving two dimensions for orthogonal vectors to span.

The dot product is the mathematical tool we use to determine if two vectors are perpendicular. Two vectors, u and v, are perpendicular if their dot product equals zero:

u ⋅ v = 0

Linear independence ensures that the two perpendicular vectors we find aren't just scalar multiples of each other, meaning they point in genuinely different directions and can form a basis for the plane orthogonal to the original vector.

Steps to Find Two Linearly Independent Vectors Perpendicular to a Given Vector

Let's outline a step-by-step method to find these vectors. We'll consider a vector v = (a, b, c) in R³.

1. Understanding the Goal

Our objective is to find two vectors, u = (x₁, y₁, z₁) and w = (x₂, y₂, z₂), such that:

• v ⋅ u = ax₁ + by₁ + cz₁ = 0
• v ⋅ w = ax₂ + by₂ + cz₂ = 0
• u and w are linearly independent.

2. Choosing Values for the First Vector (u)

To find the first vector u, we can choose arbitrary values for two of its components (x₁, y₁, or z₁) and then solve for the remaining component using the dot product equation. Let's choose x₁ and y₁ arbitrarily:

• Let x₁ = 1 and y₁ = 0.

Now, we need to solve for z₁:

a(1) + b(0) + c(z₁) = 0 a + cz₁ = 0 z₁ = -a/c (if c ≠ 0)

So, our first vector u becomes (1, 0, -a/c).

Important Considerations:

• Case 1: c ≠ 0: If c is not zero, the solution z₁ = -a/c is valid.

• Case 2: c = 0: If c is zero, we need to choose different values for x₁ and y₁ to avoid division by zero. In this case, let's try x₁ = 0 and y₁ = 1.

  Then, a(0) + b(1) + c(z₁) = 0 becomes b + c(z₁) = 0. Since c is zero, we have b = 0. If b is also zero, then v = (a, 0, 0). In this scenario, we can choose u = (0, 1, 0) and w = (0, 0, 1). If b is not zero, we proceed with b + 0(z₁) = 0, which is impossible. We need to rethink our initial choices. Let's revert to the original case and choose x₁ = 1 and y₁ = 0. The equation a(1) + b(0) + 0(z₁) = 0 becomes a = 0. Now, v = (0, b, 0). We can then select u = (1, 0, 0) and w = (0, 0, 1).

3. Choosing Values for the Second Vector (w)

Now, we need to find a second vector w = (x₂, y₂, z₂) that is perpendicular to both v and u, and is linearly independent from u. We will use a similar approach:

• Let's choose x₂ = 0 and y₂ = 1.

Now, we solve for z₂:

a(0) + b(1) + c(z₂) = 0 b + cz₂ = 0 z₂ = -b/c (if c ≠ 0)

So, our second vector w becomes (0, 1, -b/c).

Important Considerations:

• Case 1: c ≠ 0: If c is not zero, the solution z₂ = -b/c is valid.
• Case 2: c = 0: As before, if c is zero, we need to adjust our approach. Going back to v = (a, b, 0), we already handled the cases where a = 0 and b = 0. Let's assume a and b are not both zero. In this case, let's choose x₂ = 0 and z₂ = 1. Then, a(0) + b(y₂) + c(1) = 0 becomes b(y₂) + 0 = 0, so y₂ = 0. This gives us w = (0, 0, 1).

4. Verifying Perpendicularity

We need to ensure that u and w are indeed perpendicular to v by calculating the dot products:

• v ⋅ u = (a, b, c) ⋅ (1, 0, -a/c) = a(1) + b(0) + c(-a/c) = a - a = 0
• v ⋅ w = (a, b, c) ⋅ (0, 1, -b/c) = a(0) + b(1) + c(-b/c) = b - b = 0

5. Verifying Linear Independence

To verify linear independence, we need to check if there exist scalars k₁ and k₂, not both zero, such that:

k₁u + k₂w = 0

If the only solution is k₁ = 0 and k₂ = 0, then u and w are linearly independent.

Let's assume c ≠ 0:

k₁(1, 0, -a/c) + k₂(0, 1, -b/c) = (0, 0, 0) (k₁, k₂, -k₁a/c - k₂b/c) = (0, 0, 0)

This gives us the following system of equations:

• k₁ = 0
• k₂ = 0
• -k₁a/c - k₂b/c = 0

Since k₁ = 0 and k₂ = 0, the vectors u and w are linearly independent.

6. Summarizing the Results

If c is not zero, then we have found two linearly independent vectors perpendicular to v:

• u = (1, 0, -a/c)
• w = (0, 1, -b/c)

If c is zero and v = (a, b, 0), with a and b not both zero, then we have:

• u = (1, 0, 0)
• w = (0, 0, 1)

If c is zero, b is zero, and v = (a, 0, 0):

• u = (0, 1, 0)
• w = (0, 0, 1)

If c is zero, a is zero, and v = (0, b, 0):

• u = (1, 0, 0)
• w = (0, 0, 1)

Example

Let's find two linearly independent vectors perpendicular to v = (1, 2, 3).

Using our formulas:

• a = 1, b = 2, c = 3

Since c is not zero, we can use the first set of formulas:

• u = (1, 0, -a/c) = (1, 0, -1/3)
• w = (0, 1, -b/c) = (0, 1, -2/3)

To avoid fractions, we can multiply each vector by 3 without changing their direction or perpendicularity.

• u = (3, 0, -1)
• w = (0, 3, -2)

Let's verify:

• v ⋅ u = (1, 2, 3) ⋅ (3, 0, -1) = 1(3) + 2(0) + 3(-1) = 3 - 3 = 0
• v ⋅ w = (1, 2, 3) ⋅ (0, 3, -2) = 1(0) + 2(3) + 3(-2) = 6 - 6 = 0

And u and w are clearly linearly independent.

The Geometric Interpretation

Geometrically, the set of all vectors perpendicular to a given vector in R³ forms a plane passing through the origin. Finding two linearly independent vectors that are perpendicular to the given vector provides a basis for this plane. Any vector in this plane can be expressed as a linear combination of these two basis vectors. This plane is often called the null space or kernel of the linear transformation represented by the original vector as a row vector.

Alternative Method: Cross Product

Another method to find a vector perpendicular to two given vectors is using the cross product. However, in our case, we only have one vector. To use the cross product, we can arbitrarily choose another vector that is not parallel to the original vector and then take the cross product. Then, to find a second linearly independent vector perpendicular to the original vector, we need to find a vector perpendicular to both the original vector and the vector resulting from the first cross product. This can be complex and computationally intensive, and the previously explained dot product method is generally preferred.

Limitations and Special Cases

• Zero Vector: If the given vector is the zero vector, any two linearly independent vectors will be perpendicular to it.
• Two-Dimensional Space (R²): In R², finding a vector perpendicular to a given vector is simpler. If v = (a, b), then u = (-b, a) or u = (b, -a) will be perpendicular to v.
• Higher Dimensions (Rⁿ, n > 3): The same principle applies, but the number of linearly independent vectors perpendicular to a given vector increases. In Rⁿ, you can find n-1 linearly independent vectors perpendicular to a given non-zero vector.

Applications

• Computer Graphics: Determining surface normals for lighting and shading calculations.
• Physics: Finding forces or velocities that are orthogonal to a given direction.
• Engineering: Analyzing structural stability and stress distribution.
• Linear Algebra: Constructing orthogonal bases for vector spaces and solving systems of linear equations.
• Robotics: Planning robot movements and avoiding obstacles.

Common Mistakes

• Assuming any two vectors are perpendicular: Always verify perpendicularity using the dot product.
• Finding linearly dependent vectors: Ensure that the vectors you find are not scalar multiples of each other. Test this by visually inspecting or attempting to solve for scalar multipliers.
• Division by zero: Be careful when solving for the components of the perpendicular vectors, especially when dealing with cases where one or more components of the original vector are zero.
• Not considering special cases: Remember to handle cases where the original vector has zero components appropriately.

Advanced Considerations

• Gram-Schmidt Process: This process can be used to orthogonalize a set of linearly independent vectors. Starting with a set of linearly independent vectors, the Gram-Schmidt process produces a set of orthogonal vectors that span the same subspace.
• Orthogonal Matrices: Matrices whose columns (and rows) are mutually orthogonal unit vectors are called orthogonal matrices. These matrices have special properties and are used in various applications, including rotations and reflections.
• Eigenvalues and Eigenvectors: Eigenvectors of a symmetric matrix corresponding to distinct eigenvalues are orthogonal. This property is used in various applications, including principal component analysis.

Conclusion

Finding two linearly independent vectors perpendicular to a given vector is a fundamental operation in linear algebra with wide-ranging applications. By understanding the dot product and the concept of linear independence, we can systematically find these vectors. By carefully considering the cases where components of the given vector are zero, and verifying the results, we can ensure we arrive at the correct solution. Mastering this technique provides a strong foundation for understanding more advanced topics in linear algebra and its applications. Always remember the geometric interpretation – you are essentially finding a basis for the plane that is normal to your initial vector.