Introduction

Understanding how to find vectors that are orthogonal to a given vector is fundamental in vector algebra and has wide-ranging applications in physics, engineering, and computer graphics. This article will walk you through a detailed process of finding all three-dimensional vectors orthogonal to a specific vector s110, using vector cross product and standard unit vectors. We will also explore how to express these vectors in a general form and confirm their orthogonality.

Understanding Orthogonality

In vector algebra, orthogonality between two vectors refers to their dot product being zero. Formally, two vectors a and b are orthogonal if a · b 0. Given a vector v 110, we seek all vectors u that satisfy v · u 0.

Using Cross Product to Find Orthogonal Vectors

The cross product a × b of two vectors a and b yields a vector that is orthogonal to both a and b. For our problem, we start by choosing a standard unit vector and computing the cross product with the given vector v 110.

Step 1: Choosing a Standard Unit Vector

Let's start by choosing a standard unit vector. The unit vector u? in the z-direction is u? 001. This vector is orthogonal to v since their dot product is 0. Mathematically, we have:

u? · v 001 · 110 0

Step 2: Finding a Second Orthogonal Vector

To find a second vector that is also orthogonal to v, we compute the cross product of u? and v. The cross product of two vectors u? 001 and v 110 is:

This vector u? 0-11 is also orthogonal to v since:

v · u? 110 · 0-11 0

Step 3: General Form of Orthogonal Vectors

Any vector in the plane orthogonal to v can be expressed as a linear combination of u? and u?. Let u a u? b u?. Substituting the values of u? and u?, we get:

u a 001 b 0-11 a-1 b 00 b

This general form shows that any vector orthogonal to v 110 can be written with scalars a and b.

Using Standard Unit Vectors and Expressing Orthogonality

The goal is to find all solutions α?, α?, α? in the equation v · (α? e? α? e? α? e?) 0, where e?, e?, e? are the standard unit vectors. Let's consider the vector u α? e? α? e? α? e?. The orthogonality condition is:

v · (α? e? α? e? α? e?) 0

Substituting v 110, we get:

110 · (α? e? α? e? α? e?) 0

Which simplifies to:

1 × α? 1 × α? 0 × α? 0

Hence, α? -α?. This means all vectors orthogonal to v 110 are of the form:

α? e? - α? e? α? e? α? e? - α? e? α? e?

Expressing this in terms of standard unit vectors, we have:

(a-1 a 0) (0 0 c) (a-1 0 1) (0 0 c) (a-1 c 1 c)

Conclusion

By using vector cross products and standard unit vectors, we have found all three-dimensional vectors orthogonal to v 110. These vectors lie in a plane and can be expressed in the form (a-1 a 0) (0 0 c). This approach is not only insightful but also practical for a range of applications in vector algebra and beyond.