Finding a Unit Vector Parallel to the XOY-Plane and Perpendicular to a Given Vector

In this article, we will explore how to find a unit vector that is both parallel to the XOY-plane and perpendicular to a given vector. We will use the specific example of the vector

Introduction

Understanding how to find such a vector is crucial in various mathematical and practical applications, such as in computer graphics, physics, and engineering. This article will provide a detailed explanation, complete with examples, to help readers grasp the concept easily.

Step 1: Identify the Constraints

To find a vector that meets our requirements, we need to consider two key constraints:

The vector must be parallel to the XOY-plane. The vector must be perpendicular to the given vector

Constraint 1: Parallel to the XOY-Plane

A vector parallel to the XOY-plane will have a zero z-component. Therefore, any vector that is parallel to this plane can be represented as:

[mathbf{u} amathbf{i} bmathbf{j} 0mathbf{k} amathbf{i} bmathbf{j}]

Constraint 2: Perpendicular to the Given Vector

To ensure that [mathbf{u} cdot mathbf{v} (amathbf{i} bmathbf{j}) cdot (4mathbf{i} - 3mathbf{j} mathbf{k}) 0]

When we perform the dot product, we obtain:

[4a - 3b 0]

Step 2: Solve for (a) and (b)

From the equation (4a - 3b 0), we can solve for (b) in terms of (a):

[b frac{4}{3}a]

Step 3: Form the Vector (mathbf{u})

Substituting (b) back into the expression for (mathbf{u}), we get:

[mathbf{u} amathbf{i} frac{4}{3}amathbf{j} aleft(mathbf{i} frac{4}{3}mathbf{j}right)]

Step 4: Find the Unit Vector

To find the unit vector, we need to normalize (mathbf{u}). The magnitude of (mathbf{u}) is:

[|mathbf{u}| sqrt{a^2 left(frac{4}{3}aright)^2} sqrt{a^2 frac{16}{9}a^2} sqrt{frac{25}{9}a^2} frac{5}{3}a]

The unit vector (mathbf{u}_{text{unit}}) is given by:

[mathbf{u}_{text{unit}} frac{mathbf{u}}{|mathbf{u}|} frac{aleft(mathbf{i} frac{4}{3}mathbf{j}right)}{frac{5}{3}a} frac{3}{5}left(mathbf{i} frac{4}{3}mathbf{j}right) frac{3}{5}mathbf{i} frac{4}{5}mathbf{j}]

Step 5: Choose a Direction for (a)

For simplicity, we can choose (a 1). Thus, the unit vector is:

[mathbf{u}_{text{unit}} frac{3}{5}mathbf{i} frac{4}{5}mathbf{j}]

Conclusion

The unit vector that is parallel to the XOY-plane and perpendicular to the vector (4mathbf{i} - 3mathbf{j} mathbf{k}) is:

[boxed{frac{3}{5}mathbf{i} frac{4}{5}mathbf{j}}]

Related Information

Understanding the concept of finding a unit vector that is parallel to the XOY-plane and perpendicular to a vector involves a clear grasp of vector operations, including the dot product and normalization. Related topics include:

Unit Vector: A vector with a magnitude of 1, often used in various mathematical and engineering applications.

Parallel to XOY-Plane: Any vector that lies on the plane defined by the X and Y axes, having a zero z-component.

Perpendicular Vector: A vector that is perpendicular to another vector, as determined by their dot product being zero.