Exploring Unit Vectors and Their Properties

In the realm of vector analysis and trigonometry, understanding the concept of a unit vector is fundamental. A unit vector is a vector whose magnitude (or norm) is exactly one. This means that if a vector (vec{v}) is a unit vector, then its norm (|vec{v}|^2 1).

Introduction to Unit Vectors

A vector can be expressed in component form. Consider a vector (vec{v}) defined as:

(vec{v} sin(theta)hat{i} 2cos(theta)hat{j})

Here, (theta) is a parameter, (hat{i}) and (hat{j}) are the unit vectors in the x and y directions, respectively. To determine if this vector is a unit vector, we need to calculate its norm and check if it equals 1.

Calculating the Norm of the Vector

The norm (or magnitude) of the vector (vec{v}) is given by:

(|vec{v}|^2 (sin(theta))^2 (2cos(theta))^2)

Simplifying the expression:

(|vec{v}|^2 sin^2(theta) 4cos^2(theta))

Using the Pythagorean identity (sin^2(theta) cos^2(theta) 1), we can rewrite the expression:

(|vec{v}|^2 sin^2(theta) 4cos^2(theta) 1 3cos^2(theta))

For the vector to be a unit vector, this expression must equal 1. Therefore:

(1 3cos^2(theta) 1)

Subtracting 1 from both sides:

(3cos^2(theta) 0)

Which implies:

(cos(theta) 0)

This equation is true only when (theta frac{pi}{2} npi) for any integer (n). Therefore, the vector (vec{v}) is a unit vector only for these values of (theta).

Conclusion

In summary, the vector (vec{v} sin(theta)hat{i} 2cos(theta)hat{j}) is a unit vector only when (theta frac{pi}{2} npi). For other values of (theta), the norm of the vector is greater than 1 and hence it is not a unit vector.