91Ó°ÊÓ

In vector calculus, the concept of a unit vector is fundamental. A unit vector is simply a vector that has a magnitude, or length, of 1.
This is significant because unit vectors are often used to indicate direction without concern for any specific magnitude.

• The notation generally involves writing the unit vector with a hat, for instance, \( \mathbf{u} \).
• The formula for calculating the magnitude of a vector \( \mathbf{u} \) is \( \| \mathbf{u} \| = 1 \).

To express any vector \( \mathbf{v} \) as a unit vector, you simply divide \( \mathbf{v} \) by its own magnitude: \( \mathbf{u} = \frac{\mathbf{v}}{\| \mathbf{v} \|} \).
This resulting vector \( \mathbf{u} \) has a new magnitude of 1 and retains the direction of \( \mathbf{v} \). The importance of unit vectors can't be overstated in physics and engineering because they help specify only direction, which makes problems easier to handle.

The dot product is a way to multiply two vectors, resulting in a scalar quantity.
Also known as the scalar product, the dot product is important in understanding angles between vectors and projections.

• The formula for finding the dot product of two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), is \( \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos \theta \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).
• If the vectors are perpendicular, the dot product is zero since \( \cos 90^\circ = 0 \).

The dot product is also useful in projection, where one vector is projected onto another. If one of the vectors is a unit vector, such as \( \mathbf{u} \) from our problem, the dot product simplifies to the magnitude of the other vector, multiplied by \( \pm 1 \), depending on the direction.
This is why \( \mathbf{u} \cdot \mathbf{v} = \pm\|\mathbf{v}\| \) holds true when \( \mathbf{u} \) is a unit vector parallel to \( \mathbf{v} \).

Parallel vectors are vectors that have the same or exact opposite direction.
When two vectors are parallel, one can be expressed as a scalar multiple of the other.

• Mathematically, if \( \mathbf{u} \) is parallel to \( \mathbf{v} \), then \( \mathbf{u} = c \mathbf{v} \) for some scalar \( c \).
• If \( c > 0 \), the vectors have the same direction, and if \( c < 0 \), they have opposite directions.

In the context of the exercise, \( \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \) results in \( \mathbf{u} \) having a magnitude of 1 and pointing in the direction of \( \mathbf{v} \).
If we choose \( \mathbf{u} = -\frac{\mathbf{v}}{\|\mathbf{v}\|} \), then \( \mathbf{u} \) has a magnitude of 1 but an opposite direction.
This scalar relationship underpins the simplification of vector calculations, as in the problem's statement.