91Ó°ÊÓ

Projection of vectors is a fundamental concept in linear algebra and vector spaces. It involves finding the shadow or image of a vector onto another vector. This is extremely useful when determining the component of a vector that aligns with another.

In the given exercise, we need to project vector \( \mathbf{y} = \begin{pmatrix} 3 \ 1 \end{pmatrix} \) onto vector \( \mathbf{u} = \begin{pmatrix} 8 \ 6 \end{pmatrix} \).
To accomplish this, you apply the projection formula:

• \( \text{Proj}_{\mathbf{u}}\mathbf{y} = \frac{\mathbf{y} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \)

This formula essentially rescales vector \( \mathbf{u} \) to the appropriate length so that its direction matches the component of \( \mathbf{y} \) parallel to \( \mathbf{u} \).

By calculating the dot products, we find that the projection of \( \mathbf{y} \) onto \( \mathbf{u} \) is \( \begin{pmatrix} 2.4 \ 1.8 \end{pmatrix} \). This vector represents the closest point on the line with direction \( \mathbf{u} \) to the vector \( \mathbf{y} \). It helps simplify the task of finding distances between vectors and lines in the vector space.

Euclidean distance is a classic measure of the "straight line" distance between two points in space. In vector spaces, it is used to determine how far one point is from another. It's a fundamental tool for many geometric and spatial computations.

To calculate Euclidean distance between two vectors, such as \( \mathbf{y} \) and its projection, the formula is:

• \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

This formula is applied to the exercise to find the distance from \( \mathbf{y} = \begin{pmatrix} 3 \ 1 \end{pmatrix} \) to its projection \( \begin{pmatrix} 2.4 \ 1.8 \end{pmatrix} \).

By plugging in the values, we calculate:

• \( \sqrt{(3 - 2.4)^2 + (1 - 1.8)^2} = \sqrt{0.36 + 0.64} = \sqrt{1.00} = 1 \)

Hence, the Euclidean distance of 1 indicates how far the point \( \mathbf{y} \) is from the line determined by the vector \( \mathbf{u} \). This straightforward method provides an accurate measure of distance in the vector space.

In the context of vector spaces, lines through the origin are defined by vector direction rather than specific starting or ending points. These lines are infinite in nature and are fully described by a direction vector.

In this exercise, the line is determined by the vector \( \mathbf{u} = \begin{pmatrix} 8 \ 6 \end{pmatrix} \). This line includes all scalar multiples of \( \mathbf{u} \), represented mathematically as \( t\mathbf{u} = t\begin{pmatrix} 8 \ 6 \end{pmatrix} \) for any real number \( t \).

Such lines are fundamental geometrical constructs within vector spaces:

• They pass through the origin, located at \((0,0)\).
• They are defined by direction vectors like \( \mathbf{u} \).
• They contain points like \( \mathbf{0}, \mathbf{u}, 2\mathbf{u}, 3\mathbf{u}, \dots \).

Understanding these lines helps in visualizing how vectors like \( \mathbf{y} \) relate spatially to directions defined by other vectors. Lines through the origin open a window into infinite paths directions can take in vector space.