Problem 25 - Algebra and Trigonometry

Question

Decompose \(\mathbf{v}\) into two vectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), where \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\), and \(\mathbf{v}_{2}\) is orthogonal to \(\mathbf{w}\). $$ \mathbf{v}=3 \mathbf{i}+\mathbf{j}, \quad \mathbf{w}=-2 \mathbf{i}-\mathbf{j} $$

Step-by-Step Solution

Answer

\( \textbf{v}_{1} = \frac{14}{5} \textbf{i} + \frac{7}{5} \textbf{j}, \ \textbf{v}_{2} = \frac{1}{5} \textbf{i} - \frac{2}{5} \textbf{j} \).

Step 1: Calculate the dot product of \(\textbf{v}\) and \(\textbf{w}\)

To find the projection, first calculate the dot product \(\textbf{v} \cdot \textbf{w} \). \(\textbf{v} = 3 \textbf{i} + \textbf{j} \) and \(\textbf{w} = -2 \textbf{i} - \textbf{j} \): \[\textbf{v} \cdot \textbf{w} = (3 \textbf{i} + \textbf{j}) \cdot (-2 \textbf{i} - \textbf{j}) = (3)(-2) + (1)(-1) = -6 - 1 = -7. \]

Step 2: Calculate the magnitude squared of \(\textbf{w}\)

Next, calculate \(\|\textbf{w}\|^2 \): \[ \|\textbf{w}\|^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5. \]

Step 3: Find \(\textbf{v}_{1}\), the projection of \(\textbf{v}\) onto \(\textbf{w}\)

Use the projection formula \(\textbf{v}_{1} = \frac{\textbf{v} \cdot \textbf{w}}{\|\textbf{w}\|^2} \textbf{w} \): \[ \textbf{v}_{1} = \frac{-7}{5} \textbf{w} = \frac{-7}{5}(-2 \textbf{i} - \textbf{j}) = \frac{14}{5} \textbf{i} + \frac{7}{5} \textbf{j}. \]

Step 4: Find \(\textbf{v}_{2}\), the component of \(\textbf{v}\) orthogonal to \(\textbf{w}\)

Use \(\textbf{v}_{2} = \textbf{v} - \textbf{v}_{1} \): \[ \textbf{v}_{2} = (3 \textbf{i} + \textbf{j}) - \left( \frac{14}{5} \textbf{i} + \frac{7}{5} \textbf{j} \right) = \textbf{i}\left(3 - \frac{14}{5}\right) + \textbf{j}\left(1 - \frac{7}{5}\right) = \frac{1}{5} \textbf{i} - \frac{2}{5} \textbf{j}. \]

Key Concepts

Dot Product

When working with vectors, it's essential to understand the dot product, represented mathematically as \( \textbf{v} \cdot \textbf{w} \)).
The dot product is a way to multiply two vectors, resulting in a scalar (a single number).
It helps us find the angle between vectors, determine if two vectors are orthogonal, and, as seen in our example, find projections.
To compute the dot product of \( \textbf{v} = 3 \textbf{i} + \textbf{j} \) and \( \textbf{w} = -2 \textbf{i} - \textbf{j} \), we multiply corresponding components and add them:
\((3 \cdot (-2)) + (1 \cdot (-1)) = -6 - 1 = -7\).

In more general terms, if \( \textbf{a} = a_1 \textbf{i} + a_2 \textbf{j} \) and \( \textbf{b} = b_1 \textbf{i} + b_2 \textbf{j} \), then:
\(\textbf{a} \cdot \textbf{b} = a_1 b_1 + a_2 b_2 \).

Vector Projection

Vectors can be projected onto one another to understand how one vector 'falls' onto another.
The projection of \( \textbf{v} \) onto \( \textbf{w} \) is a vector parallel to \( \textbf{w} \).
It shows the component of \( \textbf{v} \) in the direction of \( \textbf{w} \).
The formula for projection is: \( \textbf{v}_1 = \frac{\textbf{v} \cdot \textbf{w}}{\textbf{\textbar{}\textbar{}w\textbar{}\textbar{}^2}} \textbf{w} \), where \( \textbf{\textbar{}\textbar{}w\textbar{}\textbar{} \) is the magnitude of \(\textbf{w}\).
In our case: \( \textbf{v}_1 = \frac{-7}{5} \textbf{w} = \frac{-7}{5}(-2 \textbf{i} - \textbf{j}) = \frac{14}{5} \textbf{i} + \frac{7}{5} \textbf{j} \).

Essentially, this gives us the part of \(\textbf{v}\) that is aligned with \( \textbf{w} \).

Orthogonal Vectors

Orthogonal vectors are at right angles to each other.
The term 'orthogonal' means their dot product is zero.
If \(\textbf{v} \cdot \textbf{w} = 0\), the vectors are orthogonal.
In our problem, to find the component of \( \textbf{v} \) orthogonal to \( \textbf{w} \), we subtract the projection \( \textbf{v}_1 \) from the original vector \( \textbf{v} \):
\( \textbf{v}_2 = \textbf{v} - \textbf{v}_1 \).
This results in a vector that is perpendicular to \( \textbf{w} \).
Following our example:
\( \textbf{v}_2 = (3 \textbf{i} + \textbf{j}) - \frac{14}{5} \textbf{i} - \frac{7}{5} \textbf{j}
= \frac{1}{5} \textbf{i} - \frac{2}{5} \textbf{j} \).

This calculated vector is orthogonal to \( \textbf{w} \).

Magnitude of a Vector

The magnitude of a vector represents its length and is typically denoted by \( \textbf{\textbar{}\textbar{}v\textbar{}\textbar{} \).
It's calculated using the Pythagorean theorem.
For a vector \( \textbf{v} = a \textbf{i} + b \textbf{j} \), the magnitude is:
\( \textbf{\textbar{}\textbar{}v\textbar{}\textbar{} = \sqrt{a^2 + b^2} \).
In our example, for \(\textbf{w} = -2 \textbf{i} - \textbf{j} \),:
\( \textbf{\textbar{}\textbar{}w\textbar{}\textbar{} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \).

This magnitude helps in normalizing vectors and is crucial when performing projections.
Understanding the magnitude is key when decomposing vectors into their components.