Problem 22 - 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 \(\mathbf{w}\). $$ \mathbf{v}=-3 \mathbf{i}+2 \mathbf{j}, \quad \mathbf{w}=2 \mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

The vectors are \(\textbf{v}_{1} = -\frac{8}{5}\textbf{i} - \frac{4}{5}\textbf{j}\) and \(\textbf{v}_{2} = -\frac{7}{5}\textbf{i} + \frac{14}{5}\textbf{j}\).

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

First, calculate the projection of \(\textbf{v}\) onto \(\textbf{w}\) using the formula: \(\text{proj}_{\textbf{w}} \textbf{v} = \frac{\textbf{v} \bullet \textbf{w}}{\textbf{w} \bullet \textbf{w}} \textbf{w}\). Compute the dot products \(\textbf{v} \bullet \textbf{w}\) and \(\textbf{w} \bullet \textbf{w}\).

Step 2: Compute the dot products

Calculate the dot product \(\textbf{v} \bullet \textbf{w} = (-3\textbf{i} + 2\textbf{j}) \bullet (2\textbf{i} + \textbf{j}) = -6 + 2 = -4\). Next, calculate the dot product of \(\textbf{w}\) with itself: \(\textbf{w} \bullet \textbf{w} = (2\textbf{i} + \textbf{j}) \bullet (2\textbf{i} + \textbf{j}) = 4 + 1 = 5\).

Step 3: Apply the projection formula

Substitute the dot products back into the projection formula: \(\text{proj}_{\textbf{w}} \textbf{v} = \frac{-4}{5} (2\textbf{i} + \textbf{j}) = -\frac{8}{5}\textbf{i} - \frac{4}{5}\textbf{j} = \textbf{v}_{1}\). Therefore, \(\textbf{v}_{1} = -\frac{8}{5}\textbf{i} - \frac{4}{5}\textbf{j}\).

Step 4: Find \(\textbf{v}_{2}\)

To find \(\textbf{v}_{2}\), use the relation \(\textbf{v} = \textbf{v}_{1} + \textbf{v}_{2}\). Rearrange it to get \(\textbf{v}_{2} = \textbf{v} - \textbf{v}_{1}\). Using the values: \(\textbf{v} = -3\textbf{i} + 2\textbf{j}\) and \(\textbf{v}_{1} = -\frac{8}{5}\textbf{i} - \frac{4}{5}\textbf{j}\), compute \(\textbf{v}_{2} = (-3\textbf{i} + 2\textbf{j}) - \bigg(-\frac{8}{5}\textbf{i} - \frac{4}{5}\textbf{j}\bigg) = -3\textbf{i} + 2\textbf{j} + \frac{8}{5}\textbf{i} + \frac{4}{5}\textbf{j} = -\frac{15}{5}\textbf{i} + \frac{8}{5}\textbf{i} + \frac{10}{5}\textbf{j} + \frac{4}{5}\textbf{j} = -\frac{7}{5}\textbf{i} + \frac{14}{5}\textbf{j}\textbf{j} = \textbf{v}_{2}\). Hence, \(\textbf{v}_{2} = -\frac{7}{5}\textbf{i} + \frac{14}{5}\textbf{j}\).

Key Concepts

Vector Projection

Let's start with the concept of vector projection. To project one vector onto another means to find a vector in the direction of the second vector. The projection of a vector \(\textbf{v}\) onto another vector \(\textbf{w}\) can be found using the formula:
\[ \text{proj}_{\textbf{w}} \textbf{v} = \frac{\textbf{v} \bullet \textbf{w}}{\textbf{w} \bullet \textbf{w}} \textbf{w} \]
This mathematical operation helps us determine the component of \(\textbf{v}\) that lies along the direction of \(\textbf{w}\).
In the given exercise, \(\textbf{v}\) is projected onto \(\textbf{w}\). First, we calculate the dot products \(\textbf{v} \bullet \textbf{w}\) and \(\textbf{w} \bullet \textbf{w}\). Then, using these dot products, substitute back into the projection formula to get the projection results.
This projection is critical to find the vector \(\textbf{v}_{1}\) that is parallel to \(\textbf{w}\).

Dot Product

The dot product is a fundamental operation in vector algebra. It is the product of the magnitudes of two vectors and the cosine of the angle between them.
The formula to compute the dot product of vectors \(\textbf{a}\) and \(\textbf{b}\) is:
\[ \textbf{a} \bullet \textbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n \]
For two-dimensional vectors, this simplifies to finding the sum of the products of their corresponding components.
In the given problem:

• The dot product \(\textbf{v} \bullet \textbf{w}\) is calculated as \(-3 \times 2 + 2 \times 1\), which equals \(-4\).
• The dot product \(\textbf{w} \bullet \textbf{w}\) equals \(2 \times 2 + 1 \times 1\), which is \5\.

Using these dot products, we can then find the vector projection of \(\textbf{v}\) onto \(\textbf{w}\).

Orthogonal Vectors

Orthogonal vectors are vectors that are perpendicular to each other. This means that the angle between them is \(90^\text{o}\).
Two vectors \(\textbf{a}\) and \(\textbf{b}\) are orthogonal if their dot product is zero, i.e., \[ \textbf{a} \bullet \textbf{b} = 0 \]
In the exercise, finding the vector \(\textbf{v}_{2}\) such that it is orthogonal to \(\textbf{w}\) means \(\textbf{v}_{2}\) should satisfy \[ \textbf{v}_{2} \bullet \textbf{w} = 0 \]
\(\textbf{v}_{2}\) is identified by subtracting the parallel component \(\textbf{v}_{1}\) from the original vector \(\textbf{v}\). This ensures that what's left of \(\textbf{v}\) after removing \(\textbf{v}_{1}\) is orthogonal to \(\textbf{w}\).

Parallel Vectors

Parallel vectors are vectors that have the same or exactly opposite direction. This means one's direction can be obtained by multiplying the other by a scalar.
In more formal terms, vectors \(\textbf{a}\) and \(\textbf{b}\) are parallel if there exists some scalar \(k\) such that \(\textbf{a} = k \textbf{b}\).
For the given problem, finding the vector \(\textbf{v}_{1}\) involves projecting \(\textbf{v}\) onto \(\textbf{w}\). This projection ensures that \(\textbf{v}_{1}\) is parallel to \textbf{w}\. Therefore, \(\textbf{v}_{1}\) is some scalar multiple of \(\textbf{w}\) which lies in the same or opposite direction as \(\textbf{w}\).