Problem 23 - 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 \(\mathbf{w}\). $$ \mathbf{v}=\mathbf{i}-\mathbf{j}, \quad \mathbf{w}=-\mathbf{i}-2 \mathbf{j} $$

Step-by-Step Solution

Answer

The vectors are \(\mathbf{v}_{1} = -\frac{1}{5}\mathbf{i} - \frac{2}{5}\mathbf{j}\) and \(\mathbf{v}_{2} = \frac{6}{5}\mathbf{i} - \frac{3}{5}\mathbf{j}\).

Step 1 - Find the projection of \(\mathbf{v}\) onto \(\mathbf{w}\)

The projection of \(\mathbf{v}\) onto \(\mathbf{w}\) is given by: \[ \text{proj}_{\mathbf{w}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} \]. Calculate the dot products: \( \mathbf{v} \cdot \mathbf{w} = [1, -1] \cdot [-1, -2] = -1 + 2 = 1 \) and \( \mathbf{w} \cdot \mathbf{w} = [-1, -2] \cdot [-1, -2] = 1 + 4 = 5 \).

Step 2 - Compute \(\mathbf{v}_{1}\)

Using the projection formula: \[ \mathbf{v}_{1} = \text{proj}_{\mathbf{w}} \mathbf{v} = \frac{1}{5} (-\mathbf{i} - 2\mathbf{j}) = -\frac{1}{5}\mathbf{i} - \frac{2}{5}\mathbf{j} \].

Step 3 - Find \(\mathbf{v}_{2}\)

Since \(\mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2}\), we can solve for \(\mathbf{v}_{2}\): \[ \mathbf{v}_{2} = \mathbf{v} - \mathbf{v}_{1} = (\mathbf{i} - \mathbf{j}) - \left(-\frac{1}{5}\mathbf{i} - \frac{2}{5}\mathbf{j}\right) = \left(1 + \frac{1}{5}\right)\mathbf{i} + \left(-1 + \frac{2}{5}\right)\mathbf{j} = \frac{6}{5}\mathbf{i} - \frac{3}{5}\mathbf{j} \].

Key Concepts

Vector Projection

Vector projection helps to find the part of one vector that points in the same direction as another vector. To project vector \(\textbf{v}\) onto vector \(\textbf{w}\), we use the formula: \[ \text{proj}_{\textbf{w}} \textbf{v} = \frac{\textbf{v} \cdot \textbf{w}}{\textbf{w} \cdot \textbf{w}} \textbf{w} \]. The dot product \(\textbf{v} \cdot \textbf{w}\) measures how much \(\textbf{v}\) 'sticks' in the direction of \(\textbf{w}\), while \(\textbf{w} \cdot \textbf{w}\) normalizes the vector to unit length. The final result is a vector parallel to \(\textbf{w}\). For example, in the exercise, the dot products were calculated as follows: \[ \textbf{v} \cdot \textbf{w} = [1, -1] \cdot [-1, -2] = 1 \] and \[ \textbf{w} \cdot \textbf{w} = [-1, -2] \cdot [-1, -2] = 5 \]. Thus, the projection formula gives us \(\text{proj}_{\textbf{w}} \textbf{v} = \frac{1}{5} \textbf{w} \), leading to the vector \(\textbf{v}_1 = -\frac{1}{5}\textbf{i} - \frac{2}{5}\textbf{j}\).

Dot Product

The dot product is a fundamental operation in vector mathematics. It combines two vectors and returns a scalar. The dot product of vectors \(\textbf{a}\) and \(\textbf{b}\), represented as \[ \textbf{a} \cdot \textbf{b} = a_1 b_1 + a_2 b_2 + ... + a_n b_n \], measures their combined magnitude. In the given exercise, the dot product \(\textbf{v} \cdot \textbf{w}\) involves calculating \[ 1 \times (-1) + (-1) \times (-2) = -1 + 2 = 1 \]. The dot product helps determine how much one vector extends in the direction of another. If the result is zero, the vectors are orthogonal; if non-zero, they are not orthogonal.

Orthogonal Vectors

Orthogonal vectors are vectors with a dot product of zero, meaning they are perpendicular to each other. For example, if \(\textbf{v}\) and \(\textbf{w}\) are orthogonal, \[ \textbf{v} \cdot \textbf{w} = 0 \]. In the exercise, \(\textbf{v}_2\) is orthogonal to \(\textbf{w}\). We found \(\textbf{v}_2\) by subtracting the parallel component \(\textbf{v}_1\) from \(\textbf{v}\). The orthogonal vector is \[ \textbf{v}_2 = \bf{v} - \text{proj}_{\textbf{w}} \textbf{v} \], and the result is \[ \textbf{v}_2 = \frac{6}{5}\textbf{i} - \frac{3}{5}\textbf{j} \]. This ensures \(\textbf{v}_2 \cdot \textbf{w} = 0\), confirming orthogonality.

Parallel Vectors

Parallel vectors are scalar multiples of each other, pointing in the same or exact opposite direction. For example, if \(\textbf{a}\) is half of \(\textbf{b}\), then \[ \textbf{a} = k \textbf{b} \] where \(k = 0.5\). In the example, we found \(\textbf{v}_1\) to be parallel to \(\textbf{w}\) using vector projection. Since \(\textbf{v}_1 = \text{proj}_{\textbf{w}} \textbf{v}\), it is a scalar multiple of \(\textbf{w}\). The vector \(\textbf{v}_1 = -\frac{1}{5}\textbf{i} - \frac{2}{5}\textbf{j} \) verifies that \(\textbf{v}_1 = \frac{1}{5} \textbf{w} \), confirming parallelism.