Problem 24 - 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 \(\mathbf{w}\). $$ \mathbf{v}=2 \mathbf{i}-\mathbf{j}, \quad \mathbf{w}=\mathbf{i}-2 \mathbf{j} $$

Step-by-Step Solution

Answer

The parallel vector \(\mathbf{v}_{1}\) is \(\frac{4}{5} \mathbf{i} - \frac{8}{5} \mathbf{j}\) and the orthogonal vector \(\mathbf{v}_{2}\) is \(\frac{6}{5} \mathbf{i} + \frac{3}{5} \mathbf{j}\).

Step 1 - Understanding Vectors

Given vectors are \(\mathbf{v} = 2\mathbf{i} - \mathbf{j}\) and \(\mathbf{w} = \mathbf{i} - 2\mathbf{j}\). We need to decompose \(\mathbf{v}\) into a parallel component \(\mathbf{v}_{1}\) and an orthogonal component \(\mathbf{v}_{2}\) with respect to \(\mathbf{w}\).

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

The vector \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\) and can be written as \(k\mathbf{w}\) for some scalar \(k\). To find \(k\), use the formula for projection: \(\mathbf{v}_{1} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w}\). Calculate the dot products \(\mathbf{v} \cdot \mathbf{w}\) and \(\mathbf{w} \cdot \mathbf{w}\).

Step 2.1 - Compute Dot Products

Calculate the dot product \(\mathbf{v} \cdot \mathbf{w}\):\( \mathbf{v} \cdot \mathbf{w} = (2\mathbf{i} - \mathbf{j}) \cdot (\mathbf{i} - 2\mathbf{j}) = 2 \cdot 1 + (-1) \cdot (-2) = 2 + 2 = 4\).Next, calculate \(\mathbf{w} \cdot \mathbf{w}\):\( \mathbf{w} \cdot \mathbf{w} = (\mathbf{i} - 2\mathbf{j}) \cdot (\mathbf{i} - 2\mathbf{j}) = 1 + 4 = 5\).

Step 3 - Calculate \(\mathbf{v}_{1}\)

Using the values from Step 2.1, \(\mathbf{v}_{1} = \frac{4}{5} \mathbf{w} = \frac{4}{5} (\mathbf{i} - 2\mathbf{j}) = \frac{4}{5} \mathbf{i} - \frac{8}{5} \mathbf{j}\).

Step 4 - Find \(\mathbf{v}_{2}\)

By definition, \(\mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2}\). Therefore, \(\mathbf{v}_{2} = \mathbf{v} - \mathbf{v}_{1}\). Substitute the values of \(\mathbf{v}\) and \(\mathbf{v}_{1}\): \(\mathbf{v}_{2} = (2\mathbf{i} - \mathbf{j}) - (\frac{4}{5} \mathbf{i} - \frac{8}{5} \mathbf{j}) = (2 - \frac{4}{5})\mathbf{i} - (1 - \frac{8}{5})\mathbf{j}\). Simplify the expression: \(\mathbf{v}_{2} = \frac{10}{5}\mathbf{i} - \frac{4}{5}\mathbf{i} - (\frac{-5 + 8}{5})\mathbf{j} = \frac{6}{5}\mathbf{i} + \frac{3}{5}\mathbf{j}\).

Key Concepts

parallel vectors

When we talk about parallel vectors, we mean that two vectors lie along the same direction. This means one vector can be expressed as a scalar multiple of the other. In the given exercise, we are asked to find a vector \(\mathbf{v}_1\) that is parallel to \(\mathbf{w}\). \(\mathbf{v}_1\) can be represented as \(k\mathbf{w}\), where \(k\) is a scalar. This is important in vector decomposition because finding the parallel component helps in dividing a vector into meaningful parts. Understanding parallel vectors helps to see how certain vector components align with a chosen direction, making analysis easier.

orthogonal vectors

Orthogonal vectors are vectors that are perpendicular to each other. Mathematically, two vectors \(\mathbf{a}\) and \(\mathbf{b}\) are orthogonal if their dot product is zero, i.e., \(\mathbf{a} \cdot \mathbf{b} = 0\). This concept is used in the given exercise when we need to find \(\mathbf{v}_2\) which is orthogonal to \(\mathbf{w}\). This means \(\mathbf{v}_2\) has no component in the direction of \(\mathbf{w}\), making it easier to analyze decomposition. Orthogonal vectors play a vital role in physics and engineering when dealing with forces, directions, and balanced systems.

dot product

The dot product, also known as the scalar product, measures the extent to which two vectors align. It is calculated as the sum of the products of their corresponding components. For instance, in the exercise, \(\mathbf{v} \cdot \mathbf{w}\) and \(\mathbf{w} \cdot \mathbf{w}\) are calculated to determine the parallel and orthogonal components. The dot product formula is \(\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + ... + a_n b_n\). The result is a scalar and can be positive, negative, or zero. Positive dot products mean vectors point in roughly the same direction, negative means opposite directions, and zero implies orthogonality.

projection of a vector

Projection of a vector \(\mathbf{v}\) onto another vector \(\mathbf{w}\) finds the component of \(\mathbf{v}\) that is in the direction of \(\mathbf{w}\). This is represented by the formula: \({\text{proj}_{\mathbf{w}}{\mathbf{v}}} = \frac{\mathbf{v} \cdot {\mathbf{w}}}{\mathbf{w} \cdot {\mathbf{w}}} \mathbf{w}\). In the given exercise, this formula helps find \(\mathbf{v}_1\), the component of \(\mathbf{v}\) parallel to \(\mathbf{w}\). Projection is essential in vector decomposition as it allows us to break a vector into orthogonal parts with respect to a given direction, simplifying analysis in physics and engineering contexts.