Problem 34 - Algebra and Trigonometry

Question

(a) Calculate proj, \(\mathbf{u}\) . (b) Resolve \(\mathbf{u}\) into \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2},\) where \(\mathbf{u}_{1}\) is parallel to \(\mathbf{v}\) and \(\mathbf{u}_{2}\) is orthogonal to \(\mathbf{v} .\) $$ \mathbf{u}=\langle 1,1\rangle, \quad \mathbf{v}=\langle 2,-1\rangle $$

Step-by-Step Solution

Answer

\(\text{proj}_{\mathbf{v}} \mathbf{u} = \langle \frac{2}{5}, -\frac{1}{5} \rangle\); components are \(\mathbf{u}_1 = \langle \frac{2}{5}, -\frac{1}{5} \rangle\), \(\mathbf{u}_2 = \langle \frac{3}{5}, \frac{6}{5} \rangle\).

Step 1: Calculate the Dot Product

To find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), we first need to calculate the dot product \(\mathbf{u} \cdot \mathbf{v}\). Use the formula: \[\mathbf{u} \cdot \mathbf{v} = (1)(2) + (1)(-1)\]This simplifies to:\[\mathbf{u} \cdot \mathbf{v} = 2 - 1 = 1\]

Step 2: Calculate the Magnitude of \(\mathbf{v}\) Squared

Next, calculate the magnitude of \(\mathbf{v}\) squared using the formula: \[\|\mathbf{v}\|^2 = (2)^2 + (-1)^2 = 4 + 1 = 5\]

Step 3: Calculate projection of \(\mathbf{u}\) on \(\mathbf{v}\)

Now, we calculate the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) using the formula: \[\text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \right) \mathbf{v}\]Substituting the values we calculated:\[\text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{1}{5} \right) \langle 2, -1 \rangle = \langle \frac{2}{5}, -\frac{1}{5} \rangle\]

Step 4: Calculate Orthogonal Vector \(\mathbf{u}_2\)

\(\mathbf{u}_2\) is the orthogonal component of \(\mathbf{u}\) relative to \(\mathbf{v}\). It can be calculated as: \[\mathbf{u}_2 = \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u} = \langle 1, 1 \rangle - \langle \frac{2}{5}, -\frac{1}{5} \rangle\]Subtracting the vectors:\[\mathbf{u}_2 = \langle 1 - \frac{2}{5}, 1 + \frac{1}{5} \rangle = \left\langle \frac{5}{5} - \frac{2}{5}, \frac{6}{5} \right\rangle = \left\langle \frac{3}{5}, \frac{6}{5} \right\rangle\]

Step 5: Resolve \(\mathbf{u}\) into \(\mathbf{u}_1\) and \(\mathbf{u}_2\)

Finally, resolve \(\mathbf{u}\) into two components: \(\mathbf{u}_1\) which is parallel to \(\mathbf{v}\) and is equal to the projection, and \(\mathbf{u}_2\) which is orthogonal to \(\mathbf{v}\):- \(\mathbf{u}_1 = \langle \frac{2}{5}, -\frac{1}{5} \rangle\)- \(\mathbf{u}_2 = \langle \frac{3}{5}, \frac{6}{5} \rangle\)

Key Concepts

Dot Product

The dot product is a fundamental operation used in vector algebra that helps us in many aspects of vector analysis. In the context of projecting one vector onto another, the dot product is the first step to understanding the angle and relationship between two vectors. Here, the dot product of vectors \( \mathbf{u} \) and \( \mathbf{v} \) is calculated using the formula:

• \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \)

For the vectors \( \mathbf{u} = \langle 1, 1 \rangle \) and \( \mathbf{v} = \langle 2, -1 \rangle \), the dot product is:

• \( \mathbf{u} \cdot \mathbf{v} = (1)(2) + (1)(-1) = 2 - 1 = 1 \)

This value is crucial as it tells us the degree of similarity or alignment between the vectors. The closer the dot product is to zero, the more orthogonal the vectors are to one another. Conversely, a larger magnitude indicates greater alignment.

Magnitude of a Vector

The magnitude of a vector, sometimes referred to as the length or norm, is a measure of how long the vector is. It is an essential concept when working with vectors as it normalizes vector interactions and assists in projecting vectors. The magnitude is calculated using the Euclidean formula:

• \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2} \)

For \( \mathbf{v} = \langle 2, -1 \rangle \), its magnitude squared is computed as:

• \( \| \mathbf{v} \|^2 = (2)^2 + (-1)^2 = 4 + 1 = 5 \)

The magnitude squared is used in the formula for vector projection to simplify calculations: \[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{v} \|^2} \right) \mathbf{v} \].Understanding the magnitude is key to interpreting the relationship between vectors when visualizing or computing projections.

Orthogonal Components

Orthogonal components in vector spaces refer to parts of vectors that are at right angles to each other. When a vector is resolved into components, such as \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), the concept of orthogonality comes into play. To achieve this, we first find the projection of one vector onto another to get the parallel component. The remainder, which cannot be projected onto the second vector, is the orthogonal component.To find the orthogonal component \( \mathbf{u}_2 \), we subtract the projection of \( \mathbf{u} \) onto \( \mathbf{v} \):

• \( \mathbf{u}_2 = \mathbf{u} - \text{proj}_{\mathbf{v}} \mathbf{u} \)

For example, calculating with the vectors given:

• \( \mathbf{u}_2 = \langle 1, 1 \rangle - \langle \frac{2}{5}, -\frac{1}{5} \rangle = \langle \frac{3}{5}, \frac{6}{5} \rangle \)

This orthogonal component \( \mathbf{u}_2 \) shows the part of \( \mathbf{u} \) that does not influence its alignment with \( \mathbf{v} \), highlighting the independence between \( \mathbf{u} \) and \( \mathbf{v} \) at right angles.