Problem 28 - Algebra and Trigonometry

Question

23-28 (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

(a) Projection: \( \left\langle \frac{2}{5}, -\frac{1}{5} \right\rangle \). (b) \( \mathbf{u}_1 = \left\langle \frac{2}{5}, -\frac{1}{5} \right\rangle \), \( \mathbf{u}_2 = \left\langle \frac{3}{5}, \frac{6}{5} \right\rangle \).

Step 1: Calculate the Dot Product

The dot product of vectors \( \mathbf{u} \) and \( \mathbf{v} \) is calculated by multiplying their corresponding components and adding the results:\[\mathbf{u} \cdot \mathbf{v} = (1)(2) + (1)(-1) = 2 - 1 = 1.\]

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

The magnitude of a vector \( \mathbf{v} = \langle a, b \rangle \) is calculated using the formula:\[\| \mathbf{v} \| = \sqrt{a^2 + b^2}.\]For \( \mathbf{v} = \langle 2, -1 \rangle \):\[\| \mathbf{v} \| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}.\]

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

The formula for the projection of \( \mathbf{u} \) on \( \mathbf{v} \) is given by:\[\text{proj}_\mathbf{v} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \right) \mathbf{v}.\]First, calculate \( \mathbf{v} \cdot \mathbf{v} \):\[\mathbf{v} \cdot \mathbf{v} = 2^2 + (-1)^2 = 4 + 1 = 5.\]Now, calculate the projection:\[\text{proj}_\mathbf{v} \mathbf{u} = \left( \frac{1}{5} \right) \langle 2, -1 \rangle = \left\langle \frac{2}{5}, -\frac{1}{5} \right\rangle.\]

Step 4: Resolve \( \mathbf{u} \) into \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \)

To resolve \( \mathbf{u} \) into \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), where \( \mathbf{u}_1 \) is the projection of \( \mathbf{u} \) on \( \mathbf{v} \) and \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \):\( \mathbf{u}_1 = \text{proj}_\mathbf{v} \mathbf{u} = \left\langle \frac{2}{5}, -\frac{1}{5} \right\rangle \).\( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \).Calculate \( \mathbf{u}_2 \):\[\mathbf{u}_2 = \langle 1, 1 \rangle - \left\langle \frac{2}{5}, -\frac{1}{5} \right\rangle = \left\langle 1 - \frac{2}{5}, 1 + \frac{1}{5} \right\rangle = \left\langle \frac{3}{5}, \frac{6}{5} \right\rangle.\]

Key Concepts

Dot Product

The dot product is a fundamental operation when working with vectors. It helps in finding the angle between two vectors or even determining if they are orthogonal. To calculate the dot product of two vectors, such as \( \mathbf{u} \) and \( \mathbf{v} \), you multiply each of their corresponding components and add those products together. For example, given \( \mathbf{u} = \langle 1, 1 \rangle \) and \( \mathbf{v} = \langle 2, -1 \rangle \), the dot product \( \mathbf{u} \cdot \mathbf{v} \) is:

• Multiply the x-components: \( 1 \times 2 = 2 \).
• Multiply the y-components: \( 1 \times -1 = -1 \).
• Add the products: \( 2 + (-1) = 1 \).

Hence, the dot product \( \mathbf{u} \cdot \mathbf{v} = 1 \). It's an indicator of how much one vector extends in the direction of another. If the result is zero, it means the vectors are orthogonal or perpendicular to each other.

Magnitude of a Vector

The magnitude of a vector is essentially its "length" or "size," and it's calculated using the Pythagorean theorem. For a vector \( \mathbf{v} = \langle a, b \rangle \), the magnitude \( \| \mathbf{v} \| \) is determined by:

• Squaring each component: \( a^2 \) and \( b^2 \).
• Adding the squares: \( a^2 + b^2 \).
• Taking the square root of the sum: \( \sqrt{a^2 + b^2} \).

For our vector \( \mathbf{v} = \langle 2, -1 \rangle \):

• Square the x-component: \( 2^2 = 4 \).
• Square the y-component: \( (-1)^2 = 1 \).
• Add the squares: \( 4 + 1 = 5 \).
• Take the square root: \( \sqrt{5} \).

The magnitude \( \| \mathbf{v} \| = \sqrt{5} \) gives us a geometric sense of the length of \( \mathbf{v} \).

Orthogonal Vectors

Orthogonal vectors are vectors that are perpendicular to each other. Two vectors are orthogonal if their dot product is zero. This property is particularly useful in various applications such as finding components of a vector that are perpendicular to a given direction.
When resolving a vector \( \mathbf{u} \) into a component that is orthogonal to another vector \( \mathbf{v} \), you can subtract the projection of \( \mathbf{u} \) on \( \mathbf{v} \) from \( \mathbf{u} \). The remaining vector, known as \( \mathbf{u}_2 \), is the orthogonal component:

• Calculate the projection \( \text{proj}_\mathbf{v} \mathbf{u} \).
• Subtract this from \( \mathbf{u} \) to find \( \mathbf{u}_2 \) which is orthogonal.

In our example, with \( \mathbf{u} = \langle 1, 1 \rangle \) and the vector projection being \( \left\langle \frac{2}{5}, -\frac{1}{5} \right\rangle \), the orthogonal component \( \mathbf{u}_2 \) is calculated as \( \langle 1, 1 \rangle - \left\langle \frac{2}{5}, -\frac{1}{5} \right\rangle = \left\langle \frac{3}{5}, \frac{6}{5} \right\rangle \). This concept is crucial for vector decomposition in vector space analysis.