Problem 29 - 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 \(\mathbf{v} .\) $$ \mathbf{u}=\langle- 2,4\rangle, \quad \mathbf{v}=\langle 1,1\rangle $$

Step-by-Step Solution

Answer

The projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is \( \langle 1, 1 \rangle \). \( \mathbf{u} \) resolves into \( \mathbf{u}_1 = \langle 1, 1 \rangle \) and \( \mathbf{u}_2 = \langle -3, 3 \rangle \).

Step 1: Calculate the Dot Product of u and v

First, calculate the dot product of \( \mathbf{u} \) and \( \mathbf{v} \). The dot product is given by \( \mathbf{u} \cdot \mathbf{v} = (-2)(1) + (4)(1) = -2 + 4 = 2 \).

Step 2: Calculate the Dot Product of v and v

Next, calculate the dot product of \( \mathbf{v} \) with itself. That is \( \mathbf{v} \cdot \mathbf{v} = (1)(1) + (1)(1) = 1 + 1 = 2 \).

Step 3: Calculate the Projection of u onto v

Now, use the formula for the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), which is \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \). Substitute the values we calculated: \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{2}{2} \langle 1, 1 \rangle = 1 \langle 1, 1 \rangle = \langle 1, 1 \rangle \).

Step 4: Identify u1 and u2

\( \mathbf{u}_1 \) is the projection of \( \mathbf{u} \) onto \( \mathbf{v} \), which is \( \langle 1, 1 \rangle \). \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \) and is calculated as \( \mathbf{u} - \mathbf{u}_1 \). Thus \( \mathbf{u}_2 = \langle -2, 4 \rangle - \langle 1, 1 \rangle = \langle -3, 3 \rangle \).

Step 5: Verify the Orthogonality of u2 to v

Finally, verify that \( \mathbf{u}_2 \) is orthogonal to \( \mathbf{v} \) by checking that their dot product is zero: \( \mathbf{u}_2 \cdot \mathbf{v} = (-3)(1) + (3)(1) = -3 + 3 = 0 \). As the result is zero, \( \mathbf{u}_2 \) is indeed orthogonal to \( \mathbf{v} \).

Key Concepts

Dot Product

The dot product is a crucial concept in vector mathematics. It involves multiplying two vectors and summing the result. You find this operation in formulas related to vector projections and sometimes in determining angles between vectors.
The dot product of two vectors, \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), is calculated as:

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

In our exercise, the dot product of \( \mathbf{u} = \langle -2, 4 \rangle \) and \( \mathbf{v} = \langle 1, 1 \rangle \) was computed as

• \((-2)(1) + (4)(1) = -2 + 4 = 2\).

The dot product resulted in 2, which is a key part of finding the projection and understanding orthogonality.

Orthogonal Vectors

Orthogonality in vectors refers to the scenario when two vectors are at right angles (90 degrees) to each other. This relationship results in their dot product being zero. When you identify a vector as orthogonal to another, it means they do not contribute to one another in terms of direction.
To check if two vectors, say \( \mathbf{a} \) and \( \mathbf{b} \), are orthogonal, calculate their dot product:

• If \( \mathbf{a} \cdot \mathbf{b} = 0 \), they are orthogonal.

In the resolution of the vector \( \mathbf{u} \) into components \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), the orthogonal vector \( \mathbf{u}_2 \) resulted in a dot product of zero with \( \mathbf{v} \), confirming its perpendicular nature.

Vector Components

Breaking down a vector into components involves splitting it into parts that add up to the original vector. This is a common and useful technique in physics and engineering, allowing the simplification of complex vector operations.
In this exercise, the vector \( \mathbf{u} = \langle -2, 4 \rangle \) was split into two components:

• \( \mathbf{u}_1 \), which is parallel to \( \mathbf{v} \), and
• \( \mathbf{u}_2 \), which is orthogonal to \( \mathbf{v} \).

The process involved finding the projection \( \mathbf{u}_1 = \langle 1, 1 \rangle \) and then subtracting it from \( \mathbf{u} \) to find \( \mathbf{u}_2 = \langle -3, 3 \rangle \). These components effectively describe the influence of \( \mathbf{u} \) in the directions parallel and perpendicular to \( \mathbf{v} \).

Scalar Projection

The scalar projection of a vector \( \mathbf{u} \) onto another vector \( \mathbf{v} \) represents the "shadow" or "component" of \( \mathbf{u} \) that points in the direction of \( \mathbf{v} \). This is vital in applications where only the effective part of one vector in the direction of another matters.
The formula for scalar projection is expressed as:

• \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|} \)

However, by using vector projection, you would employ the same formula result but keep \( \mathbf{v} \) directional influence intact.

• Thus, \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \).

The aforementioned formula was used in the calculation, ultimately identifying the parallel component as \( \langle 1, 1 \rangle \). This concept highlights the significance of vector directions in projections.