Problem 19 - Advanced Engineering Mathematics

Question

Verify that the vector $$ \mathbf{c}=\mathbf{b}-\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^{2}} \mathbf{a} $$ is orthogonal to the vector \(\mathbf{a}\).

Step-by-Step Solution

Answer

\( \mathbf{c} \cdot \mathbf{a} = 0 \), hence \( \mathbf{c} \) is orthogonal to \( \mathbf{a} \).

Step 1: Understand the Orthogonality Condition

To show that vector \( \mathbf{c} \) is orthogonal to \( \mathbf{a} \), we need to verify that their dot product is zero, i.e., \( \mathbf{c} \cdot \mathbf{a} = 0 \).

Step 2: Express Vector \( \mathbf{c} \)

Given \( \mathbf{c} = \mathbf{b} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^{2}} \mathbf{a} \), we rewrite it in terms of dot products and magnitudes.

Step 3: Compute the Dot Product \( \mathbf{c} \cdot \mathbf{a} \)

Calculate the dot product:\[ \mathbf{c} \cdot \mathbf{a} = \left( \mathbf{b} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^{2}} \mathbf{a} \right) \cdot \mathbf{a}\]Using the distributive property of dot products, this becomes:\[ \mathbf{b} \cdot \mathbf{a} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^{2}} (\mathbf{a} \cdot \mathbf{a})\]

Step 4: Simplify the Expression

Notice that \( \mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^{2} \), so the expression becomes:\[ \mathbf{b} \cdot \mathbf{a} - \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^{2}} \|\mathbf{a}\|^{2}\]This simplifies further to:\[ \mathbf{b} \cdot \mathbf{a} - (\mathbf{a} \cdot \mathbf{b}) = 0\]

Step 5: Conclusion

Since \( \mathbf{b} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} = 0 \), we have shown that \( \mathbf{c} \cdot \mathbf{a} = 0 \), confirming that \( \mathbf{c} \) is orthogonal to \( \mathbf{a} \).

Key Concepts

Dot Product

In vector algebra, the dot product is a fundamental operation that helps us determine how two vectors relate to each other geometrically. The dot product of two vectors, \( \mathbf{u} \) and \( \mathbf{v} \), is given by \( \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos \theta \), where \( \theta \) is the angle between the vectors.
The dot product is a scalar, not a vector, which means it only has magnitude, not direction. This operation is useful in finding the angle between vectors, checking for orthogonality, and projecting one vector onto another. If \( \mathbf{u} \cdot \mathbf{v} = 0 \), it indicates that the vectors are orthogonal (perpendicular).
Additionally, there is an algebraic way to compute the dot product using vector components: \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + \ldots + u_nv_n \). This approach involves multiplying each corresponding component of the vectors and summing the results.

Vector Algebra

Vector algebra is a branch of mathematics focused on vector quantities and the rules governing their manipulation and computation. Vectors are objects that have both magnitude and direction and can be represented as arrows in space. In vector algebra, operations such as addition, subtraction, and scalar multiplication are fundamental.
- **Addition and Subtraction**: To add or subtract vectors, we perform these operations component-wise. If \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), then \( \mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2) \) and \( \mathbf{a} - \mathbf{b} = (a_1 - b_1, a_2 - b_2) \).
- **Scalar Multiplication**: Multiplying a vector by a scalar changes its magnitude without altering its direction. If \( k \) is a scalar and \( \mathbf{a} = (a_1, a_2) \), then \( k\mathbf{a} = (ka_1, ka_2) \).
Vector algebra is essential in physics and engineering, particularly for analyzing quantities like force, velocity, and displacement. By employing the rules of vector algebra, we can simplify and resolve complex vector operations.

Orthogonal Vectors

Orthogonal vectors are an important concept in vector spaces. Two vectors are orthogonal if their dot product is zero. This means that the two vectors are perpendicular to each other, which can be visualized as forming a right angle in space.
Orthogonal vectors have unique properties that make them useful in various applications, such as simplifying systems of equations, optimizing solutions in linear algebra, and developing coordinate systems. For example, in the exercise involving vectors \( \mathbf{c} \) and \( \mathbf{a} \), the goal was to show that \( \mathbf{c} \) is orthogonal to \( \mathbf{a} \).
By finding that their dot product was zero, it was confirmed that the vectors are orthogonal. This concept is not only theoretical but also practical, as orthogonality is fundamental in designing algorithms, especially in fields like computer graphics and machine learning.