Problem 33 - Essential Calculus Early Transcendentals

Question

Show that the vector orth \(_{\mathrm{a}} \mathbf{b}=\mathbf{b}-\) proja \(_{\mathrm{a}} \mathbf{b}\) is orthogonal to a. (It is called an orthogonal projection of \(\mathbf{b} . )\)

Step-by-Step Solution

Answer

The vector \( \text{orth}_a \mathbf{b} \) is orthogonal to \( \mathbf{a} \) because their dot product is zero.

Step 1: Write the formula for orthogonal and projection

We start with the two given components of the problem. The vector \( \text{proj}_a \mathbf{b} \) (projection of \( \mathbf{b} \) onto \( \mathbf{a} \)) is given by the formula: \[ \text{proj}_a \mathbf{b} = \frac{\mathbf{b} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a} \]The orthogonal component of \( \mathbf{b} \) to \( \mathbf{a} \) is then computed as: \[ \text{orth}_a \mathbf{b} = \mathbf{b} - \text{proj}_a \mathbf{b} \]

Step 2: Use dot product properties

To show that \( \text{orth}_a \mathbf{b} \) is orthogonal to \( \mathbf{a} \), we need to show that their dot product is zero, i.e.,\[ \text{orth}_a \mathbf{b} \cdot \mathbf{a} = 0 \]This is based on the property that two vectors are orthogonal if and only if their dot product is zero.

Step 3: Compute \( \text{orth}_a \mathbf{b} \cdot \mathbf{a} \)

Substitute \( \text{orth}_a \mathbf{b} = \mathbf{b} - \text{proj}_a \mathbf{b} \) into the dot product:\[(\mathbf{b} - \text{proj}_a \mathbf{b}) \cdot \mathbf{a} = \mathbf{b} \cdot \mathbf{a} - (\frac{\mathbf{b} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a}) \cdot \mathbf{a}\]

Step 4: Simplify the expression

Simplify the right-hand side of the equation:\[ \mathbf{b} \cdot \mathbf{a} - \frac{\mathbf{b} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} (\mathbf{a} \cdot \mathbf{a}) = \mathbf{b} \cdot \mathbf{a} - (\mathbf{b} \cdot \mathbf{a})\]This simplifies further to:\[ 0\]

Step 5: Conclude the proof

Since \((\mathbf{b} - \text{proj}_a \mathbf{b}) \cdot \mathbf{a} = 0 \), it follows that \( \text{orth}_a \mathbf{b} \) is orthogonal to \( \mathbf{a} \) by definition.

Key Concepts

Dot Product

The dot product is a fundamental operation when working with vectors. It's a way of multiplying two vectors together to get a single number, which is known as a scalar. This can be calculated by multiplying corresponding components of the two vectors and summing the results. For example, for vectors \( \mathbf{u} = (x_1, y_1, z_1) \) and \( \mathbf{v} = (x_2, y_2, z_2) \), the dot product \( \mathbf{u} \cdot \mathbf{v} \) is computed as:

• \( x_1x_2 + y_1y_2 + z_1z_2 \)

This concept is crucial as it provides a measure of how much one vector extends in the direction of another vector. A key property of the dot product is that it can tell us when two vectors are orthogonal, which means they form a right angle with each other. This is because the dot product of two orthogonal vectors is zero. This property is used to determine the orthogonality of the vector \( \text{orth}_a \mathbf{b} \) in our original exercise. Calculating the dot product and finding it to be zero confirms that the vectors are orthogonal.

Vector Projection

Vector projection is a way of projecting one vector onto another. It allows us to decompose a vector into two components: one that is parallel to the vector onto which we are projecting, and one that is orthogonal. The formula for the projection of vector \( \mathbf{b} \) onto vector \( \mathbf{a} \), denoted as \( \text{proj}_a \mathbf{b} \), is:

• \[ \text{proj}_a \mathbf{b} = \frac{\mathbf{b} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a} \]

This formula essentially scales the vector \( \mathbf{a} \) by how much \( \mathbf{b} \) "aligns" with it, which is calculated using the dot product. The projection answers the question, "How much of \( \mathbf{b} \) lies in the direction of \( \mathbf{a} \)?"
In the context of the original exercise, vector projection was used to identify the parallel component of \( \mathbf{b} \) relative to \( \mathbf{a} \). Subtracting this projection from \( \mathbf{b} \) gives us \( \text{orth}_a \mathbf{b} \), the component orthogonal to \( \mathbf{a} \). This makes it a vital step in determining orthogonal projections.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at right angles, forming an angle of 90 degrees with each other. This geometric property is characterized by their dot product being zero. When two vectors are orthogonal, it implies that they do not influence each other in their respective directions.
In our exercise, we were tasked with proving that the orthogonal component of \( \mathbf{b} \) relative to \( \mathbf{a} \), \( \text{orth}_a \mathbf{b} \), is orthogonal to \( \mathbf{a} \). This is established by demonstrating that the dot product of \( \text{orth}_a \mathbf{b} \) with \( \mathbf{a} \) equals zero:

• \( \text{orth}_a \mathbf{b} \cdot \mathbf{a} = 0 \)

Understanding the notion of orthogonal vectors is crucial in vector mathematics as it pertains to various applications such as vector decompositions, orthogonal projections, and even in scenarios like computer graphics where such perpendicularly aligned vectors are used in rendering.