Problem 32 - University Calculus: Early Transcendentals

Question

A force \(\mathbf{F}=2 \mathbf{i}+\mathbf{j}-3 \mathbf{k}\) is applied to a spacecraft with velocity vector \(\mathbf{v}=3 \mathbf{i}-\mathbf{j} .\) Express \(\mathbf{F}\) as a sum of a vector parallel to \(\mathbf{v}\) and a vector orthogonal to \(\mathbf{v}\).

Step-by-Step Solution

Answer

\( \mathbf{F} = (\frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j}) + (\frac{1}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k}) \).

Step 1: Calculate the Dot Product of F and v

To start, calculate the dot product of vectors \( \mathbf{F} \) and \( \mathbf{v} \). This is given by \( \mathbf{F} \cdot \mathbf{v} = (2)(3) + (1)(-1) + (-3)(0) = 6 - 1 = 5 \).

Step 2: Calculate the Magnitude Squared of v

Next, find the magnitude squared of \( \mathbf{v} \). This is given by \( |\mathbf{v}|^2 = 3^2 + (-1)^2 = 9 + 1 = 10 \).

Step 3: Find Vector Component of F Parallel to v

The component of \( \mathbf{F} \) that is parallel to \( \mathbf{v} \) is found using the formula: \( \text{proj}_{\mathbf{v}} \mathbf{F} = \frac{\mathbf{F} \cdot \mathbf{v}}{|\mathbf{v}|^2} \cdot \mathbf{v} = \frac{5}{10} \cdot (3\mathbf{i} - \mathbf{j}) = \frac{1}{2}(3\mathbf{i} - \mathbf{j}) = \frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j} \).

Step 4: Find Vector Component of F Orthogonal to v

The orthogonal component is found by subtracting the parallel component from \( \mathbf{F} \): \( \mathbf{F}_{\perp} = \mathbf{F} - \text{proj}_{\mathbf{v}} \mathbf{F} = (2\mathbf{i} + \mathbf{j} - 3\mathbf{k}) - (\frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j}) = (2 - \frac{3}{2})\mathbf{i} + (1 + \frac{1}{2})\mathbf{j} - 3\mathbf{k} = \frac{1}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k} \).

Step 5: Conclusion: Express F as Sum of Both Components

Thus, \( \mathbf{F} \) can be written as a sum of a vector parallel and a vector orthogonal to \( \mathbf{v} \): \( \mathbf{F} = (\frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j}) + (\frac{1}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k}) \).

Key Concepts

Understanding the Dot Product

The dot product is a fundamental operation in vector algebra. It involves multiplying two vectors, resulting in a scalar value. This operation is crucial as it helps determine the relationship between two vectors. To compute the dot product, such as between vectors \( \mathbf{F} \) and \( \mathbf{v} \):

• Multiply corresponding components.
• Add the results together.

The formula for the dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is:\[\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z\]This calculation can tell us how aligned the vectors are. If the dot product is positive, the vectors point in roughly the same direction. A dot product of zero indicates that the vectors are orthogonal, meaning they are perpendicular to each other, which brings us to our next topic.

Exploring Orthogonal Vectors

Orthogonal vectors are vectors that meet at a right angle. This is a special case in vector space where the dot product of the vectors equals zero. If you calculate the dot product and it results in zero:

• The vectors form a perpendicular angle with each other.
• Orthogonal vectors imply no parallel component.

Consider two vectors, \( \mathbf{A} \) and \( \mathbf{B} \). If they are orthogonal, then:\[\mathbf{A} \cdot \mathbf{B} = 0\]This property is useful in vector decomposition, where you often need a vector to be expressed in terms of a component that is orthogonal to another vector, just like the component of \( \mathbf{F} \) that is orthogonal to \( \mathbf{v} \). Such decomposition often facilitates simplifications in engineering and physics applications.

Understanding Vector Projection

Vector projection involves expressing one vector onto another, providing insight into how much of one vector lies along the direction of another. This is particularly useful when decomposing forces. The projection of vector \( \mathbf{F} \) onto vector \( \mathbf{v} \) is calculated using the formula:\[\text{proj}_{\mathbf{v}} \mathbf{F} = \frac{\mathbf{F} \cdot \mathbf{v}}{|\mathbf{v}|^2} \times \mathbf{v}\]Here’s how you do it:

• Calculate the dot product \( \mathbf{F} \cdot \mathbf{v} \).
• Find the magnitude squared of \( \mathbf{v} \), which is \( |\mathbf{v}|^2 \).
• Divide the dot product by the magnitude squared, multiplying by the vector \( \mathbf{v} \).

This process helps in splitting a vector into two components: one parallel and another orthogonal to a given vector. This is essential in many real-world applications, such as resolving forces in mechanics, to simplify problem-solving.