Problem 30 - Thomas Calculus

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

\( \mathbf{F} = \frac{3}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} + \frac{1}{2} \mathbf{i} + \frac{3}{2} \mathbf{j} - 3 \mathbf{k} \), with parallel and orthogonal components.

Step 1: Calculate the dot product of F and v

The dot product of two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \). Here, we calculate the dot product of \( \mathbf{F} = 2 \mathbf{i} + \mathbf{j} - 3 \mathbf{k} \) and \( \mathbf{v} = 3 \mathbf{i} - \mathbf{j} \):\[ \mathbf{F} \cdot \mathbf{v} = (2)(3) + (1)(-1) + (-3)(0) = 6 - 1 + 0 = 5. \]

Step 2: Compute the magnitude of v squared

The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is given by \( \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \). However, for projection purposes, we need \( \|\mathbf{v}\|^2 = a^2 + b^2 + c^2 \). For \( \mathbf{v} = 3 \mathbf{i} - \mathbf{j} \), we have:\[ \|\mathbf{v}\|^2 = 3^2 + (-1)^2 = 9 + 1 = 10. \]

Step 3: Find the projection of F onto v

The projection of \( \mathbf{F} \) onto \( \mathbf{v} \) is given by the formula:\[ \text{proj}_{\mathbf{v}} \mathbf{F} = \frac{\mathbf{F} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v}. \]Substituting the known values:\[ \text{proj}_{\mathbf{v}} \mathbf{F} = \frac{5}{10}(3\mathbf{i} - \mathbf{j}) = \frac{1}{2}(3\mathbf{i} - \mathbf{j}) = \frac{3}{2} \mathbf{i} - \frac{1}{2} \mathbf{j}. \]

Step 4: Determine the orthogonal component of F

The vector \( \mathbf{F} \) can be expressed as the sum of two components: one parallel to \( \mathbf{v} \) and one orthogonal to \( \mathbf{v} \). The orthogonal component is:\[ \mathbf{F}_{\perp} = \mathbf{F} - \text{proj}_{\mathbf{v}} \mathbf{F}. \]Substituting the components:\[ \mathbf{F}_{\perp} = (2\mathbf{i} + \mathbf{j} - 3\mathbf{k}) - \left(\frac{3}{2} \mathbf{i} - \frac{1}{2} \mathbf{j}\right). \]Simplifying, we get:\[ \mathbf{F}_{\perp} = \left(2 - \frac{3}{2}\right)\mathbf{i} + \left(1 + \frac{1}{2}\right)\mathbf{j} - 3\mathbf{k} = \frac{1}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k}. \]

Step 5: Express F as the sum of the parallel and orthogonal vectors

Now that we have the parallel and orthogonal components:\[ \mathbf{F} = \text{proj}_{\mathbf{v}} \mathbf{F} + \mathbf{F}_{\perp}. \]Substituting the known components:\[ \mathbf{F} = \left( \frac{3}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} \right) + \left( \frac{1}{2} \mathbf{i} + \frac{3}{2} \mathbf{j} - 3\mathbf{k} \right). \]

Key Concepts

Dot Product

The dot product is a fundamental operation between two vectors that results in a scalar. This scalar is a measure of how much one vector extends in the direction of another. To calculate the dot product, multiply the corresponding components of the two vectors and add them all together. Mathematically, the dot product of vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\) is given by:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]In the given problem, to find the dot product of \(\mathbf{F} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}\) and \(\mathbf{v} = 3\mathbf{i} - \mathbf{j}\), only the \(\mathbf{i}\) and \(\mathbf{j}\) components contribute, because \(\mathbf{v}\) has no \(\mathbf{k}\) component:

• \(2 \times 3 = 6\)
• \(1 \times (-1) = -1\)
• \(-3 \times 0 = 0\)

Adding these up gives \(\mathbf{F} \cdot \mathbf{v} = 5\). This value shows that there's some alignment between \(\mathbf{F}\) and \(\mathbf{v}\). By understanding the dot product, we grasp how these vectors interact directionally.

Vector Projection

Vector projection takes one vector and effectively "projects" it onto another, showing how much of one vector lies in the direction of the other. You calculate the vector projection of \(\mathbf{F}\) onto \(\mathbf{v}\) using the formula:\[ \text{proj}_{\mathbf{v}} \mathbf{F} = \frac{\mathbf{F} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} \]In practical terms, this formula takes the dot product and scales it by the magnitude squared of \(\mathbf{v}\). For our example, \(\mathbf{F}\cdot \mathbf{v} = 5\) and \(\|\mathbf{v}\|^2 = 10\). Thus,\[ \text{proj}_{\mathbf{v}} \mathbf{F} = \frac{5}{10}(3\mathbf{i} - \mathbf{j}) = \frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j} \]This projection tells us the component of \(\mathbf{F}\) that is in the same direction as \(\mathbf{v}\). It's useful in problems where we want to decompose a force or vector into parts that lie along some specific directions.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at a right angle, making their dot product zero. This property allows for easy separation of forces and other vector quantities into components that do not influence each other.In this context, the orthogonal component \(\mathbf{F}_{\perp}\) is what remains when \(\mathbf{F}\) is subtracted by its projection onto \(\mathbf{v}\). It is calculated as:\[ \mathbf{F}_{\perp} = \mathbf{F} - \text{proj}_{\mathbf{v}} \mathbf{F} \]For our example:

• \(\mathbf{F} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}\)
• \(\text{proj}_{\mathbf{v}} \mathbf{F} = \frac{3}{2}\mathbf{i} - \frac{1}{2}\mathbf{j}\)

Subtracting these, we get:\[ \mathbf{F}_{\perp} = \left(2 - \frac{3}{2}\right)\mathbf{i} + \left(1 + \frac{1}{2}\right)\mathbf{j} - 3\mathbf{k} \]This results in \(\mathbf{F}_{\perp} = \frac{1}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k}\). This decomposition ensures that the effects of the parallel and orthogonal components of \(\mathbf{F}\) do not interfere with one another.

Magnitude of a Vector

The magnitude of a vector, often viewed as the "length" or "size" of the vector, is a crucial component in understanding vector behavior and relationships.To find the magnitude of a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), use the formula:\[ \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \]However, in projection or other vector arrangements, knowing \(\|\mathbf{v}\|^2\) is often more useful. Considering the current problem, \(\mathbf{v} = 3\mathbf{i} - \mathbf{j}\) has:

• \(a = 3\)
• \(b = -1\)
• \(c = 0\)

This yields\[ \|\mathbf{v}\|^2 = 3^2 + (-1)^2 = 9 + 1 = 10 \]The squared magnitude is pivotal in the calculations for projection and helps isolate directional components effectively.