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Chapter 22: Problem 12

heitsvektor \(\widehat{\boldsymbol{n}}=\frac{1}{\sqrt{6}}(1,-1,2)^{T}\) und zerlegen Sie den Vektor \(\boldsymbol{v}=(1,1,1)^{T}\) in zwei orthogonale Komponenten \(\boldsymbol{v}^{\prime}\) und \(\boldsymbol{v}^{\prime \prime}\), wobei \(\boldsymbol{v}^{\prime}\) zu \(\widehat{n}\) parallel ist.

Short Answer

Expert verified
Answer: The two orthogonal components of the vector (1,1,1) with respect to the unit normal vector (1/sqrt(6), -1/sqrt(6), 2/sqrt(6)) are: 1) The parallel component: (1/3, -1/3, 2/3) 2) The orthogonal component: (2/3, 4/3, 1/3)

Step by step solution

01

Find the parallel component

To find the component of \(\boldsymbol{v}\) parallel to \(\widehat{\boldsymbol{n}}\), we can use the projection formula: \(\boldsymbol{v}^{\prime} = (\boldsymbol{v} \cdot \widehat{\boldsymbol{n}}) \widehat{\boldsymbol{n}}\) First, calculate the dot product of \(\boldsymbol{v}\) and \(\widehat{\boldsymbol{n}}\): \(\boldsymbol{v} \cdot \widehat{\boldsymbol{n}} = (1,1,1) \cdot \frac{1}{\sqrt{6}}(1,-1,2) = \frac{1}{\sqrt{6}}(1 - 1 + 2) = \frac{2}{\sqrt{6}}\) Now, multiply this result by \(\widehat{\boldsymbol{n}}\) to get the parallel component: \(\boldsymbol{v}^{\prime} = \frac{2}{\sqrt{6}}\frac{1}{\sqrt{6}}(1,-1,2)^{T} = \frac{1}{3}(1,-1,2)^{T}\)
02

Find the orthogonal component

To find the component of \(\boldsymbol{v}\) orthogonal to \(\widehat{\boldsymbol{n}}\), we can subtract the parallel component \(\boldsymbol{v}^{\prime}\) from the original vector \(\boldsymbol{v}\): \(\boldsymbol{v}^{\prime\prime} = \boldsymbol{v} - \boldsymbol{v}^{\prime} = (1,1,1)^{T} - \frac{1}{3}(1,-1,2)^{T}\) Calculate the difference to get the orthogonal component: \(\boldsymbol{v}^{\prime\prime} = \left(1 - \frac{1}{3}, 1 + \frac{1}{3}, 1 - \frac{2}{3}\right)^{T} = \left(\frac{2}{3}, \frac{4}{3}, \frac{1}{3}\right)^{T}\)
03

Verify the orthogonality and parallelism

To make sure our solution is correct, we need to verify that \(\boldsymbol{v}^{\prime}\) is parallel to \(\widehat{\boldsymbol{n}}\), and that \(\boldsymbol{v}^{\prime}\) and \(\boldsymbol{v}^{\prime\prime}\) are orthogonal to each other. First, check the parallelism. We found \(\boldsymbol{v}^{\prime} = \frac{1}{3}(1,-1,2)^{T}\). Since this is just a scalar multiple of \(\widehat{\boldsymbol{n}}\), they are parallel. Next, check the orthogonality by finding the dot product of \(\boldsymbol{v}^{\prime}\) and \(\boldsymbol{v}^{\prime\prime}\): \(\boldsymbol{v}^{\prime} \cdot \boldsymbol{v}^{\prime\prime} = \left(\frac{1}{3}, -\frac{1}{3}, \frac{2}{3}\right) \cdot \left(\frac{2}{3}, \frac{4}{3}, \frac{1}{3}\right) = \frac{1}{9}(2 - 4 + 2) = 0\) Since their dot product is 0, \(\boldsymbol{v}^{\prime}\) and \(\boldsymbol{v}^{\prime\prime}\) are orthogonal. Thus, we have successfully decomposed the vector \(\boldsymbol{v}\) into two orthogonal components: \(\boldsymbol{v}^{\prime} = \frac{1}{3}(1,-1,2)^{T}\) (parallel to \(\widehat{\boldsymbol{n}}\)) \(\boldsymbol{v}^{\prime\prime} = \left(\frac{2}{3}, \frac{4}{3}, \frac{1}{3}\right)^{T}\) (orthogonal to \(\widehat{\boldsymbol{n}}\))

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orthogonal Components
When discussing vectors, "orthogonal components" refer to parts of a vector that are perpendicular to each other. This concept comes into play when we break down a vector into specific sections that are related to another vector, like the process of decomposing the vector \( \boldsymbol{v} \) into \( \boldsymbol{v}^{\prime} \) and \( \boldsymbol{v}^{\prime\prime} \). Here, \( \boldsymbol{v} = (1,1,1)^T \) is split into two parts:
  • \( \boldsymbol{v}^{\prime} \) which is parallel to the reference direction given by \( \widehat{\boldsymbol{n}} \).
  • \( \boldsymbol{v}^{\prime\prime} \) which is orthogonal (or perpendicular) to \( \widehat{\boldsymbol{n}} \).
These two components combined still give us the original vector, as one projects the vector onto the direction of \( \widehat{\boldsymbol{n}} \) while the other covers the "left over" part that remains perpendicular.
Projection Formula
The 'projection formula' is a key tool in vector decomposition used to find the component of one vector along the direction of another. In our example, to determine \( \boldsymbol{v}^{\prime} \) (the parallel component), we use the formula:

\[ \boldsymbol{v}^{\prime} = (\boldsymbol{v} \cdot \widehat{\boldsymbol{n}}) \widehat{\boldsymbol{n}} \]
This formula tells us to first find the dot product of \( \boldsymbol{v} \) and \( \widehat{\boldsymbol{n}} \). This numerical result represents how much of \( \boldsymbol{v} \) aligns with \( \widehat{\boldsymbol{n}} \).
  • Multiply this scalar with the unit direction vector \( \widehat{\boldsymbol{n}} \) to gain the exact directional component \( \boldsymbol{v}^{\prime} \).
Thus, the projection formula helps isolate the portion of the vector that points along a specific direction.
Parallel and Orthogonal Vectors
In vector algebra, two vectors are parallel if they point in the same or exactly opposite directions. This is checked if one vector is a scalar multiple of the other. In our case, \( \boldsymbol{v}^{\prime} = \frac{1}{3}(1,-1,2)^T \) is a scalar multiple of \( \widehat{\boldsymbol{n}} \), confirming that these vectors are parallel.

On the contrary, vectors are orthogonal if their dot product equals zero. When solving our exercise, we confirmed that \( \boldsymbol{v}^{\prime} \) and \( \boldsymbol{v}^{\prime\prime} \) are orthogonal by calculating:\[ \boldsymbol{v}^{\prime} \cdot \boldsymbol{v}^{\prime\prime} = 0 \]
This calculation ensures that the vectors are indeed perpendicular to each other, a crucial property when resolving a vector into components.
Dot Product Calculation
The 'dot product' is a fundamental operation in vector mathematics that gives a scalar value. It serves multiple purposes, such as determining the angle between vectors or projecting one vector onto another. In our exercise, the calculation \( \boldsymbol{v} \cdot \widehat{\boldsymbol{n}} \) was essential. This yielded:

\[ \boldsymbol{v} \cdot \widehat{\boldsymbol{n}} = \frac{2}{\sqrt{6}} \]
Why is this important? Because it indicates how much of \( \boldsymbol{v} \) lies in the direction of \( \widehat{\boldsymbol{n}} \). The result is used in the projection formula to find \( \boldsymbol{v}^{\prime} \). This dot product also enables us to verify the orthogonality between \( \boldsymbol{v}^{\prime} \) and \( \boldsymbol{v}^{\prime\prime} \), illustrating the broad application of this operation in vector analysis.

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Most popular questions from this chapter

Warum sind die Symmetrie und die Antisymme-trie eines kartesischen Tensors \(t_{i j}\) zweiter Stufe koordinateninvariante Eigenschaften.

Warum verschwindet das doppelt-verj眉ngendeProdukt \(s^{i j} a_{i j}\), wenn der Tensor \(s^{i j}\) symmetrisch und der Tensor \(a_{i j}\) alternierend ist? Was ist \(s^{i j} a_{\vec{i}}\) ?

sei \(B\) eine weitere Basis mit $$ b_{1}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right), \quad b_{2}=\left(\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right), \quad b_{3}=\left(\begin{array}{r} -1 \\ -1 \\ 1 \end{array}\right) $$ Berechnen Sie die Vektoren \(\boldsymbol{b}^{1}, \boldsymbol{b}^{2}\) und \(\boldsymbol{b}^{3}\) der Dualbasis \(B^{*}\), den kovarianten Metriktensor \(g_{i j}\), den kontravarianten Metriktensor \(g^{i j}\) durch Invertieren der Matrix \(\left(g_{i j}\right)\) und 眉berpr眉fen Sie die Gleichung (22.8).

Sind \(e_{i}\) die Komponenten eines Einheitsvektorsund \(v_{i}\) jene eines beliebigen Vektors des \(\mathbb{R}^{3}\), so gilt die Identit盲t $$ v_{i}=v_{j} e_{j} e_{i}+\varepsilon_{i j k} e_{j} \varepsilon_{k l m} v_{l} e_{m} $$ Begr眉nden Sie diese Identit盲t. Was bedeutet sie in Vektorform?

Berechnen Sie im \(\mathbb{R}^{3}\) die nachstehend angef眉hrtenAusdr眉cke, die alle das Kronecker-Delta enthalten: $$ \begin{array}{llll} \delta_{i i}, & \delta_{i j} \delta_{j k}, & \delta_{i j} \delta_{i i}, & \delta_{i j} \delta_{j k} \delta_{k i} . \end{array} $$
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