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Chapter 5: Problem 4

Gegeben sei eine Ebene \(E=v+\mathbb{R} w_{1}+\mathbb{R} w_{2} \subset \mathbb{R}^{3}\). Zeige: Setzt man \(a:=w_{1} \times w_{2}\) und \(b:=\langle v, a\rangle\), so gilt $$ E=\left\\{x \in \mathbb{R}^{3}:\langle x, a\rangle=b\right\\} $$

Short Answer

Expert verified
For any point \(x\) on the plane \(E = v + \mathbb{R} w1 + \mathbb{R} w2\), it follows that \(\langle x, a \rangle = b\), thus showing the equivalence of the two definitions of the plane.

Step by step solution

01

Choose an arbitrary point from plane E

Choose an arbitrary point \(x\) in plane \(E = v + \mathbb{R} w1 + \mathbb{R} w2\). This point can be expressed as \(x = v + r1 w1 + r2 w2\) for some \(r1, r2 \in \mathbb{R}\).
02

Compute the dot product

We define vector \(a\) as the cross product of \(w1\) and \(w2\), and scalar \(b\) as the dot product of \(v\) and \(a\). Find the dot product of \(x\) and \(a\) i.e., \(\langle x, a \rangle\). Substitute the expression for \(x\) from step 1 to get: \(\langle v + r1 w1 + r2 w2, a \rangle\). By using properties of dot product, this simplifies to \(\langle v, a \rangle + r1 \langle w1, a \rangle + r2 \langle w2, a \rangle\).
03

Simplify the dot product

Notice that \(a = w1 \times w2\). Because the dot product of a vector with the cross product of two vectors is zero, it follows that: \(\langle w1, a \rangle = \langle w2, a \rangle = 0\). Thus the entire expression simplifies to \(\langle v, a \rangle\), which by definition is \(b\). We have hence shown that \(\langle x, a \rangle = b\) for any point \(x\) on the plane.

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Key Concepts

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

Cross Product
The cross product is a key operation in vector algebra, particularly helpful in three-dimensional space. It takes two vectors and returns another vector that is perpendicular to the plane containing the original vectors. This is quite unique compared to many other operations in vector spaces, such as the dot product, which results in a scalar.

Given two vectors \( \mathbf{w_1} \) and \( \mathbf{w_2} \), their cross product \( \mathbf{w_1} \times \mathbf{w_2} \) is defined as:\[\mathbf{w_1} \times \mathbf{w_2} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\mathbf{w_{1x}} & \mathbf{w_{1y}} & \mathbf{w_{1z}} \\mathbf{w_{2x}} & \mathbf{w_{2y}} & \mathbf{w_{2z}} \end{vmatrix}\]This determinant gives a vector that is perpendicular to both \( \mathbf{w_1} \) and \( \mathbf{w_2} \).

The magnitude of this resulting vector is equal to the area of the parallelogram formed by the original two vectors. This property is useful in many applications, including computer graphics, physics, and engineering.
Dot Product
The dot product, also known as the scalar product, is another fundamental operation in vector spaces. Unlike the cross product, the dot product of two vectors results in a scalar. This operation essentially measures how much of one vector goes in the direction of another, giving insight into the angle between the two vectors.

For vectors \( \mathbf{v} \) and \( \mathbf{a} \), where \( \mathbf{a} \) is the cross product discussed previously, the dot product is defined as:\[\langle \mathbf{v}, \mathbf{a} \rangle = v_x a_x + v_y a_y + v_z a_z\]Here, \( v_x, v_y, v_z \) are the components of \( \mathbf{v} \) and \( a_x, a_y, a_z \) are the components of \( \mathbf{a} \).

The dot product can help determine whether vectors are perpendicular. If the dot product is zero, the vectors are orthogonal. This property is exploited in the solution, using the fact that the cross product \( \mathbf{w_1} \times \mathbf{w_2} \) is perpendicular to both \( \mathbf{w_1} \) and \( \mathbf{w_2} \). Thus, their dot products with the result of the cross product will equal zero.
Plane Equation
In three-dimensional geometry, a plane can be described using vectors. A common form of the equation of a plane involves a point on the plane and a normal vector. The given problem describes the plane using vector expressions and the properties of cross and dot products.

The standard form of a plane equation is given by:\[\langle \mathbf{x}, \mathbf{n} \rangle = d\]where \( \mathbf{n} \) is a normal vector to the plane, and \( d \) is a constant.
  • For this exercise, \( \mathbf{n} \) is found as the cross product of \( \mathbf{w_1} \) and \( \mathbf{w_2} \), making \( \mathbf{a} = \mathbf{w_1} \times \mathbf{w_2} \) the normal vector.
  • \( d \) is calculated as the dot product of vector \( \mathbf{v} \) and the normal vector \( \mathbf{a} \), equal to \( b \).
This simplification to the plane's equation illustrates how these vector operations interact:
  • The cross product finds a normal to the plane.
  • The dot product confirms the scalar relation that defines any point \( \mathbf{x} \) lying on the plane.
Understanding the plane equation provides insight into vector relationships and geometric solutions, which is crucial for fields like computer graphics and physics.

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Most popular questions from this chapter

$$ \text { Sei } s \text { die symmetrische Bilinearform auf dem } \mathbb{R}^{3} \text {, die gegeben ist durch die Matrix } $$ $$ \left(\begin{array}{rrr} 3 & -2 & 0 \\ -2 & 2 & -2 \\ 0 & -2 & 1 \end{array}\right) $$ Bestimme eine Basis \(\mathcal{A}\) des \(\mathbb{R}^{3}\), so da脽 \(\mathrm{M}_{\mathcal{A}}(s)\) Diagonalgestalt hat und eine weitere Basis \(\mathcal{B}\), so da脽 $$ \mathrm{M}_{\mathcal{B}}(s)=\left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right) $$

Das Vektorprodukt zweier Vektoren im \(\mathbb{R}^{3}\) l盲脽t sich f眉r \(n \geq 3\) folgenderma脽en zu einem Produkt von \(n-1\) Vektoren im \(\mathbb{R}^{n}\) verallgemeinern: Sind \(x^{(1)}, \ldots, x^{(n-1)} \in \mathbb{R}^{n}\), so sei $$ x^{(1)} \times \ldots \times x^{(n-1)}:=\sum_{i=1}^{n}(-1)^{i+1}\left(\operatorname{det} A_{i}\right) \cdot e_{i} $$ wobei \(A \in \mathrm{M}((n-1) \times n ; \mathbb{R})\) die Matrix ist, die aus den Zeilen \(x^{(1)}, \ldots, x^{(n-1)}\) besteht und \(A_{i}\) aus \(A\) durch Streichen der \(i\)-ten Spalte entsteht. Wie im Fall \(n=3\) entsteht \(x^{(1)} \times \ldots \times x^{(n-1)}\) also durch formales Entwickeln von $$ \operatorname{det}\left(\begin{array}{cccc} e_{1} & e_{2} & \ldots & e_{n} \\ x_{1}^{(1)} & x_{2}^{(1)} & \ldots & x_{n}^{(1)} \\ \vdots & \vdots & & \vdots \\ x_{1}^{(n-1)} & x_{2}^{(n-1)} & \ldots & x_{n}^{(n-1)} \end{array}\right) $$ nach der ersten Zeile. Zeige, da脽 f眉r das verallgemeinerte Vektorprodukt gilt: $$ \begin{aligned} &\text { a) } x^{(1)} \times \ldots \times x^{(i-1)} \times(x+y) \times x^{(i+1)} \times \ldots \times x^{(n-1)}= \\ &x^{(1)} \times \ldots \times x^{(i-1)} \times x \times x^{(i+1)} \times \ldots \times x^{(n-1)}+ \\ &x^{(1)} \times \ldots \times x^{(i-1)} \times y \times x^{(i+1)} \times \ldots \times x^{(n-1)}, \\ &x^{(1)} \times \ldots \times x^{(i-1)} \times(\lambda x) \times x^{(i+1)} \times \ldots \times x^{(n-1)}= \\ &\lambda\left(x^{(1)} \times \ldots \times x^{(i-1)} \times x \times x^{(i+1)} \times \ldots \times x^{(n-1)}\right) \end{aligned} $$ b) \(x^{(1)} \times \ldots \times x^{(n-1)}=0 \Leftrightarrow x^{(1)}, \ldots, x^{(n-1)}\) linear abh盲ngig. c) \(\left\langle x^{(1)} \times \ldots \times x^{(n-1)}, y\right\rangle=\operatorname{det}\left(\begin{array}{cccc}y_{1} & y_{2} & \ldots & y_{n} \\ x_{1}^{(1)} & x_{2}^{(1)} & & x_{n}^{(1)} \\ \vdots & \vdots & & \vdots \\ x_{1}^{(n-1)} & x_{2}^{(n-1)} & \ldots & x_{n}^{(n-1)}\end{array}\right)\) d) \(\left\langle x^{(1)} \times \ldots \times x^{(n-1)}, x^{(i)}\right\rangle=0\), f眉r \(i=1, \ldots, n-1\)

Zeige, \(\mathrm{da} \mathfrak{}\) die schiefsymmetrische Bilinearform (vgl. 5.4.1) \(\omega: \mathbb{R}^{2 n} \times \mathbb{R}^{2 n} \rightarrow \mathbb{R}\) aus 5.3.2 nicht-entartet ist, d. h. : Ist \(\omega(v, w)=0\) f眉r alle \(w \in \mathbb{R}^{2 n}\), so ist \(v=0\).

Wir wollen zeigen, da脽 auf einem \(\mathbb{R}\)-Vektorraum nicht jede Metrik aus einer Norm und nicht jede Norm aus einem Skalarprodukt entsteht. (Zur Erinnerung: Eine Norm auf einem \(\mathbb{R}\)-Vektorraum \(V\) ist eine Abbildung \(V \rightarrow \mathbb{R}_{+}\)mit den Eigenschaften N1, N2, \(\mathrm{N} 3\) aus 5.1.2, eine Metrik auf \(V\) ist eine Abbildung \(V \times V \rightarrow \mathbb{R}_{+}\)mit den Eigenschaften D1, D2, D3 aus 5.1.2.) a) Zeige, da脽 f眉r \(n \geq 2\) auf dem \(\mathbb{R}^{n}\) durch \(\|x\|:=\max \left\\{\left|x_{i}\right|: 1 \leq i \leq n\right\\}\) eine Norm definiert ist, f眉r die kein Skalarprodukt \(\langle,\),\(rangle auf \mathbb{R}^{n}\) existiert mit \(\|x\|=\sqrt{\langle x, x\rangle}\). b) Sei \(V=\mathcal{C}(\mathbb{R}: \mathbb{R})\) der Vektorraum der stetigen Funktionen, und f眉r \(k \in \mathbb{N}, f \in V\) sei \(\left.\|f\|_{k}:=\max \| f(x) \mid: x \in[-k, k]\right\\} .\) Zeige, \(\mathrm{da} \beta\) durch $$ d(f, g):=\sum_{k=0}^{\infty} 2^{-k} \frac{\|f-g\|_{k}}{1+\|f-g\|_{k}} $$ eine Metrik auf \(V\) definiert ist, f眉r die keine Norm \|\|\(: V \rightarrow \mathbb{R}_{+}\)existiert, so da脽 \(\|f-g\|=d(f, g)\)

Sei \(K\) ein K枚rper mit char \(K \neq 2\) und \(V\) ein \(K\)-Vektorraum. Zeige, da脽 sich jede Bilinearform auf \(V\) in eindeutiger Weise als Summe einer symmetrischen und einer schiefsymmetrischen Bilinearform darstellen l盲脽t.
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