Problem 11 a. Show that the volume $V$ of... [FREE SOLUTION] (2024)

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Chapter 1: Problem 11

a. Show that the volume $V$ of a parallelepiped generated by three linearlyindependent vectors $u, v, w \in R^{3}$ is given by $V=|(u \wedge v) \cdotw|$, and introduce an oriented volume in $R^{3}$. b. Prove that $$ V^{2}=\left|\begin{array}{ccc} u \cdot u & u \cdot v & u \cdot w \\ v \cdot u & v \cdot v & v \cdot w \\ w \cdot u & w \cdot v & w \cdot w \end{array}\right| $$

Short Answer

Expert verified

The volume V is $ V = |(\textbf{u} \times \textbf{v}) \boldsymbol{\bullet} \textbf{w}| \. For \( V^2 $, it is shown that \text{det} \begin{vmatrix} \textbf{u} \boldsymbol{\bullet} \textbf{u} & \textbf{u} \boldsymbol{\bullet} \textbf{v} & \textbf{u} \boldsymbol{\bullet} \textbf{w} \ \textbf{v} \boldsymbol{\bullet} \textbf{u} & \textbf{v} \boldsymbol{\bullet} \textbf{v} & \textbf{v} \boldsymbol{\bullet} \textbf{w} \ \textbf{w} \boldsymbol{\bullet} \textbf{u} & \textbf{w} \boldsymbol{\bullet} \textbf{v} & \textbf{w} \boldsymbol{\bullet} \textbf{w} \end{vmatrix} = V^2 \).

Step by step solution

Understanding the Parallelepiped Volume Formula

The volume of a parallelepiped generated by three linearly independent vectors $\textbf{u}, \textbf{v}, \textbf{w}$ in $ \textbf{R}^3 $ is given by the absolute value of the scalar triple product. The scalar triple product is defined as $ (\textbf{u} \times \textbf{v}) \boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\bullet}}}} \textbf{w} $, where $ \times $ is the cross product and $ \boldsymbol{\boldsymbol{\boldsymbol{\bullet}}} $ is the dot product.

Finding the Cross Product $ \textbf{u} \times \textbf{v} $

Compute the cross product of vectors $\textbf{u} $ and $ \textbf{v} $. This yields another vector perpendicular to both $ \textbf{u} $ and $ \textbf{v}$: \[ \textbf{u} \times \textbf{v} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \]

Calculating the Dot Product with $\textbf{w} $

Take the dot product of $ \textbf{u} \times \textbf{v} $ with $ \textbf{w} $. This gives the scalar value: \[((u \times v) \boldsymbol{\bullet} w = (u_2v_3 - u_3v_2)w_1 + (u_3v_1 - u_1v_3)w_2 + (u_1v_2 - u_2v_1)w_3) \]. The absolute value of this scalar is the volume $ V $ of the parallelepiped.

Defining Oriented Volume

By taking the scalar triple product without the absolute value, we can introduce oriented volume, which considers the sign of the computed scalar. Therefore, the oriented volume is given by $ (\textbf{u} \times \textbf{v}) \boldsymbol{\bullet} \textbf{w} $.

Proving the Determinant Formula for $ V^2 $

To show $ V^{2} = \begin{vmatrix} u \boldsymbol{\bullet} u & u \boldsymbol{\bullet} v & u \boldsymbol{\bullet} w \ v \boldsymbol{\bullet} u & v \boldsymbol{\bullet} v & v \boldsymbol{\bullet} w \ w \boldsymbol{\bullet} u & w \boldsymbol{\bullet} v & w \boldsymbol{\bullet} w \end{vmatrix}$, start from the definition. The inner product matrix $A$ entries are given by the dot products of vectors. Then, use the Gram determinant to connect this matrix to the volume squared: $V^2 = \text{det}(A)$.

Using Vector Properties to Derive Determinants

Recall that for any set of vectors, the determinant of their Gram matrix provides the square of the volume of the parallelepiped formed by these vectors: $ V^2 = \text{det} \begin{vmatrix} u \boldsymbol{\bullet} u & u \boldsymbol{\bullet} v & u \boldsymbol{\bullet} w \ v \boldsymbol{\bullet} u & v \boldsymbol{\bullet} v & v \boldsymbol{\bullet} w \ w \boldsymbol{\bullet} u & w \boldsymbol{\bullet} v & w \boldsymbol{\bullet} w \end{vmatrix} $. Therefore, it's shown that $ V^2 = \text{det}(A) $.

Key Concepts

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

Scalar Triple Product

The Scalar Triple Product is essential in calculating the volume of a parallelepiped. It combines both the cross product and dot product. To compute it, first find the cross product of two vectors, say $\textbf{u}$ and $\textbf{v}$, symbolized as $(\textbf{u} \times \textbf{v})$. Next, take the dot product of this result with the third vector $\textbf{w}$. Mathematically, it's expressed as $(\textbf{u} \times \textbf{v}) \bullet \textbf{w}$. This scalar value's absolute value gives the volume of the parallelepiped formed by the three vectors. This value indicates how much space is enclosed within the parallelepiped's boundaries.
Understanding the role of each product here helps significantly in geometrical interpretations and applications in physics.

Cross Product

The Cross Product is a vector arithmetic operation fundamental in finding the volume via the scalar triple product. Given two vectors, $\textbf{u} = [u_1, u_2, u_3]$ and $\textbf{v} = [v_1, v_2, v_3]$, their cross product $\textbf{u} \times \textbf{v}$ yields a new vector that's perpendicular to both $\textbf{u}$ and $\textbf{v}$. The formula to compute it is:

\[ \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \ \end{vmatrix} \]
This determinant expands to:
\[ \textbf{u} \times \textbf{v} = (u_2 v_3 - u_3 v_2)\textbf{i} - (u_1 v_3 - u_3 v_1)\textbf{j} + (u_1 v_2 - u_2 v_1)\textbf{k} \]
The cross product's result is crucial as it serves a critical component in the scalar triple product and thus in determining the volume.

Dot Product

The Dot Product is another significant vector operation that plays an important role in the scalar triple product. Given vectors $\textbf{a}$ and $\textbf{b}$, their dot product is computed as:
\[ \textbf{a} \bullet \textbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]
This operation results in a scalar. For our problem, once we have the cross product $\textbf{u} \times \textbf{v}$, we use the dot product with $\textbf{w}$ to get:
\[ (\textbf{u} \times \textbf{v}) \bullet \textbf{w} \]
Essentially, you're projecting the perpendicular vector $\textbf{u} \times \textbf{v}$ onto $\textbf{w}$. This particular scalar indicates the volume when the absolute value is taken. This is because the dot product represents how much one vector extends in the direction of another.

Gram Determinant

The Gram Determinant is a formula used to find the volume squared of a parallelepiped formed by three vectors $\textbf{u}$, $\textbf{v}$, and $\textbf{w}$. The Gram matrix is constructed by computing inner (dot) products of all possible pairs of vectors as follows:
\[ A = \begin{vmatrix} \textbf{u} \bullet \textbf{u} & \textbf{u} \bullet \textbf{v} & \textbf{u} \bullet \textbf{w} \ \textbf{v} \bullet \textbf{u} & \textbf{v} \bullet \textbf{v} & \textbf{v} \bullet \textbf{w} \ \textbf{w} \bullet \textbf{u} & \textbf{w} \bullet \textbf{v} & \textbf{w} \bullet \textbf{w} \ \end{vmatrix} \]
The determinant of this Gram matrix ultimately computes the square of the volume of the parallelepiped. This method builds on the idea that the square of the volume of a geometric figure can be derived from orthogonal projections of the involved vectors. Hence, computing the determinant of matrix $\textbf{A}$ gives us:
\[ V^2 = \text{det}(A) \]
This determinant formula reinforces the theory connecting vector products with geometric volume.

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$Problem 11 a. Show that the volume $V$ of... [FREE SOLUTION] (3)$

Most popular questions from this chapter

Find all unit vectors which are perpendicular to the vector $(2,2,1)$ andparallel to the plane determined by the points $(0,0,0),(1,-2,1)$, $(-1,1,1)$.Show that an equation of a plane passing through three noncolinear points$p_{1}=\left(x_{1}, y_{1}, z_{1}\right), p_{2}=\left(x_{2}, y_{2},z_{2}\right), p_{3}=\left(x_{3}, y_{3}, z_{3}\right)$ is given by $$ \left(p-p_{1}\right) \wedge\left(p-p_{2}\right) \cdot\left(p-p_{3}\right)=0, $$ where $p=(x, y, z)$ is an arbitrary point of the plane and $p-p_{1}$, forinstance, means the vector $\left(x-x_{1}, y-y_{1}, z-z_{1}\right)$.Show that the tangent lines to the regular parametrized curve $\alpha(t)=$$\left(3 t, 3 t^{2}, 2 t^{3}\right)$ make a constant angle with the line $y=0,z=x$.Let $\alpha: I \rightarrow R^{2}$ be a regular parametrized plane curve(arbitrary parameter), and define $n=n(t)$ and $k=k(t)$ as in Remark 1. Assumethat $k(t) \neq 0, t \in I$. In this situation, the curve $$ \beta(t)=\alpha(t)+\frac{1}{k(t)} n(t), \quad t \in I $$ is called the evolute of $\alpha$ (Fig. 1-17). a. Show that the tangent at $t$ of the evolute of $\alpha$ is the normal to$\alpha$ at $t$. b. Consider the normal lines of $\alpha$ at two neighboring points $t_{1},t_{2}$, $t_{1} \neq t_{2}$. Let $t_{1}$ approach $t_{2}$ and show that theintersection points of the normals converge to a point on the trace of theevolute of $\alpha$. Figure 1-17Given two planes $a_{i} x+b_{i} y+c_{i} z+d_{i}=0, i=1,2$, prove that anecessary and sufficient condition for them to be parallel is $$ \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}, $$ where the convention is made that if a denominator is zero, the correspondingnumerator is also zero (we say that two planes are parallel if they eithercoincide or do not intersect).

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Problem 11 a. Show that the volume \(V\) of... [FREE SOLUTION] (2024)

Short Answer

Step by step solution

Understanding the Parallelepiped Volume Formula

Finding the Cross Product \( \textbf{u} \times \textbf{v} \)

Calculating the Dot Product with \(\textbf{w} \)

Defining Oriented Volume

Proving the Determinant Formula for \( V^2 \)

Using Vector Properties to Derive Determinants

Key Concepts

Scalar Triple Product

Cross Product

Dot Product

Gram Determinant

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Decision Maths

Probability and Statistics

Geometry

Theoretical and Mathematical Physics

Statistics

Study anywhere. Anytime. Across all devices.

Privacy Overview