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Chapter 1: Problem 8

Prove that the distance \(\rho\) between the nonparallel lines $$ \begin{array}{ll} x-x_{0}=u_{1} t, & y-y_{0}=u_{2} t, \quad z-z_{0}=u_{3} t \\ x-x_{1}=v_{1} t, & y-y_{1}=v_{2} t, \quad z-z_{1}=v_{3} t \end{array} $$ is given by $$ \rho=\frac{|(u \wedge v) \cdot r|}{|u \wedge v|} $$ where \(u=\left(u_{1}, u_{2}, u_{3}\right), v=\left(v_{1}, v_{2}, v_{3}\right), r=\left(x_{0}-x_{1}, y_{0}-y_{1}, z_{0}-z_{1}\right)\).

Short Answer

Expert verified
The distance \( \rho \) is \( \frac{|(u \times v) \bullet r|}{|u \times v|} \).

Step by step solution

01

- Understand the Given Lines

The first line can be parameterized as:\(x - x_{0} = u_{1} t\),\(y - y_{0} = u_{2} t\),\(z - z_{0} = u_{3} t\)\Similarly, the second line can be parameterized as:\( x - x_{1} = v_{1} t\),\( y - y_{1} = v_{2} t\),\(z - z_{1} = v_{3} t\)
02

- Define Vector Notations

Let \(u = [u_{1}, u_{2}, u_{3}]\) and \(v = [v_{1}, v_{2}, v_{3}]\).\ Define the position vector \( r = [x_{0} - x_{1}, y_{0} - y_{1}, z_{0} - z_{1}]\).
03

- Cross Product of Direction Vectors

Compute the cross product of vectors \(u\) and \(v\):\[ u \times v = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \ u_{1} & u_{2} & u_{3} \ v_{1} & v_{2} & v_{3} \end{vmatrix} \ Rightarrow u \times v = [(u_{2}v_{3} - u_{3}v_{2}), (u_{3}v_{1} - u_{1}v_{3}), (u_{1}v_{2} - u_{2}v_{1})] \]
04

- Magnitude of Cross Product

Calculate the magnitude of this cross product: \[|u \times v| = \sqrt{(u_{2}v_{3} - u_{3}v_{2})^2 + (u_{3}v_{1} - u_{1}v_{3})^2 + (u_{1}v_{2} - u_{2}v_{1})^2}\]
05

- Dot Product of Distances

Compute the dot product \((u \times v) \bullet r\):\[ (u \times v) \bullet r = (u_{2}v_{3} - u_{3}v_{2})(x_{0}-x_{1}) + (u_{3}v_{1} - u_{1}v_{3})(y_{0}-y_{1}) + (u_{1}v_{2} - u_{2}v_{1})(z_{0}-z_{1}) \]
06

- Calculate the Distance

Finally, the distance \( \rho \) between the nonparallel lines is calculated using the formula: \[\rho = \frac{|(u \times v) \bullet r|}{|u \times v|}\]

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Key Concepts

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

Vector Cross Product
The cross product of two vectors is essential in finding the distance between skew lines. Let's break it down. The cross product of vectors \(\mathbf{u}\) and \(\mathbf{v}\), written as \(\mathbf{u} \times \mathbf{v}\), results in a new vector that is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\). This is useful for finding the shortest distance between two lines because we need a vector perpendicular to both lines. The formula to compute the cross product is:
\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \]
This determinant expands to:
\[ \mathbf{u} \times \mathbf{v} = \left[ (u_2v_3 - u_3v_2), (u_3v_1 - u_1v_3), (u_1v_2 - u_2v_1) \right] \]
The resulting vector (components labeled in respective i, j, k directions) helps us compute the distance between two nonparallel lines. Remember, this new vector is orthogonal to both original vectors.
Vector Dot Product
Understanding the dot product is crucial for measuring how aligned two vectors are. The dot product of two vectors \(\mathbf{a} = [a_1, a_2, a_3]\) and \(\mathbf{b} = [b_1, b_2, b_3]\), written as \(\mathbf{a} \cdot \mathbf{b}\), is calculated as:
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
For our exercise, we need to find the distance between skew lines, involving the dot product of the cross product of \(\mathbf{u}\) and \(\mathbf{v}\), and the vector \(\mathbf{r}\). The expression for this specific dot product is:
\[ (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{r} = (u_2v_3 - u_3v_2)(x_0- x_1) + (u_3v_1 - u_1v_3)(y_0 - y_1) + (u_1v_2 - u_2v_1)(z_0 - z_1) \]
This result is a scalar value that helps determine the numerator in the distance formula between the skew lines.
Magnitude of a Vector
To find the distance between skew lines, we need the magnitude of the cross product vector. The magnitude (or length) of a vector \(\mathbf{v} = [v_1, v_2, v_3]\) is given by the formula:
\[ | \mathbf{v} | = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
When dealing with the vector resulting from the cross product \(\mathbf{u} \times \mathbf{v}\), the magnitude is:
\[ |\mathbf{u} \times \mathbf{v}| = \sqrt{ (u_2v_3 - u_3v_2)^2 + (u_3v_1 - u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 } \]
This magnitude serves as the denominator in the distance formula. Thus, the complete distance formula between non-parallel (skew) lines is:
\[ \rho = \frac{|( \mathbf{u} \times \mathbf{v}) \cdot \mathbf{r}|}{| \mathbf{u} \times \mathbf{v} |} \]
This formula allows us to compute the shortest distance between two skew lines using vector operations effectively.

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Most popular questions from this chapter

Let \(\alpha: I \rightarrow R^{3}\) be a regular parametrized curve (not necessarily by \(\operatorname{arc}\) length and let \(\beta: J \rightarrow R^{3}\) be a reparametrization of \(\alpha(I)\) by the arc length \(s=s(t)\), measured from \(t_{0} \in I\) (see Remark 2). Let \(t=t(s)\) be the inverse function of \(s\) and set \(d \alpha / d t=\alpha^{\prime}, d^{2} \alpha / d t^{2}=\alpha^{\prime \prime}\), etc. Prove that a. \(d t / d s=1 /\left|\alpha^{\prime}\right|, d^{2} t / d s^{2}=-\left(\alpha^{\prime} \cdot \alpha^{\prime \prime} /\left|\alpha^{\prime}\right|^{4}\right)\). b. The curvature of \(\alpha\) at \(t \in I\) is $$ k(t)=\frac{\left|\alpha^{\prime} \wedge \alpha^{\prime \prime}\right|}{\left|\alpha^{\prime}\right|^{3}} . $$ c. The torsion of \(\alpha\) at \(t \in I\) is $$ \tau(t)=-\frac{\left(\alpha^{\prime} \wedge \alpha^{\prime \prime}\right) \cdot \alpha^{\prime \prime \prime}}{\left|\alpha^{\prime} \wedge \alpha^{\prime \prime}\right|^{2}} $$ d. If \(\alpha: I \rightarrow R^{2}\) is a plane curve \(\alpha(t)=(x(t), y(t))\), the signed curvature (see Remark 1) of \(\alpha\) at \(t\) is $$ k(t)=\frac{x^{\prime} y^{\prime \prime}-x^{\prime \prime} y^{\prime}}{\left(\left(x^{\prime}\right)^{2}+\left(y^{\prime}\right)^{2}\right)^{3 / 2}} . $$

Show that the knowledge of the vector function \(b=b(s)\) (binormal vector) of a curve \(\alpha\), with nonzero torsion everywhere, determines the curvature \(k(s)\) and the absolute value of the torsion \(\tau(s)\) of \(\alpha\).

Let \(\alpha: I \rightarrow R^{3}\) be a curve parametrized by arc length with curvature \(k(s) \neq 0, s \in I\). Let \(P\) be a plane satisfying both of the following conditions: 1\. \(P\) contains the tangent line at \(s\). 2\. Given any neighborhood \(J \subset I\) of \(s\), there exist points of \(\alpha(J)\) in both sides of \(P\). Prove that \(P\) is the osculating plane of \(\alpha\) at \(s\).

The natural orientation of \(R^{2}\) makes it possible to associate a sign to the area \(A\) of a parallelogram generated by two linearly independent vectors \(u, v \in R^{2}\). To do this, let \(\left\\{e_{i}\right\\}, i=1,2\), be the natural ordered basis of \(R^{2}\), and write \(u=u_{1} e_{1}+u_{2} e_{2}, v=v_{1} e_{1}+v_{2} e_{2}\). Observe the matrix relation $$ \left(\begin{array}{cc} u \cdot u & u \cdot v \\ v \cdot u & v \cdot v \end{array}\right)=\left(\begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}\right)\left(\begin{array}{cc} u_{1} & v_{1} \\ u_{2} & v_{2} \end{array}\right) $$ and conclude that $$ A^{2}=\left|\begin{array}{ll} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}\right|^{2} $$ Since the last determinant has the same sign as the basis \(\\{u, v\\}\), we can say that \(A\) is positive or negative according to whether the orientation of \(\\{u, v\\}\) is positive or negative. This is called the oriented area in \(R^{2}\).

In general, a curve \(\alpha\) is called a helix if the tangent lines of \(\alpha\) make a constant angle with a fixed direction. Assume that \(\tau(s) \neq 0, s \in I\), and prove that: * a. \(\alpha\) is a helix if and only if \(k / \tau=\) const. *b. \(\alpha\) is a helix if and only if the lines containing \(n(s)\) and passing through \(\alpha(s)\) are parallel to a fixed plane. *c. \(\alpha\) is a helix if and only if the lines containing \(b(s)\) and passing through \(\alpha(s)\) make a constant angle with a fixed direction. d. The curve $$ \alpha(s)=\left(\frac{a}{c} \int \sin \theta(s) d s, \frac{a}{c} \int \cos \theta(s) d s, \frac{b}{c} s\right), $$ where \(c^{2}=a^{2}+b^{2}\), is a helix, and that \(k / \tau=a / b\).
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