Chapter 11, Problem 19PS

a.

To determine

Write the distance between the point and the line as a function of $s$

Answer to Problem 19PS

  $\boxed{\frac{\sqrt{10s^2-18s+194}}{2.29}units}$

Explanation of Solution

Given information:

Consider the line given by the parametric equations

  $x=-t+3,y=\frac{1}{2}t+1,z=2t-1$

and the point $(4,3,s)$ for any real number $s$

Write the distance between the point and the line as a function of $s$

Calculation:

Consider the vectors in space $a=x_{1}i+y_{1}j+z_{1}k$ and $b=x_{2}i+y_{2}j+z_{2}k$ then the cross product of $a=x_{1}i+y_{1}j+z_{1}k$ and $b=x_{2}i+y_{2}j+z_{2}k$ will be the vector

  $a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|$

  $=i[y_1(z_2)-y_2(z_1)]-j[x_1(z_2)-x_2(z_1)]+k[x_1(y_2)-x_2(y_1)]$

Distance between a point Q and line in space with direction vector $u$ is

  $D=\frac{\Vert \overrightarrow{PQ}\times u\Vert }{\Vert u\Vert }$

The coordinates of point Q is $(4,3,s)$ where $s$ is a real number. The parametric equation of line is

  $\begin{array}{l}x=-t+3\\ y=\frac{1}{2}t+1\\ z=2t-1\end{array}$

Hence the direction vector of line will be

  $u=\langle -1,\frac{1}{2},2\rangle$

Hence the coordinates of a general point on the line will be

  $\begin{array}{l}x=-\lambda +3\\ y=\frac{1}{2}\lambda +1\\ z=2\lambda -1\end{array}$

Therefore the coordinates of point P is $\left(-\lambda +3,\frac{1}{2}\lambda +1,2\lambda -1\right)$ .

Hence vector $\overrightarrow{PQ}$ is

  $\overrightarrow{PQ}=\langle -\lambda -3-4,\frac{1}{2}\lambda +1-3,2\lambda -1-s\rangle$

  $\langle -\lambda -7,\frac{1}{2}\lambda -2,2\lambda -1-s\rangle$

Now cross product of two vectors and is

  $\overrightarrow{PQ}\times u=\left[\begin{array}{ccc}i& j& k\\ -\lambda -7& \frac{1}{2}\lambda -2& 2\lambda -1-s\\ 4& 3& s\end{array}\right]$

  $\begin{array}{l}=i\left(\left(\frac{1}{2}\lambda -2\right)s-3\left(2\lambda -1-s\right)\right)-j\left(\left(-\lambda -7\right)s-4\left(2\lambda -1-s\right)\right)\\ +k\left(\left(-\lambda -7\right)3-\left(\frac{1}{2}\lambda -2\right)4\right)\\ =\langle \left(\frac{1}{2}s-6\right)\lambda +s+3,\left(s+8\right)\lambda +3s-4,-5\lambda -13\rangle \end{array}$

Distance vector never consist terms of $\lambda$ . So eliminate them by taking suitable values of $s$ .

Therefore

  $D=\frac{\Vert \overrightarrow{PQ}\times u\Vert }{\Vert u\Vert }$

  $\begin{array}{l}=\frac{\sqrt{{\left(s+3\right)}^2+{\left(3s-4\right)}^2+{\left(-13\right)}^2}}{2.29}\\ =\frac{\sqrt{10s^2-18s+194}}{2.29}\end{array}$

Hence the distance of given point from the line is $\boxed{\frac{\sqrt{10s^2-18s+194}}{2.29}units}$ .

b.

To determine

Use the graph to find the value of s.

Answer to Problem 19PS

  $\boxed{5.95units}$

Explanation of Solution

Given information:

Consider the line given by the parametric equations

  $x=-t+3,y=\frac{1}{2}t+1,z=2t-1$

and the point $(4,3,s)$ for any real number $s$

Use a graphing utility to graph the function from part (a). Use the graph to find the value of s such that the distance between the point and the line is a minimum.

Calculation:

Consider the vectors in space $a=x_{1}i+y_{1}j+z_{1}k$ and $b=x_{2}i+y_{2}j+z_{2}k$ then the cross product of $a=x_{1}i+y_{1}j+z_{1}k$ and $b=x_{2}i+y_{2}j+z_{2}k$ will be the vector

  $a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|$

  $=i[y_1(z_2)-y_2(z_1)]-j[x_1(z_2)-x_2(z_1)]+k[x_1(y_2)-x_2(y_1)]$

Distance between a point Q and line in space with direction vector $u$ is

  $D=\frac{\Vert \overrightarrow{PQ}\times u\Vert }{\Vert u\Vert }$

Using $TI-83$ ; the graph of distance can be plotted.

Press $"Y="$ button and write the equation.

  

Press $"Window"$ button and make such readings.

  

Press $"Graph"$ button and graph is like

  

To find the minimum value of function, press “Trace” button and use arrow keys to find the lowest point. Following screen is received.

  

Hence, the minimum distance line and point is $\boxed{5.95units}$

c.

To determine

Does it appear that the graph has slant asymptotes?

Answer to Problem 19PS

The function does not have any slant asymptote.

Explanation of Solution

Given information:

Consider the line given by the parametric equations

  $x=-t+3,y=\frac{1}{2}t+1,z=2t-1$

and the point $(4,3,s)$ for any real number $s$

Use the zoom feature of the graphing utility to zoom out several times on the graph in part (b). Does it appear that the graph has slant asymptotes? Explain. If it appear to have slant asymptotes, find them.

Calculation:

Consider the vectors in space $a=x_{1}i+y_{1}j+z_{1}k$ and $b=x_{2}i+y_{2}j+z_{2}k$ then the cross product of $a=x_{1}i+y_{1}j+z_{1}k$ and $b=x_{2}i+y_{2}j+z_{2}k$ will be the vector

  $a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|$

  $=i[y_1(z_2)-y_2(z_1)]-j[x_1(z_2)-x_2(z_1)]+k[x_1(y_2)-x_2(y_1)]$

Distance between a point Q and line in space with direction vector $u$ is

  $D=\frac{\Vert \overrightarrow{PQ}\times u\Vert }{\Vert u\Vert }$

Slant asymptote means the range in x-axis where graph have almost infinite slope. In TI-83;

Press “Zoom” button and select “0:ZoomFit”. The following screen appears.

  

On zooming the graph, there is no such range is found.

Hence the function does not have any slant asymptote.