Chapter 11, Problem 17PS

a.

To determine

Find the distance between the point and the line given by each set of parametric equations.

Answer to Problem 17PS

  $\boxed{2.96units}$

Explanation of Solution

Given information:

The distance between a point $Q$ and a line in space is given by

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

where $u$ is a direction vector for the line and $P$ is a point on the line. Find the distance between the point and the line given by each set of parametric equations.

  $\left(1,5,-2\right)$

  $x=-2+4t,y=3,z=1-t$

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 $\left(1,5,-2\right)$ . The parametric equation of line is

  $\begin{array}{l}x=-2+4t\\ y=3\\ z=1-t\end{array}$

Direction vector of line is $u=\langle 4,0,-1\rangle$ . Now the coordinates of a general point on the line will be

  $\begin{array}{l}x=4\lambda -2\\ y=3\\ z=1-\lambda \end{array}$

Hence coordinates of point $P$ is $\left(4\lambda -2,3,1-\lambda \right)$ .

  $\begin{array}{l}\overrightarrow{PQ}=\langle 4\lambda -2-1,3-5,1-\lambda +2\rangle \\ =\langle 4\lambda -3,-2,3-\lambda \rangle \end{array}$

Now cross product of two vectors and is

  $\overrightarrow{PQ}\times u=\left[\begin{array}{ccc}i& j& k\\ 4\lambda -3& -2& 3-\lambda \\ 4& 0& -1\end{array}\right]$

  $\begin{array}{l}=i(2-0)-j(-4\lambda +3-12+4\lambda )+(0+8)\\ =i(2)-j(-9)+k(8)\\ =\langle 2,9,8\rangle \end{array}$

Therefore

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

  $\begin{array}{l}=\frac{\sqrt{2^2+9^2+8^2}}{\sqrt{4^2+{(-1)}^2}}\\ =2.96\end{array}$

Hence the distance of given point from the line is $\boxed{2.96units}$ .

b.

To determine

Find the distance between the point and the line given by each set of parametric equations.

Answer to Problem 17PS

  $\boxed{0.745units}$

Explanation of Solution

Given information:

The distance between a point $Q$ and a line in space is given by

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

where $u$ is a direction vector for the line and $P$ is a point on the line. Find the distance between the point and the line given by each set of parametric equations.

  $(1,-2,4)$

  $x=2t,y=-3+t,z=2+2t$

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 $(1,-2,4)$ . The parametric equation of line is

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

Direction vector of line is $u=\langle 2,1,2\rangle$ . Now the coordinates of a general point on the line will be

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

Therefore coordinates of point $P$ is $\left(2\lambda ,-3+\lambda ,2+2\lambda \right)$ . Hence vector $\overrightarrow{PQ}$ is

  $\begin{array}{l}\overrightarrow{PQ}=\langle 2\lambda -1,-3+\lambda +2,2+2\lambda -4\rangle \\ =\langle 2\lambda -1,-1+\lambda ,-2+2\lambda \rangle \end{array}$

Now cross of two vectors and is

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

  $\begin{array}{l}=i(-2+2\lambda +2-2\lambda )-j(4\lambda -2+4-4\lambda )+k(2\lambda -1+2-2\lambda )\\ =i\left(0\right)-j(-2)+k(1)\\ =\langle 0,-2,1\rangle \end{array}$

Therefore

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

  $\begin{array}{l}=\frac{\sqrt{{(-2)}^2+1^2}}{\sqrt{2^2+1^2+2^2}}\\ =0.745\end{array}$

Hence the distance of point $(1,-2,4)$ from the line $PQ$ is $\boxed{0.745units}$ .