Chapter 11, Problem 18PS

a.

To determine

Find the shortest distance between the point $Q(2,0,0)$ .

Answer to Problem 18PS

  $\boxed{2.12units}$

Explanation of Solution

Given information:

Find the shortest distance between the point $Q(2,0,0)$ and the line determined by the points $P_1(0,0,1)$ and $P_2(0,1,2)$

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 $(2,0,0)$ . The line is determined by the points $P_1(0,0,1)$ and $P_2(0,1,2)$

Hence the direction vector of line will be

  $\begin{array}{l}u=\langle 0-0,1-0,2-1\rangle \\ =\langle 0,1,1\rangle \end{array}$

Equation of line is given as $P_{{}_1}+ut$

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

  $\begin{array}{l}x=0\\ y=\lambda \\ z=1+\lambda \end{array}$

Therefore the coordinates of point P is $(0,\lambda ,1+\lambda )$ . Hence vector $\overrightarrow{PQ}$ is

  $\overrightarrow{PQ}=\langle 2,-\lambda ,-1-\lambda \rangle$

Now cross product of two vectors and is

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

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

Therefore

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

  $\begin{array}{l}=\frac{\sqrt{1^2+{(-2)}^2+2^2}}{\sqrt{1^2+1^2}}\\ =2.12\end{array}$

Hence the distance of given point from the line is $\boxed{2.12units}$ .

b.

To determine

Find the shortest distance between the points.

Answer to Problem 18PS

  $\boxed{2.12units}$

Explanation of Solution

Given information:

Find the shortest distance between the point $Q(2,0,0)$ and the line segment joining the points $P_1(0,0,1)$ and $P_2(0,1,2)$

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 $(2,0,0)$ . The line segment is formed by joining the points $P_1(0,0,1)$ and $P_2(0,1,2)$

Hence the vector of line will be

  $\begin{array}{l}u=\langle 0-0,1-0,2-1\rangle \\ =\langle 0,1,1\rangle \end{array}$

Line segment lies on yz-plane as $x$ coordinates remains zero. Thus the equation of line is

  $\begin{array}{l}y=y_1=\frac{y_2-y_1}{z_2-z_1}(z-z_1)\\ y-0=\frac{1-0}{2-1}\left(z-1\right)\\ y=z-1\end{array}$

Therefore the coordinates of point P is $(0,\lambda ,1+\lambda )$ . Hence vector $\overrightarrow{PQ}$ is

  $\overrightarrow{PQ}=\langle 2,-\lambda ,-1-\lambda \rangle$

Now cross product of two vectors and is

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

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

Therefore

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

  $\begin{array}{l}=\frac{\sqrt{1^2+{(-2)}^2+2^2}}{\sqrt{1^2+1^2}}\\ =2.12\end{array}$

Hence the distance of given point from the line segment is $\boxed{2.12units}$ .