Chapter 11, Problem 17PS

A.)

To determine

Find the shortest distance between the given point and the line passing through the points $P_{1}and{P}_2$ .

Answer to Problem 17PS

The distance of given point from the line is 2.12 units.

Explanation of Solution

Given:

The coordinates of points Q is $\left(2,0,0\right)$ . The line is determined by the points $P_1\left(0,0,1\right)and{P}_2\left(0,1,2\right)$

Hence the direction vector of the line will be

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

Consider the vectors in space $\text{a}=x_1\text{i}+y_1\text{j}+z_1\text{k}$ and $\text{b}=x_2\text{i}+y_2\text{j}+z_2\text{k}$ then the cross product of $\text{a}=x_1\text{i}+y_1\text{j}+z_1\text{k}$ and $\text{b}=x_2\text{i}+y_2\text{j}+z_2\text{k}$ will be the vector

  $\begin{array}{l}a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ =\text{i}\left[y_1\left(z_2\right)–y_2\left(z_1\right)\right]–\text{j}\left[x_1\left(z_2\right)–x_2\left(z_1\right)\right]+\text{k}\left[x_1\left(y_2\right)–x_2\left(y_1\right)\right]\end{array}$

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

  $\begin{array}{l}D=\frac{\parallel PQ\times u\parallel }{\parallel u\parallel }\hfill \\ \hfill \end{array}$

The coordinates of points Q is $\left(2,0,0\right)$ . The line is determined by the points $P_1\left(0,0,1\right)and{P}_2\left(0,1,2\right)$

Hence the direction vector of the line will be

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

Equation of line is given as $P_1+\text{u}t$

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

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

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

  $\overline{PQ}=2,–\lambda ,–1–\lambda$

Now cross product of two vectors and is

  $\begin{array}{l}\overline{PQ}\times u=\left[\begin{array}{ccc}i& j& k\\ 2& -\lambda & -1-\lambda \\ 0& 1& 1\end{array}\right]\\ =i(-\lambda +1+\lambda )-j(-2+0)+k(0-0)\\ =i(1)-j(2)+k(2)\\ =\langle 1,-2,2\rangle \end{array}$

Therefore

  $\begin{array}{cc}D& =\frac{\parallel \overline{PQ}\times u\parallel }{\parallel u\parallel }\\ & =\frac{\sqrt{1^2+{(-2)}^2+2^2}}{\sqrt{1^2+1^2}}\\ & =2.12\end{array}$

Hence the distance of given point from the line is 2.12 units.

B.)

To determine

Find the shortest distance between the given point and the line passing through the points $P_{1}and{P}_2$ .

Answer to Problem 17PS

The distance of given point from the line is 2.12 units.

Explanation of Solution

Given:

The coordinates of points Q is $\left(2,0,0\right)$ . The line is determined by the points $P_1\left(0,0,1\right)and{P}_2\left(0,1,2\right)$

Hence the direction vector of the line will be

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

Consider the vectors in space $\text{a}=x_1\text{i}+y_1\text{j}+z_1\text{k}$ and $\text{b}=x_2\text{i}+y_2\text{j}+z_2\text{k}$ then the cross product of $\text{a}=x_1\text{i}+y_1\text{j}+z_1\text{k}$ and $\text{b}=x_2\text{i}+y_2\text{j}+z_2\text{k}$ will be the vector

  $\begin{array}{l}a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ =\text{i}\left[y_1\left(z_2\right)–y_2\left(z_1\right)\right]–\text{j}\left[x_1\left(z_2\right)–x_2\left(z_1\right)\right]+\text{k}\left[x_1\left(y_2\right)–x_2\left(y_1\right)\right]\end{array}$

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

  $\begin{array}{l}D=\frac{\parallel PQ\times u\parallel }{\parallel u\parallel }\hfill \\ \hfill \end{array}$

The coordinates of points Q is $\left(2,0,0\right)$ . The line is determined by the points $P_1\left(0,0,1\right)and{P}_2\left(0,1,2\right)$

Hence the direction vector of the line will be

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

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

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

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

  $\overline{PQ}=2,–\lambda ,–1–\lambda$

Now cross product of two vectors and is

  $\begin{array}{l}\overline{PQ}\times u=\left[\begin{array}{ccc}i& j& k\\ 2& -\lambda & -1-\lambda \\ 0& 1& 1\end{array}\right]\\ =i(-\lambda +1+\lambda )-j(-2+0)+k(0-0)\\ =i(1)-j(2)+k(2)\\ =\langle 1,-2,2\rangle \end{array}$

Therefore

  $\begin{array}{cc}D& =\frac{\parallel \overline{PQ}\times u\parallel }{\parallel u\parallel }\\ & =\frac{\sqrt{1^2+{(-2)}^2+2^2}}{\sqrt{1^2+1^2}}\\ & =2.12\end{array}$

Hence the distance of given point from the line is 2.12 units.