Chapter 11, Problem 16PS

Solution

A.)

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

Hence the distance of given point from the line is 2.96 units

Given:

The coordinates of points Q is $\left(1,5,–2\right)$ . The parametric equation of line is

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

Explanation:

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

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

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

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

Directionvector of line is $\text{u}=\langle 4,0,–1\rangle$ . Now the coordinates of a general point on the line will be

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

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

Hence vector $\overline{PQ}$ is

  $\begin{array}{l}\overline{PQ}=\left(4\lambda –2–1,3–5,1–\lambda +2\right)\hfill \\ =\left(4\lambda –3,–2,3–\lambda \right)\hfill \end{array}$

Now cross product of two vectors and is

  $\begin{array}{l}\overline{PQ}\times u=\left[\begin{array}{ccc}i& j& k\\ 4\lambda -3& -2& 3-\lambda \\ 4& 0& -1\end{array}\right]\\ =i(2-0)-j(-4\lambda +3-12+4\lambda )+k(0+8)\\ =i(2)-j(-9)+k(8)\\ =\langle 2,9,8\rangle \end{array}$

Therefore,

  $\begin{array}{cc}D& =\frac{\parallel \overline{PQ}\times u\parallel }{\parallel u\parallel }\\ & =\frac{\sqrt{2^2+9^2+8^2}}{\sqrt{4^2+{(-1)}^2}}\\ & =2.96\end{array}$

Hence the distance of given point from the line is 2.96 units.

B.)

The distance of given point from the line PQ is 0.745 units

Given:

The coordinates of points Q is $\left(1,–2,4\right)$ . The parametric equation of line is

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

Explanation:

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

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

ThecoordinatesofpointsQis $\left(1,–2,4\right)$ .Theparametricequationoflineis

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

Directionvectoroflineis $\text{u}=\langle 2,1,2\rangle$ .Nowthecoordinatesofageneralpointonthelinewillbe

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

ThereforecoordinatesofpointPis $\left(2\lambda ,–3+\lambda ,2+2\lambda \right).$

Hencevector $\overline{PQ}$ is

  $\begin{array}{l}\overline{PQ}=2\lambda –1,–3+\lambda +2,2+2\lambda –4 \\ =2\lambda –1,–1+\lambda ,–2+2\lambda \end{array}$

Nowcrossproductoftwovectorsandis

  $\begin{array}{l}\overline{PQ}\times u=\left[\begin{array}{ccc}i& j& k\\ 2\lambda -1& -1+\lambda & -2+2\lambda \\ 2& 1& 2\end{array}\right]\\ \begin{array}{l}=\text{i}\left(–2+2\lambda +2–2\lambda \right)–\text{j}\left(4\lambda –2+4–4\lambda \right)+\text{k}\left(2\lambda –1+2–2\lambda \right)\hfill \\ =\text{i}\left(0\right)–\text{j}\left(–2\right)+\text{k}\left(1\right)\hfill \\ =\langle 0,–2,1\rangle \hfill \end{array}\end{array}$

Therefore,

  $\begin{array}{cc}D& =\frac{\parallel \overline{PQ}\times u\parallel }{\parallel u\parallel }\\ & =\frac{\sqrt{-2^2+1^2}}{\sqrt{2^2+{(1)}^2+2^2}}\\ & =0.745\end{array}$

Hencethedistanceofgivenpointfromtheline PQis0.745units.