Chapter 7.6, Problem 25E

To determine

To find: the length of the normal segment

  $\left(p\right)$ and angle the normal segment makes with the positive $x$ axis $\left(\varphi \right)$ by writing the equation in normal form.

Explanation of Solution

Given information:

  $x=3$

Calculation:

Rearrange the equation in the standard form of line $\left(Ax+By+C=0\right)$ as shown below.

  $\begin{array}{c}x=3\end{array}$

  $\begin{array}{c}x-3=0\left[\text{Subtract 3 from each side}\right]\end{array}$

Thus the value of $A,B$ and $C$ is $1,0$ and $-3$ respectively.

First evaluate $\sqrt{A^2+B^2}$ as shown below.

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{{\left(1\right)}^2+{\left(0\right)}^2}\end{array}$

  $\begin{array}{c}=\sqrt{1}\end{array}$

  $\begin{array}{c}=1\end{array}$

Since $C$ is negative, divide each term of the standard equation by $+\sqrt{A^2+B^2}$ as shown below to determine the normal form of the equation.

  $\begin{array}{c}\frac{1}{1}x+\frac{0}{1}y-\frac{3}{1}=0\end{array}$

  $\begin{array}{c}x+0y-3=0\end{array}$

Compare the above equation with the general normal form equation, $x\cos \varphi +y\sin \varphi -p=0$ to get $\cos \varphi =1,\sin \varphi =0\text{and }p=3$ .

Since cosine and sine both are poisitive so $\varphi$ must lie in the first quadrant. Now evaluate $\varphi$ as shown below.

  $\begin{array}{c}\tan \varphi =\frac{\sin \varphi }{\cos \varphi }\end{array}$

  $\begin{array}{c}=\frac{0}{1}\left[\text{Put the value of sine and cosine}\right]\end{array}$

  $\begin{array}{c}\tan \varphi =0\end{array}$

  $\begin{array}{c}\varphi \approx 0°\end{array}$

Thus the value of $p$ and $\varphi$ is $3$ and $0°$ respectively.