Chapter 10.4, Problem 40E

To determine

To find: The set of parametric equations of the line passes through $\left(-4,5,2\right)$ and perpendicular to the plane $-x+2y+z=5$ .

Answer to Problem 40E

The set of parametric equations of the line passes through $\left(-4,5,2\right)$ and perpendicular to the plane $-x+2y+z=5$ are $x=-4$ , $y=5-t$ and $z=2$ .

Explanation of Solution

Given information:

The line passes through $\left(-4,5,2\right)$ and it is perpendicular to the plane $-x+2y+z=5$ .

Calculation:

The directional vector to the line can be computed by the cross product of the normal vectors to the $xy$ and $yz$ planes respectively. There are $u=\langle 0,0,1\rangle$ and $v=\langle 1,0,0\rangle$ .

Calculate the cross product of the vector $u$ and $v$ to find the directional vector on the plane.

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 0& 1\\ 1& 0& 0\end{array}\right|\\ =-\text{j}\\ \text{=}\langle a,b,c\rangle \end{array}$

The equation of the line are $x=x_1+at$ , $y=y_1+bt$ and $z=z_1+ct$ and with $\left(x_1,y_1,z_1\right)=\left(-4,5,2\right)$ .

Find the equations of line as follows.

  $\begin{array}{c}x=-4+0\left(t\right)\\ =-4\\ y=5-1\left(t\right)\\ =5-t\end{array}$

and,

  $\begin{array}{c}z=2+0\left(t\right)\\ =2\end{array}$

Therefore, the set of parametric equations of the line passes through $\left(-4,5,2\right)$ and is perpendicular to the plane $-x+2y+z=5$ are $x=-4$ , $y=5-t$ and $z=2$ .