Chapter 12, Problem 24CLT

To determine

Find parametric equations of the line passing through the points.

Answer to Problem 24CLT

  $\boxed{\begin{array}{l}x=-1+2t\\ y=2-4t\\ z=t\end{array}}$

Explanation of Solution

Given information:

Find sets of parametric equations of the line passing through the points $\left(-1,2,0\right)$ and and perpendicular to $2x-4y+z=8$ .

Calculation:

Consider the parametric equations of the line passing through the points

  $\left(-1,2,0\right)$ and equation of plane $2x-4y+z=8$

Find the normal vector of the plane using standard results

  $2x-4y+z=8$

  $\begin{array}{l}ax+by+cz=d\\ a=2\\ b=-4\\ c=1\end{array}$

Now point $a_1,a_2,a_3$

Parallel vector $b_1,b_2,b_3$

The vector equation is $r=\left(a_1,a_2,a_3\right)+t\left(b_1,b_2,b_3\right)$

Hence the vector equation of required line is,

  $r=\left(-1,2,0\right)+t\left(2,-4,1\right)$

Now use equality of vector and express the vector equation in to parametric equation,

  $\begin{array}{l}r=\left(-1,2,0\right)+t\left(2,-4,1\right)\\ x,y,z=\left(-1,2,0\right)+t\left(2,-4,1\right)\\ =\left(-1+2t,2-4t,t\right)\\ so,\\ x=-1+2t\\ y=2-4t\\ z=t\end{array}$

Hence the parametric equation for the line is,

  $\boxed{\begin{array}{l}x=-1+2t\\ y=2-4t\\ z=t\end{array}}$