# Find parametric equations for the line

• Jan 4th 2007, 04:45 PM
Jenny20
Find parametric equations for the line
Question
Lsub1 : x=1+2t, y=2-t, z=4-2t
Lsub2 : x=9+t , y=5+3t, z=-4-t

Lsub1 and Lsub2 intersect at the point (7,-1,-2).

Find parametric equations for the line that is perpendicular to Lsub1 and Lsub2 passes through their point of intersection.

* if possible, please draw me the picture. I can't figure out how this line can perpendicular to Lsub1 and Lsub2.

Thank you very much.
• Jan 4th 2007, 05:10 PM
ThePerfectHacker
Quote:

Originally Posted by Jenny20
Question
Lsub1 : x=1+2t, y=2-t, z=4-2t
Lsub2 : x=9+t , y=5+3t, z=-4-t

Lsub1 and Lsub2 intersect at the point (7,-1,-2).

Find parametric equations for the line that is perpendicular to Lsub1 and Lsub2 passes through their point of intersection.

* if possible, please draw me the picture. I can't figure out how this line can perpendicular to Lsub1 and Lsub2.

Thank you very much.

Consider the xyz plane.
The x-axis and y-axis intersect.
And z-axis is perpendicular to both of them.
That is how the visuallization looks.

Line 1) Is algined with the vector,
$\displaystyle \bold{u}=2\bold{i}-\bold{j}-2\bold{k}$

Line 2) Is algined with vector,
$\displaystyle \bold{v}=\bold{i}+3\bold{j}-\bold{k}$

The line that we seek is perpendicular to both, that is a vector that is orthogonal to both these non-zero vectors.

We can use the cross product,
$\displaystyle \bold{u}\times \bold{v} =\left| \begin{array}{ccc}\bold{i}&\bold{j}&\bold{k} \\ 2&-1&-2\\ 1&3&-1 \end{array} \right| =7\bold{i}+7\bold{k}$.

We can remove the 7 multiple, because if it is algined with $\displaystyle <7,0,7>$ it is certainly aligned with $\displaystyle <1,0,1>$.

Thus, the parametric equations are,
$\displaystyle \left\{ \begin{array}{c} x=7+t\\y=-1\\z=-2+t \end{array} \right\}$