Answer
$x=2-2t,y=3+4t, z=-2t$
Work Step by Step
The parametric equations of a straight line can be found by knowing the value of a vector, such as $v=v_1i+v_2j+v_3k$, passing through a point $P(x_0,y_0,z_0)$ as follows:
$x=x_0+t v_1,y=y_0+t v_2; z=z_0+t v_3$
Here, we have a line perpendicular to the two vectors u and v; this means that it will be parallel to the cross product of $u \times v$
Thus, $n=u \times v=-2i+4j-2k$
Therefore, we have the vector $v=\lt -2,4,-2 \gt$ and $P=(2,3,0)$ .
Thus, we get the parametric equations:
$x=2-2t,y=3+4t, z=0-2t$
Hence, $x=2-2t,y=3+4t, z=-2t$
