Question

Find a vector equation and parametric equations for the line. (use the parameter t.) the line through the point (1, 0, 6) and perpendicular to the plane x 4y z = 6

Answer 1

The plane $x+4y+z=6$ has normal vector $\vec n = \vec\imath+4\,\vec\jmath+\vec k$. Any line perpendicular to this plane is parallel to $\vec n$. The one that passes through the origin is $\vec n\,t$, where $t\in\Bbb R$. Translate this line by the vector $\vec\imath+6\,\vec k$ to get the line we want,

$\vec r(t) = \vec n\,t + \vec\imath + 6\,\vec k = (t+1)\,\vec\imath + 4t\,\vec\jmath + (t + 6)\,\vec k$

In parametric form, we have

$\begin{cases} x(t) = t + 1 \\ y(t) = 4t \\ z(t) = t + 6 \end{cases}$