# Find a vector equation and parametric equations for the line. The

Find a vector equation and parametric equations for the line.
The line through the point $\left(1,0,8\right)$ and perpendicular to the plane $x+3y+z=3$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Nancy Johnson
The line passing through the point (1,0,8) and perpendicular to the plane $x+3y+z=3$
Let ${r}_{0}=i+0j+8k$
A line perpendicular to the plane $x+3y+z=3$ has the same direction as a normal vector to the plane, which is $n=<1,3,1>$
So, take $v=i+3j+k$
By using the vector eqaution for the line $r={r}_{0}+tv$
$r=\left(i+0j+8k\right)+t\left(i+3j+k\right)$
$=\left(1+t\right)i+\left(0+3t\right)j+\left(8+t\right)k$
$=\left(1+t\right)i\left(3t\right)j+\left(8+t\right)k$
Thus, the vector equation of the line is $r\left(t\right)=\left(1+t\right)+i\left(3t\right)j+\left(8+t\right)k$
Obtain the parametric equation of line as follows.
Let ${P}_{0}=\left({x}_{0},{y}_{0},{z}_{0}\right)=\left(1,0,8\right)$ and $v=$
By the parametric equation for the line through the point $\left({x}_{0},{y}_{0},{z}_{0}\right)$ and parallel to the direction vector $$ are
That implies,

Thus, the parametric equation of the line is