Respuesta :
Answer:
The parametric equations of line are x=-8-t, y=5+8t and z=2+t.
Step-by-step explanation:
It is given that the line passes through the point (−8, 5, 2) and is perpendicular to the plane given by −x + 8y + z = 5.
The coordinate of point are (−8, 5, 2).
Normal vector = <-1,8,1>
The position vector of a line is
[tex]\overrightarrow{r}=(x_0,y_0,z_0)+t<a,b,c>[/tex]
then the parametric equations of line are [tex]x=x_0+at,y=y_0+bt,z=z_0+ct[/tex].
Where, (x_0,y_0,z_0) are the coordinate of point and <a,b,c> is normal vector.
The position vector of given line is
[tex]\overrightarrow{r}=(-8,5,2)+t<-1,8,1>[/tex]
The parametric equations of line are
[tex]x=-8-t[/tex]
[tex]y=5+8t[/tex]
[tex]z=2+t[/tex]
Therefore the parametric equations of line are x=-8-t, y=5+8t and z=2+t.
It can be deduced that the parametric equation will be x = 8 - t, y= 5 + 8t, and z= 2 + t.
How to find the parametric equations
From the given information, it was stated that the line passes through the point (−8, 5, 2) and is perpendicular to the plane given by −x + 8y + z = 5.
In this case, the coordinate of the point include (-8, 5, 2). The normal vector is <-1,8,1>. The position vector of the line will be: (-8, 5, 2) + t ( -1, 8, 1).
Hence, the parametric equation will be x= 8 - t, y= 5 + 8t, and z= 2 + t.
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