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

klytamnestra9a 2021-11-11 Answered
Find a vector equation and parametric equations for the line.
The line through the point (1,0,8) and perpendicular to the plane x+3y+z=3
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Expert Answer

Nancy Johnson
Answered 2021-11-12 Author has 17 answers
The line passing through the point (1,0,8) and perpendicular to the plane x+3y+z=3
Let r0=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=r0+tv
r=(i+0j+8k)+t(i+3j+k)
=(1+t)i+(0+3t)j+(8+t)k
=(1+t)i(3t)j+(8+t)k
Thus, the vector equation of the line is r(t)=(1+t)+i(3t)j+(8+t)k
Obtain the parametric equation of line as follows.
Let P0=(x0,y0,z0)=(1,0,8) and v=<a,b,c<1,3,1>
By the parametric equation for the line through the point (x0,y0,z0) and parallel to the direction vector <a,b,c> are x=x0+at, y=y0+bt, z=z0+ct
That implies,
x=1+t, y=0+3t, z=8+t
x=1+t, y=3t,z=8+t
Thus, the parametric equation of the line is
x(t)=1+t, y(t)=3t, z(t)=8+t
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