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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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

Nancy Johnson · Accepted Answer

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=&amp;lt;1,3,1&amp;gt;  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=&amp;lt;a,b,c≥&amp;lt;1,3,1&amp;gt;  By the parametric equation for the line through the point (x0,y0,z0) and parallel to the direction vector &amp;lt;a,b,c&amp;gt; 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

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

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