Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −8) and parallel to the line x = −1 + 4t, y = 6 − 3t, z = 3 + 7t

Answer:[tex]r=(0,11,-8)+(4,-3,7)t[/tex]x=4ty=11-3tz=-8+7tStep-by-step explanation:The line is parallel to x=-1+4t, y=6-3t, z=3+7t.Two lines are parallel if they have the same direction, and in the parametric form, the direction of a line is always the vector of constants that multiply t (or the parameter). So in this case, the direction is (4,-3,7); this is also the direction of the missing line because they are parallel.The vector equation of a line is given by:[tex]r=r_{0} +tv[/tex]where v is the direction vector, and [tex]r_{0}[/tex] is a point of the line. So, for this case, the line pass for the point (0, 11, −8):[tex]r_{0}=(0,11,-8)[/tex]With the direction, the vector equation is:[tex]r=(0,11,-8)+(4,-3,7)t[/tex]The parametric equations are just the simplification of the vector equation:x=4ty=11-3tz=-8+7t