The distance between the line r =3 i -2 j + k + ( i -2 j ) and the plane r (2 i + j + k )=4 is

Step-by-step solution

The distance between a line and a plane is determined by their relative orientation. For the line r =3 i -2 j + k + ( i -2 j ) and the plane r (2 i + j + k )=4 , the direction vector b = i -2 j is perpendicular to the normal vector n =2 i + j + k since their dot product is zero: b n = (1)(2)+(-2)(1)+(0)(1)=0 Thus, the line is parallel to the plane, and the distance equals the perpendicular distance from any point on the line to the plane. Using point a =3 i -2 j + k , compute: a n =3 2+(-2) 1+1 1=5 The magnitude of