Given the line r1 3 8t2 5 7 and the plane 3x4y2z1 what is th

Given the line r=(1, 3, 8)+t(-2, 5, 7) and the plane 3x+4y-2z=1, what is the relationship or this line with respect to the plane.

Solution

The normal vector for plane p =(3,4,-2)

Line is parallel to v=(-2,5,7)

Dot product=p.v=3*(-2)+5*4+7*(-2)=0

Thus the line and the plane are parallel to each other.

At t=0 w=(1,3,8)

Dot product=p.w=3*1+4*3-2*8=-1 and not equal to 1

Thus Line and the pane donot intersect