This is incorrect. That gives |PS| which is the distance from P to S, not the distance from P to thenearestpoint on plane.

Well, yes, of course. Otherwise, the two would intersect so the distance would be 0. There was really no reason to do this.my second attempt ::

the vector normal to the line, n1 = i+j-1/2k

the vector normal to the plane, n2 = i+2j+6k

n1.n2 = 0 ==> the two normal are perpendicular ===> the line and the plane are parallel

What? This is simply the distance from the the given point to (0, 0, 0). What does that have to do with the plane?thus the distance for the line to the plane = sqrt(2^2+1^2+(-1/2)^2) = sqrt(21)/2 unit

I don't see whyWHICH ATTEMPT IS CORRECT SINCE I GOT TWO DIFFERENT ANSWERS??? PLEASE HELP ME..eitherattempt should be correct. Both are based on incorrect logic.

As you say, a perpendicular vector to the plane is i+ 2j+ 6k and a point on the line is (2, 1, -1/2). A line through (2, 1, -1/2) in the direction of i+ 2j+ 6k is x= 2+ t, y= 1+ 2t, z= -(1/2)+ 6t. Since that line passes through (2, 1, -1/2) and is perpendicular to the plane, it will intersect the plane at the point closest to (2, 1, -1/2). That line will intersect the given plane when (2+ t)+ 2(1+ 2t)+ 6(-(1/2)+ 6t)= (2+ 2- 3)+ (t+ 4t+ 36t)= 1+ 41t= 10 or t= 9/41. Put t into the equation of the perpendicular line to determine that point and then find the distance between those two points.