Find the shortest distance from (6,-4,4) to the line joining (2,1,2) and (3,-1,4).

Hi

Find the vector equation of the line joining pt (2,1,2) & (3,-1,4) using r(t) = a + t (b - a ). hence eq wud be r =(2,1,2) +t(1, -2,2).

Suppose the position vector of P from the origin O is OP = p = (6,-4,-4).

The shortest distance of P from L is given by the length of the perpendicular from P to L. Suppose this perpendicular meets L at H. Then we want to find the length of PH. 

  Using the dot product

Since H lies on L we can say that   OH = h = (2,1,2) +t(1, -2,2). = (2+t , 1-2t , 2+2t)
for some value of t which we need to find.
Also, vector PH = -p + h = -(6 , -4, -4) + (2+t , 1-2t , 2+2t)  = (t-4, 5-2t, 6+2t)

But PH is perpendicular to the direction vector (1, -2,2) of line L.
So the dot product of vector PH and (4, 1, -2) is zero.
So t-4, 5-2t, 6+2t).(1, -2,2) =  0   so   solve for t, and find pt h, then find the distance between the two points.