Find the distance from the point P(3,5,6) to the line (x-1)/2 =(y+1)/3 =(z-1)/3.
Make a sketch of the line and the point A
Let A be a point on the given line, why not A(1,-1,1)
vector AP = <2, 6, 5> <---- vector u
a vector in the direction of the line is <2,3,3> <---- vector v
the projection of u on v = u dot v / |v|
= (4+18+15)/√(4+9+9) = 37/√22
If B is the point on the line closest to A, we now have
AB = 37/√22 , and
PB^2 + AB^2 = AP^2
PB^2 + 1369/22 = (2^2 + 6^2 + 5^2) = 65
PB^2 = 61/22
PB = √61/√22
= appr 1.66515... check my arithmetic
other way, same sketch
we can find the angle at A
u dot v = |u| |v| cosA
37 = √22√65cosA
cosA = 35/√1430
A = 11.91946...°
then PB/AP = sin 11.91946
PB = APsin11.91946...°
= √65 sin11.91946.. = 1.66515 same as above