so you want the line through P0 perpendicular to L. That is, it lies in the plane perpendicular to L.
That plane is (1)(x+2) + (-2)(y+2) + (3)(z-2) = 0
Or, x-2y+3z = 8
The distance is thus |(1)(-4)+(-2)(-4)+(3)(-5)|/√(1^2+2^2+3^2)
Now just find Q.
Let L be the line passing through the point P(−2, −2, 2) with direction vector d=[1, −2, 3]T. Find the shortest distance d from the point P0(−4, −4, −5) to L, and the point Q on L that is closest to P0. Use the square root symbol '√' where needed to give an exact value for your answer.
d=?
Q=(?,?,?)
2 answers
how do you find Q?