Linear Algebra question

Question

Find the shortest distance from the point P=(-3,-2,-3) to a point on the line given by 1: (x,y,z) = (-7t, 1t, 1t). Please help ASAP! Thanks!

Answer 1

A unit vector tangent to the line is: T = (-7, 1, 1) / √51

The projection of the vector P on T is found with dot product:

P . T = (-3,-2,-3) . (-7, 1, 1) / √51 = 16 / √51

The component vector of P along T is:

P_|| = (16 / √51) T

The component vector of P, that is perpendicular to the line is the remainder:

P_⊥ = P - P_||

The magnitude of P_⊥ is the shortest distance to the line.