# If the position vectors of points T and S are 3a, - 2a, +a, and 4a, + 6a, + 2a, respectively, find (a) the coordinates of T and S, (b) the distance vector from T to S, (c) the distance between T and S.

If the position vectors of points T and S are 3a, - 2a, +a, and 4a, + 6a, + 2a, respectively, find (a) the coordinates of T and S, (b) the distance vector from T to S, (c) the distance between T and S.
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gutsy8v
Taking the coordinate system with ${A}_{X},{A}_{Y},{A}_{Z}$ as unit vectors in 3 noncoplanar directions,.
a. the coordinates of t and s are
t is [3,-2,1] and s is [4,6,2]
b.distance vector from t to s is $s-t=\left[4-3,6+2,2-1\right]=\left[1,8,1\right]={A}_{X}+8{A}_{Y}+{A}_{Z}$
c.distance from t to $s=\left[\sqrt{\left({A}_{X}+8{A}_{Y}+{A}_{Z}\right).\left({A}_{X}+8{A}_{Y}+{A}_{Z}\right)}\right]$
If ${A}_{X},{A}_{Y},{A}_{Z}$ are mutually orthogonal unit vectors then we get
distance from t to $s=\sqrt{1+64+1}=\sqrt{66}$