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.

Hayley Bernard
2022-07-30
Answered

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gutsy8v

Answered 2022-07-31
Author has **14** answers

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 $s-t=[4-3,6+2,2-1]=[1,8,1]={A}_{X}+8{A}_{Y}+{A}_{Z}$

c.distance from t to $s=[\sqrt{({A}_{X}+8{A}_{Y}+{A}_{Z}).({A}_{X}+8{A}_{Y}+{A}_{Z})}]$

If ${A}_{X},{A}_{Y},{A}_{Z}$ are mutually orthogonal unit vectors then we get

distance from t to $s=\sqrt{1+64+1}=\sqrt{66}$

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 $s-t=[4-3,6+2,2-1]=[1,8,1]={A}_{X}+8{A}_{Y}+{A}_{Z}$

c.distance from t to $s=[\sqrt{({A}_{X}+8{A}_{Y}+{A}_{Z}).({A}_{X}+8{A}_{Y}+{A}_{Z})}]$

If ${A}_{X},{A}_{Y},{A}_{Z}$ are mutually orthogonal unit vectors then we get

distance from t to $s=\sqrt{1+64+1}=\sqrt{66}$

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Let $B=\{2x,3x+{x}^{2},-1\},{B}^{\prime}=\{1,1+x,1+x+{x}^{2}\}$

Need to find the transformation matrix from $B$ to ${B}^{\prime}$.

I know that:

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How to proceed using this info in order to find the transformation matrix?

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Is