Finding relative coordinates of a vector

AI Thread Summary
To find the coordinates of the vector [7, 4, 1] relative to the vectors [-1, 2, -9], [1, 5, 1], and [47, -8, -7], the correct approach involves setting up an augmented matrix. The user confirmed that the goal is to express the vector as a linear combination of the three given vectors, leading to the equations -a + b + 47c = 7, 2a + 5b - 8c = 4, and 9a + b - 7c = 1. After setting up the matrix, the user initially calculated values for a, b, and c but later realized a mistake that would affect these results. The discussion emphasizes the importance of correctly interpreting the problem and verifying calculations. The method of using an augmented matrix for solving the system of equations is validated.
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Homework Statement



I'm trying to find the coordinates of a vector [ 7 4 1 ] relative to these three vectors:
[ -1 2 -9 ] , [ 1 5 1 ] , [ 47 -8 -7 ]

Homework Equations



none

The Attempt at a Solution



Do I get all these vectors into an augumented matrix?
like:
[ -1 1 47 | 7 ]
[ 2 5 -8 | 4 ]
[ -9 1 -7 | 1 ]
and then row reduce completely? I got very awkward numbers and just wanted to make sure I'm using the correct method.
 
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I take it that "relative to these three vectors" means write it as a linear combination of those vectors. What you want are three numbers, a, b, c such that
a<-1, 2, 9>+ b<1, 5, 1>+ c<47, -8, -7>= <7, 4, 1>. Looking at x, y, and z components separately that means -a+ b+ 47c= 7, 2a+ 5b- 8c= 4, 9a+ b- 7c= 1 which gives exactly the augmented matrix you have.

Are you sure about that "47"? That number looks out of place.
 
yes that 47 must be there because it was part of another question where I had to find a third vector orthogonal to the other two (involving the cross product) and I'm confident about that.

If you say that my setup is right, then my more awkward numbers must be right.
I got
a = -4/43
b = 28/27
c = 145/1161
 
oups, I made a mistake and it'll affect my a b c values. But thanks for confirming my method Ivy.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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