Writing parameters in terms of given vectors

jj364
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Homework Statement




Four 3-vectors a, b, c, and d are related by the equation
ax + by + cz = d;
where x, y, and z are real parameters. Using a suitable combination of scalar and vector
products, findd x, y, and z in terms of the vectors

Homework Equations





The Attempt at a Solution



So I tried to eliminate one vector at a time by doing the cross product of d and b etc so:

d^b=(a^b)x + (c^b)z and
d^c=(a^c)x + (b^c)y the other term cancelling as b^b=0

Therefore:
z=(d^b-(a^b)x)/(c^b)
y=(d^c-(a^c)x)/(b^c)

And substituting into the equation in the question everything seems to cancel out and I get:

xa=d, not sure if my approach is just entirely wrong. Any help would be greatly appreciated.
 
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jj364 said:

Homework Statement

Four 3-vectors a, b, c, and d are related by the equation
ax + by + cz = d;
where x, y, and z are real parameters. Using a suitable combination of scalar and vector
products, findd x, y, and z in terms of the vectors

Homework Equations


The Attempt at a Solution



So I tried to eliminate one vector at a time by doing the cross product of d and b etc so:

d^b=(a^b)x + (c^b)z and
d^c=(a^c)x + (b^c)y the other term cancelling as b^b=0

Therefore:
z=(d^b-(a^b)x)/(c^b)
y=(d^c-(a^c)x)/(b^c)

And substituting into the equation in the question everything seems to cancel out and I get:

xa=d, not sure if my approach is just entirely wrong. Any help would be greatly appreciated.

I find your choice of symbols and your notation very confusing and hard to read. Your choice to label a,b,c as vectors and x,y,z as real constants doesn't help, when common notation would be the other way. Neither does your notation for cross product help. So I'm going to change your notation and give you a hint that might help. Let's say you are given vectors ##\vec u,\, \vec v,\, \vec w,\, \vec z## and real numbers ##a,\, b,\, c## such that$$
a\vec u + b\vec v + c \vec w =\vec z$$Now consider what would happen if you dot ##\vec u \times \vec v## into both sides. See if that helps you solve for one of the constants.
 
Last edited:
Hi, the notation used for defining vectors and parameters is just the one in the question and I didn't want to confuse the vector x with a cross product, so I used the other notation for cross product, nevertheless I accept that it was a bit confusing, apologies.

Ok I see this totally, makes perfect sense. Obviously 2 out of 3 of the terms would be reduced to zero with the triple scalar product leaving only the one without repeated vectors. I tried using a matrix method in the end but this was a bit useless without actually knowing what the vectors are. Thanks, I think this has totally solved it for me but I'll just give it a go and come back later. Thank you very much.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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