the conditions vec(r)* vec(s) =0, and vec(r) * vec(x) =c, and vec(r) xx vec(x) = vec(s) . Find x in each of the three mutually orthogonal directions, vec(r), vec(s) , and vec(r) xx vec(s).

kjukks1234531 2022-09-25 Answered
The conditions r s = 0, and r x = c, and r × x = s . Find x in each of the three mutually orthogonal directions, r , s , and r × s
So far x r = x r | r | r | r | = c | r | 2 r , and x s = x s | s | s | s | = x ( r × x ) | s | s | s | = 0 Since x ( r × x ) is the volumn of a degenerate parallelepiped. Where I'm having most difficulty is in...
x r × s = x ( r × s ) | r × s | 2 r × s . Since r and s are orthogonal does that mean | r × s | = | r | | s | . And also can I calculate x ( r × s )using the triple product to be | r | | s | | x s | = | r | | s | | x x s | = | r | | s | | x | . Is there a simpler simplification of this?
So does x r × s = | x | | r | | s | x ?
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Answers (2)

Lorenzo Acosta
Answered 2022-09-26 Author has 13 answers
Just to simplify the notation a bit, let t = r × s
Because x s = 0, we know x = x r + x t . Therefore
r × x = s r × ( x r + x t ) = s r × x r + r × x t = s .
But r × x r = 0, so we conclude that r × x t = s . Moreover, r and x t are orthogonal, so | r | | x t | = | s |
This tells us | x t | = | s | | r | , so x t = ± | s | | r | t | t | = ± | s | | r | | t | t . We also know | t | = | r | | s | , further simplifying x t to ± 1 | r | 2 t
Which sign is it? Well, we want r × x t = s , but r × t is proportional to s . Therefore x t = 1 | r | 2 t
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Marcus Bass
Answered 2022-09-27 Author has 2 answers
consider r × x = s
taking cross product from both sides
r × ( r × x ) = r × s
you can use vector triple cross product formula
then
( r . x ) r ( r . r ) x = r × s
( r . x ) r | r | 2 x = r × s
rearranging
x = c | r | 2 r 1 | r | 2 r × s
since r , s and r × s are mutually orthogonal they are linearly independent therefore you can get x , as a linear combination of these r , s and r × s vectors
x = c | r | 2 r 1 | r | 2 ( r × s ) 0 s
so you can take component as
x r = c | r | 2
x r × s = 1 | r | 2
and
x s = 0
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