Vector equation involving linearly independent vectors of the three dimensional space

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  • #1
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Homework Statement


Solve the following vector equation for ##\vec{y}##. ##\vec{a}##, and ##\vec{b}## are linearly independent vectors of the three dimensional space.

##\vec{a} \times (8\vec{y}+\vec{b}) = \vec{b}\times(-5\vec{y}+\vec{a})##

Homework Equations




The Attempt at a Solution


First I developed the brackets
##\vec{a}\times8\vec{y}+\vec{a}\times\vec{b} = \vec{b}\times(-5\vec{y})+\vec{b}\times\vec{a}##
I subtracted ##\vec{b}\times(-5\vec{y})## and ##\vec{a}\times\vec{b}## from both sides, so I get
##\vec{a}\times8\vec{y}-\vec{b}\times(-5\vec{y})=\vec{b}\times\vec{a}-\vec{a}\times\vec{b}## which also equals to
##\vec{y}\times(-8\vec{a}-5\vec{b}) = 2\vec{b}\times\vec{a}##

Then I expressed ##\vec{y}## as the linear combination of ##\vec{a}##, ##\vec{b}## and ##\vec{b}\times\vec{a}##
##\vec{y} = \alpha\vec{a}+\beta\vec{b}+\gamma\vec{b}\times\vec{a}## then I multiplied it (cross product) with ##-8\vec{a}-5\vec{b}##
##\vec{y}\times(-8\vec{a}-5\vec{b}) = -8\alpha(\vec{a}\times\vec{a})-5\alpha(\vec{a}\times\vec{b})-8\beta(\vec{b}\times\vec{a})-5\beta(\vec{b}\times\vec{b}) + \gamma(\vec{b}\times\vec{a})\times(-8\vec{a}-5\vec{b})##
##=(5\alpha-8\beta)(\vec{b}\times\vec{a})-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})]##
From the equation this has to be equal to ##2\vec{b}\times\vec{a}## so ##5\alpha-8\beta = 2## and ##\gamma = 0##

But if I make up an arbitrary ##\vec{y}## vector of this form and choose random ##\vec{a}## and ##\vec{b}## (linearly independent of course) it doesn't satisfy the equation. Where did I mess up?
 

Answers and Replies

  • #2
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So if ##\gamma = 0 ## then doesn't that mean that ##\vec{y}## is a linear combination of ##\vec{a}## and ##\vec{b}## ?

When I computed it it seemed that ##\gamma## can be anything since the resultant ##\gamma## component term vectors are both perpendicular to ##\vec{a}## x ##\vec{b}##
 
  • #3
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So if ##\gamma = 0 ## then doesn't that mean that ##\vec{y}## is a linear combination of ##\vec{a}## and ##\vec{b}## ?

When I computed it it seemed that ##\gamma## can be anything since the resultant ##\gamma## component term vectors are both perpendicular to ##\vec{a}## x ##\vec{b}##
How did you get that ##\gamma## can be anything?
 
  • #4
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Both a and b are perpendicular to a x b , right? Assuming that neither a nor b are the zero vector.
 
  • #5
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Yes
 
  • #6
12,505
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so their cross products are zero and thus ##\gamma## can be any value, right?
 
  • #7
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Did you test it by choosing an a and b vector, construct a y vector and evaluate your original equation?

Perhaps there's a mistake in the check...
 
  • #8
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Let ##\vec{a} =
\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix},
\vec{b} =
\begin{bmatrix}
1 \\
2 \\
1 \\
\end{bmatrix},
\vec{y} =
\begin{bmatrix}
2 \\
1 \\
0 \\
\end{bmatrix},
\vec{y}## has to be a good choice because ##5\cdot 2 - 8\cdot1 = 2##, and ##\vec{a},\vec{b}## are independendent.
##-8\vec{a}-5\vec{b}=
\begin{bmatrix}
-13 \\
-18 \\
-13 \\
\end{bmatrix}, \vec{y}\times(-8\vec{a}-5\vec{b}) =
\begin{bmatrix}
-13 \\
26 \\
-23 \\
\end{bmatrix}
##, but ##2\vec{b}\times\vec{a} =
\begin{bmatrix}
2 \\
0 \\
-2 \\
\end{bmatrix}
##
 
  • #9
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Wouldn't you compute y from the the ##\alpha, \beta, \gamma## values and the a and b vectors? and then show that a, b and y satisfy the original formula?
 
  • #10
DrClaude
Mentor
7,554
3,894
##\vec{y}## has to be a good choice because ##5\cdot 2 - 8\cdot1 = 2##
If ##\alpha = 2## and ##\beta=1##, then you have the wrong ##\vec{y}##.
 
  • #11
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16
Wouldn't you compute y from the the ##\alpha, \beta, \gamma## values and the a and b vectors? and then show that a, b and y satisfy the original formula?
If ##\alpha = 2## and ##\beta=1##, then you have the wrong ##\vec{y}##.
Aah what a mistake I made there... I should have written the final form of y, I've mistaken ##\alpha,\beta,\gamma## for its components, but they are coefficients of ##\vec{a},\vec{b}##. It works now, thank you!

But ##\gamma## has to be 0, otherwise it doesn't work for me.
 
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  • #12
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Aah what a mistake I made there... I should have written the final form of y, I've mistaken ##\alpha,\beta,\gamma## for its components, but they are coefficients of ##\vec{a},\vec{b}##. It works now, thank you!

But ##\gamma## has to be 0, otherwise it doesn't work for me.
My feeling is that ##\gamma## can be any value but perhaps @DrClaude can comment on it.
 
  • #13
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16
Could you show it with equations, please?
 
  • #14
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let a = (1,0,0) and b=(0,1,0) and so a x b = ( 0,0,1)

then y=(8,-5,10) where ##\beta = (-5/8) \alpha##.

I chose ##\alpha = 8## and so ##\beta = -5## and arbitrarily chose ##\gamma = 10##

a x (8y) + a x b = (0,80,-40) + (0,0,1) = (0,80,-39)

b x (-5y) + b x a = (-50,0,40) + (0,0,-1) = (-50,0,39)

Well, I didn't succeed! I see where I got ##\gamma## wrong now and I think you're right, it must be zero.

----------------------

Let y = ##\alpha## a + ##\beta## b + ##\gamma## (a x b)

then ##\alpha## a x (8a + 5b) = 0 + 5 ##\alpha## ( a x b)

and ##\beta## b x ( 8a + 5b) = -8 ##\beta## (a x b) + 0

and ##\gamma## (a x b) x (8a + 5b) = 8##\gamma## (a x b) x a + 5##\gamma## (a x b) x b

...
 
Last edited:
  • #15
DrClaude
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##\vec{y}\times(-8\vec{a}-5\vec{b}) = -8\alpha(\vec{a}\times\vec{a})-5\alpha(\vec{a}\times\vec{b})-8\beta(\vec{b}\times\vec{a})-5\beta(\vec{b}\times\vec{b}) + \gamma(\vec{b}\times\vec{a})\times(-8\vec{a}-5\vec{b})##
##=(5\alpha-8\beta)(\vec{b}\times\vec{a})-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})]##
From the equation this has to be equal to ##2\vec{b}\times\vec{a}## so ##5\alpha-8\beta = 2## and ##\gamma = 0##
You found
$$
-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})] = 0
$$
from which either ##\gamma = 0## or
$$
8[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]+5[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})] = 0
$$
With a bit of algebra, you can transform this into the equation ##c_1 \vec{a} = c_2 \vec{b}##, which contradicts ##\vec{a}## and ##\vec{b}## being linearly independent (except if ##c_1 = c_2 = 0##, but you can eliminate that possibility for real vectors).
 
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  • #17
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16
You found
$$
-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})] = 0
$$
from which either ##\gamma = 0## or
$$
8[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]+5[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})] = 0
$$
With a bit of algebra, you can transform this into the equation ##c_1 \vec{a} = c_2 \vec{b}##, which contradicts ##\vec{a}## and ##\vec{b}## being linearly independent (except if ##c_1 = c_2 = 0##, but you can eliminate that possibility for real vectors).
Aah, so this would have been the precise way of saying this, I should have known!

Thank you all for you help!
 

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